Chapter - 11_Heat and Thermodynamics

CHAPTER-11 HEAT AND THERMODYNAMICS FILL IN THE BLANKS 1. 2.

One mole of a monoatomic ideal gas is mixed with one mole of a diatomic ideal gas. The molar specific heat of the mixture at constant volume is ..... (1984, 2M) The variation of temperature of a material as heat is given to it at constant rate as shown in the figure. The material is in solid state at the point O. The state of the material at the point P is .... (1985, 2M)

C T

D

A P B

O

Heat added

During an experiment an ideal gas is found to obey an additional law P2V = constant. The gas is initially at a temperature T and volume V. When it expands to a volume 2V, the temperature becomes .... (1987, 2M) 4. 300 g of water at 25°C is added to 100 g of ice at 0°C. The final temperature of the mixture is ....°C (1989, 2M) 5. The earth receives at its surface radiation from the sun at the rate of 1400 Wm–2. The distance of the centre of the sun from the surface of the earth is 1.5 × 1011 m and the radius of the sun is 7 × 108 m. Treating the sun as a black body, it follows from the above data that its surface temperature is ....K. (1989, 2M) 6. A solid copper sphere (density ρ and specific heat c) of radius r at an initial temperature 200 K is suspended inside a chamber whose walls are at almost 0 K. The time required for the temperature of the sphere to drop to 100 K is ...... (1991, 2M) 7. A point source of heat of power P is placed at the centre of a spherical shell of mean radius R. The material of the shell has thermal conductivity K. If the temperature difference between the outer and inner surface of the shell is not to exceed T, the thickness of the shell should not be less than. (1991, 1M) 8. A substance of mass M kg requires a power input of P watts to remain in the molten state at its melting point. When the power source is turned off, the simple completely solidifies in time t seconds. The latent heat of fusion of the substances is .... (1992, 1M) 9. A container of volume 1 m3 is divided into two equal parts by a partition. One part has an ideal gas at 300 K and the other part is vacuum. The whole system is thermally isolated from the surroundings.When the partition is removed, the gas expands to occupy the whole volume. Its temperature will now be ..... (1993; 1M) 10. An ideal gas with pressure P, volume V and temperature T is expanded isothermally to a volume 2V and a final pressure P1. If the same gas is expanded adiabatically to a volume 2V, the final pressure is Pa. The ratio of the specific 3.

Pa is ..... P1

(1994, 2M)

11. Two metal cubes A and B of a same size are arranged as shown in figure. The extreme ends of the combination are maintained at the indicated temperatures. The arrangement is thermally insulated. The coefficients of thermal conductivity of A and B are 300 W/m°C and 200 W/mºC, respectively. After steady state is reached the temperature of the interface will be ....... (1996, 2M)

100°C

heats of the gas is 1.67. The ratio

12. A ring shaped tube contains two ideals gases with equal masses and relative molar masses M1 = 32 and M2 = 28. The gases are separated by one fixed partition and another movable stopper S which can move freely without friction inside the ring. The angle α as shown in the figure is .... degrees. (1997, 2M)

A

B 0°C T M2

M1 α

S

144

13. A gas thermometer is used as a standard thermometer for measurement of temperature. When the gas container of the thermometer is immersed in water at its triple point 273.16 K, the pressure in the gas thermometer reads 3.0 × 104N/m2. When the gas container of the same thermometer is immersed in another system the gas pressure reads 3.5 × 104N/m2. The temperature of this system is therefore......°C. (1997, 1M) 14. Earth receives 1400 W/m2 of solar power. If all the solar energy falling on a lens of area 0.2 m2 is focussed onto a block of ice of mass 280g, the time taken to melt the ice will be .... minutes. (Latent heat of fusion of ice = 3.3 × 105 J/kg) (1997, 2M)

TRUE FALSE 1. 2.

The root mean square speeds of the molecules of different ideal gases, maintained at the same temperature are the same. (1981; 2M) The volume V versus temperature T graphs for a certain amount of a perfect gas at two pressure p 1 and p 2 are as shown in figure. It follows from the graphs that p 1 is greater than p 2 . (1982; 2M) V P1

P2

T

3. 4.

Two different gases at the same temperature have equal root mean square velocities. (1982; 2M) The curves A and B in the figure shown P-V graphs for an isothermal and an adiabatic process for an ideal gas. The isothermal process is represeented by the curve A. (1985; 3M) P

A B V

5.

6. 7.

The root mean square (rms) speed of oxgen molecules (O2 ) at a certain temperaure T (degree absolute) is V. If the temperature is doubled and oxygen gas dissociates into atomc oxygen, the rms speed remains unchanged. (1987; 2M) At a given temperature, the specific heat of a ideal gas at constant pressure is always greater than its specific heat at constant volume. (1987; 2M) Two spheres of the same material have radii 1 m and 4 m temperature 4000 K and 2000 K respectively. The energy radiated per second by the first sphere is greater than that by the second. (1988; 2M)

OBJECTIVE QUESTIONS

2.

At room temperature, the rms speed of the molecules of a certain diatomic gas is found to be 1930 m/s. The gas is : (1984; 2M) (a) H2 (b) F2 (c) O2 (d) Cl2

3.

70 calories of heat required to raise the temperature of 2 moles of an ideal diatomic gas at constant pressure from 30°C to 35°C. The amount of heat required (in calories) to raise the temperature of the same gas through the same range (30°C to 35°C) at constant volume is : (1985; 2M) (a) 30 (b) 50 (c) 70 (d) 90

Only One option is correct : 1. An ideal monoatomic gas is taken round the cycle ABCDA as shown in the P-V diagram (see figure). The work done during the cycle is : (1983; 1M) P 2P, V B

A

(a) PV (c)

1 PV 2

P, V

2P, 2V C

P, 2V

D V

(b) 2PV (d) zero 145

4.

5.

6.

7.

Steam at 100°C is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 15°C till the temperature of the calorimeter and its contents rises to 80°C. The mass of the steam condensed in kg is : (1986; 2M) (a) 0.130 (b) 0.065 (c) 0.260 (d) 0.135 If one mole of a monoatomic gas (γ = 5/3) is mixed with one mole of a diamonic gas (γ = 7/5), the value of γ for the mixture is : (1988; 2M) (a) 1.40 (b) 1.50 (c) 1.53 (d) 3.07 A cylinder of radius R made of a material of thermal conductivity K1 is surrounded by a cylindrical shell of inner radius R and outer radus 2R made of a material of thermal conductivity K2 . The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is : (1988; 2M) (a) K1 + K2 (b) K1 K2 /(K1 + K2 ) (c) (K1 + 3K 2 )/4 (d) (3K1 + K2 )/4 When an ideal diatomic gas is heated at constant pressure the fraction of the heat energy supplied which increases the internal energy of the gas is : (1990; 2M)

2 (a) 5 (c) 8.

3 7

3 (b) 5 (d)

5 7

Three closed vessels A, B and C at the same temperature T and contain gases which obey the Maxwellian distribuion of velocities. Vessel A contains only O2 , B only N2 and C a mixture of equal\quantities of O2 and N2. If the average speed of the O2 molecules in vessel A is v1 , that of the N2 molecules in vessel B is v2 , the average speed of the O2 molecules in vessel C is : (1992; 2M) (a) (v1 + v2 )/2 (b) v1 (c) (v1 v2 )1/2

(d)

3kT / M where M is the mass of an oxygen molecules. 9.

Three rods of identical cross-sectional area and made from the same metal form the sides of an isosceles triangle ABC, right angled at B. The points A and B are maintained at temperature T and

( 2 )T

respectively.

In the seady state, the temperature of the point C is Tc.

Assuming that only heat conduction takes place, Tc/ T is : (1995; 2M)

1 (a)

3 (b)

2( 2–1)

(c)

3

(

1 2–1

)

(d)

2 +1 1

(

2 +1

)

10. Two metallic spheres S 1 and S 2 are made of the same material and have got identical surface finish. The mass of S 1 is thrice that of S 2 . Both the spheres are heated to the same high temperature and placed in the same room having lower temperature but are thermally insultated from each other. The ratio of the initial rate of cooling of S 1 to that S 2 is : (1996; 2M) (a)

(c)

1 3

3 1

1 (b)

3 1/3

1 (d)    3

11. The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K the root mean square velocity of the gas molecules is v, at 480 K it becomes : (1996; 2M) (a) 4 v (b) 2 v (c) v/2 (d) v/4 12. The average translational KE and the rms speed of molecules in a sample of oxygen gas at 300 K are 6.21 × 10 – 2 1 J and 484 m/s respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour) : (1997; 1M) (a) 12.42 × 10–21 J, 968 m/s (b) 8.78 × 10–21 J, 684 m/s (c) 6.21 × 10–21 J, 968 m/s (d) 12.42 × 10–21 J, 684 m/s 13. The intensity of radiation emitted by the sun has its maximum value at a wavelength of 510 nm and that emitted by the north star has the maximum value at 350 nm. If these starts behave like blackbodies, then the ratio of the surface temperature of the sun and the north star is : (1997; 1M) (a) 1.46 (b) 0.69 (c) 1.21 (d) 0.83 14. The average translational kinetic energy of O2 (molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N2 (molar mass 28) molecules in eV at the same temperature is : (1997; 1M) (a) 0.0015 (b) 0.003 (c) 0.048 (d) 0.768

146

17. A vessels contains a mixture of one mole of oxygen and two moles of nitrogen at 300 K. The ratio of the average rotational kinetic energy per O2 molecules to per N2 molecule is : (1998; 2M) (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) depends on the moment of inertia of the two molecules 18. Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V. The mass of the gas in A is mA and that in B is mB . the gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The changes in the pressure in A and B are found to be ∆P and 1.5 ∆P respectively. Then : (1998; 2M) (a) 4 mA = 9 mB (b) 2 mA = 3 mB (c) 3 mA = 2 mB (d) 9 mA = 4 mB 19. Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move, while that of B is held fixed. The same rise in temperature of the gas in A is 30 K, then the rise in temperature of the gas in B is : (1998; 2M) (a) 30 K (b) 18 K (c) 50 K (d) 42 K

(a)

( 2 / 7)

(b)

(1/7 )

(c)

( 3 ) /5

(d)

( 6 ) /5

22. A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is: (1999; 2M) (a) 4 RT (b) 15 RT (c) 9 RT (d) 11 RT 23. A monoatomic ideal gas, initially at temperature T1, is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If L1 and L2 are the lengths of the gas column before and after expansion respectively, then T1 /T2 is given by: (2000; 2M) (a) (L1 /L2 )2/3 (b) (L1 /L2 ) (c) (L2 /L1 ) (d) (L2 /L1 )2/3 24. The plots of intensity versus wavelength for three black bodies at temperatures T1, T2 and T3 respectively are as shown. Their temperature are such that : (2000; 2M) I

T3 T2

T1

λ

(a) T1 > T2 > T3 (c) T2 > T3 > T1

(b) T1 > T3 > T2 (d) T3 > T2 > T1

25. A block of ice at – 10°C is slowly heated and converted to steam at 100°C. Which of the following curves represents the phenomena qualitatively? (2000; 2M)

Temp.

16. A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be : (1997; 1M) (a) 225 (b) 450 (c) 900 (d) 1800

21. The ratio of the speeds of sound in nitrogen gas to that in helium gas, at 300 K is : (1999; 2M)

Temp.

15. A vessel contains 1 mole of O2 gas (molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessels containing one mole of the gas (molar mass 4) at a temperature 2T has a pressure of : (1997; 1M) (a) P/8 (b) P (c) 2P (d) 8 P

(a)

(b) Heat supplied

Temp.

(c)

(d) Heat supplied

147

Heat supplied

Temp.

20. A black body is at a temperature of 2880 K. The energy of radiation emitted by this object with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien constant, b = 2.88 × 10–6 nm-K. Then: (2003; 2M) (a) U1 = 0 (b) U3 = 0 (c) U1 > U2 (d) U2 > U1

Heat supplied

26. Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three diferent ways, the work done by the gas is W1 if the process is purely isothermal W2 if purely isobaric and W3 if purely adiabatic, then : (2000; 2M) (a) W2 > W1 > W3 (b) W2 > W3 > W1 (c) W1 > W2 > W3 (d) W1 > W3 > W2 27. An ideal gas is initially at temperature T and volume V. Its volume is increased by ∆V due to an increase in temperature ∆T, pressure remaining constant. The quantity δ = ∆V/V∆T varies with temperature as : (2000; 2M) δ

δ

(a) T+∆T

T

T

T+ ∆T

T

δ

δ

(c) T+∆T

(a) He and O2 (c) He and Ar

m1 m2

m1 (c) m 2

T

T+∆T

T

V

(b) O2 and He (d) O2 and N2

C

2

B

V ( m3 ) A 1

2 P (N/m)

(2002; 2M)

33. Which of the following graphs correctly represent the variation of β = − dV / dP with P for an ideal gas at V consant temperature

(2002; 2M) β

β

m2 m1

(a)

m2 (d) m 1

(b) P

P

β

β

90°C

29. Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends

1 2

by the gas in the process C → A is : (a) –5 J (b) – 10 J (c) –15 J (d) –20 J

T

(b)

P

10

28. Two monoatomic ideal gases 1 and 2 of molecular masses m1 and m2 respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas 1 to the gas 2 is given by : (2000; 2M) (a)

temperature will decrease volume will increase pressure will remain constant temperature will increase

31. P-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to : (2001; 2M)

(d) T

the the the the

32. An ideal gas is taken through the cyclie A → B → C → A, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done

(b) T

(a) (b) (c) (d)

(c)

0°C

(d) P

90°C

are kept at 0°C and 90°C respectively. The temperature of junction of the three rods will be : (2001; 2M) (a) 45°C (b) 60°C (c) 30°C (d) 20°C 30. In a given process of an ideal gas, dW = 0 and dQ < 0. Then for the gas : (2001; S)

P

34. An ideal block body at room temperature is thrown into a furnace. It is observed that : (2002; 2M) (a) Initially it is the darkest body and at later times the brightest (b) it is the darkest body at all times (c) it cannot be distinguished at all times (d) initially it is the darkest body and at later times it cannot be distinguished 148

35. The graph, shown in the diagram, represents the variation of temperature (T) of the bodies, x and y having same surface area, with time (t) due to the emission of radiation. find the correct relation between the emissivity and bodies : (a) Ex > Ey and a x (b) Ex < Ey and a x (c) Ex > Ey and a x (d) Ex < Ey and a x

T

P

P

A

A

y x (c)

(d) B

t

C

C V

absorptivity power of the two (2003; 2M) < ay > ay > ay < ay

36. Two rods one of aluminium and the other made of steel, having initial length l1 and l2 are connected together to form a single rod of length l1 + l2. The coefficients of linear expansion for aluminium and steel are αa and αs respectively. If the length of each rod is increased by the same amount when their temperature

l1 are raised by t°C, then find the ratio l + l . 1 2

B V

38. 2 kg of ice at – 20°C is mixed with 5 kg of water at 20°C in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice are 1 kcal/kg/°C and 0.5 kcal/kg/°C while the latent heat of fusion of ice is 80 kcal/kg : (2003; 2M) (a) 7 kg (b) 6 kg (c) 4 kg (d) 2 kg 39. Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating is constant. Which of the following graphs represent the variation of temperature with time? (2004; 2M) Temp.

Temp.

(2000; 2M)

αs (a) α a (c)

αa (b) α s αs

(αα + α s )

(d)

(a)

Time (b)

Time

αa

(αα + α s ) Temp.

Temp.

37. The P-T diagram for an ideal gas is shown in the figure, where AC is an adiabatic process, find the corresponding P-V diagram : (2003; 2M) (c) P

Time

(d)

Time

A

B

40. An ideal gas expands isothermally from a volume V1 to V2 and then compressed to original volume V 1 adiabatically. Initial pressure is P1 and final pressure is P3. The total work done is W. Then : (2004; 2M) (a) P3 > P2 , W > 0 (b) P3 > P2 , W < 0 (c) P3 > P1 , W > 0 (d) P3 = P1 , W = 0

C T

41. Two identical conducting rods are first connected independently to two vessels, one containing water at 100°C and the other containing ice at 0°C. In the second case, the rods are joined end to end and connected to the same vessels. Let q 1 and q 2 g/s be the rate of melting of ice in the two cases respectively.

P

P

A

A

(a)

(b) B

C V

C

B

The ratio q1 is : q2

V

149

(2004; 2M)

1 (a) 2 (c)

(a) (b) (c) (d)

2 (b) 1

4 1

(d)

1 4

42. Three discs, A, B and C having radii 2 m, 4 m and 6 m respectively are coated with carbon black on their outer surfaces. The wavelengths corresponding to maximum intensit are 300 nm, 400 nm, and 500 nm, respectively. The power radiated by them are QA : QB and QC respectively : (2004; 2M) (a) QA is maximum (b) QB is maximum (c) QC is maximum (d) QA = QB = QC 43. Water of volume 2 L in a containder is heated with a coil of 1 kW at 27°C. The lid of the containder is open and energy dissipates at rate of 160 J/s. In how much time temperature will rise from 27°C to 77°C? (Given specific heat of water is 4.2 kJ/kg] : (2005; 2M) (a) 8 min 20 s (b) 6 min 2 s (c) 7 min (d) 14 min 44. In which of the following process, convection does not take place primarily? (2005; 2M) (a) Sea and land breeze (b) Boiling of water (c) Warming of glass of bulb due to filament (d) Heating air around a furnace

47. A body with area A and temperature T and emissivity e = 0.6 is kept inside a spherical black body. What will be the maximum energy radiated by body(2005; 2M) (a) 0.60 σ AT4 (b) 0.80 σ AT4 4 (c) 1.00 σ AT (d) 0.40 σ AT4 48. An ideal gas is expanding such that pT = constant. The coefficient of volume expansion of the gas is (2008; 3M) (a) 1/T (b) 2/T (c) 3/T (d) 4/T 49. Two rods of different materials having coefficients of thermal expansionα1 , α2 and Young's moduli Y1 , Y2 respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the rods. If α1 : α2 = 2 : 3, the thermal stresses developed in the two rods are equal provided Y1 : Y2 is equal to : (1989; 2M) (a) 2 : 3 (b) 1 : 1 (c) 3 : 2 (d) 4 : 9

OBJECTIVE QUESTIONS More than one options are correct? 1.

For an ideal gas : (1989; 2M) (a) the change in internal energy in a constant pressure process from temperature T1 and T2 is equal to nC V (T2 – T1 ), where CV is the molar heat capacity at constant volume and n the number of moles of the gas. (b) the change in internal energy of the gas and work done by the gas are equal in magnitude in an adiabatic process (c) the internal energy does not change in an isothermal process (d) no heat is added or removed in an adiabatic process

2.

An ideal gas is taken from the state A (pressure P, volume V) to the state B (pressure P/2, volume 2V) along a straight line path in the P-V diagram. Select the correct statements from the following : (1993; 2M) (a) The work done by the gas in the process A to B exceeds the work that would be done by it if the system were taken from A to B along an isotherm (b) In the T-V diagram, the path AB becomes a part of a parabola

45. Variation of radiant energy emitted by sun, filament of tungsten lamp and welding are as a function of its wavelength is shown in figure. Which of the following option is the correct match? (2005; 2M)

E1 T3 T2 T1 l (a) (b) (c) (d)

Sun-T1 , Sun-T2 , Sun-T3 , Sun-T1 ,

tungsten tungsten tungsten tungsten

filament-T2 , filament-T1 , filament-T2 , filament-T3 ,

welding welding welding welding

From1 14.5°C to 15.5°C at 760 mm of Hg From 98.5°C to 99.5°C at 760 mm of Hg From 13.5°C to 14.5° at 76 mm of Hg From 3.5°C to 4.5°C at 76 mm of Hg

are-T3 are-T3 are-T1 are-T2

46. Calorie is defined as the amount of heat required to raise temperature of 1 g of water by 1°C and it is defined under which of the following conditions? (2000; 2M) 150

(c) In the P-T diagram, the path AB becomes a part of a hperbola (d) In going from A to B, the temperature T of the gas first increase to a maximum value and then decrease 3.

4.

5.

Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface area of the two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength λB corresponding to maximum spectral radiancy in the radiation from B shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A by 1.00 µm. If the temperature of A is 5802 K : (1994; 2M) (a) the temperature of B is 1934 K (b) λB = 1.5 µm (c) the temperature of B is 11604 K (d) the temperature of B is 2901 K From the following statements concerning ideal gas at any given temperature T, select the correct one (s) : (1995; 2M) (a) The coefficient of volume expansion at constant pressure is the same for all ideal gases (b) The average translational kinetic energy per molecule of oxygen gas is 3 kT, k being Boltzmann constant (c) The mean-free path of molecules increases with increase in the pressure (d) In a gaseous mixture, the average translational kinetic energy of the molecules of each component is different.

8.

2vrms

T0

(a) (b) (c) (d) 9.

Black Black Black Black by it

body will absorb more radiation body will absorb less radiation body emit more energy body emit energy equal to energy absorbed

CV and CP denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then (2009; M) (a) CP - CV is large for a diatomic ideal gas than for a monoatomic ideal gas (b) CP + CV is large for a diatomic ideal gas than for a monoatomic ideal gas (c) CP / CV is large for a diatomic ideal gas than for a monoatomic ideal gas (d) CP . CV is large for a diatomic ideal gas than for a monoatomic ideal gas

10. The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then, (2009; M)

(b) no molecule can have speed less vp / 2 (c) v mp < v < vrms (d) the average kinetic energy of a molecule is

A bimetallic strip is formed out of two identical strips one of copper and the other of brass. The coefficients of linear expansion of the two metals are αC and αB. On heating, the temperature of the strip goes up by ∆T and the strip bends to form an arc of radius of curvature R. Then R is : (1999; 3M) (a) proportional to ∆T (b) inversley proportional to ∆T (c) proportional to | αB – αC| (d) inversely proportional to | αB – αC| A black body of temperature T is inside chamber of temperature T0. Now the closed chamber is slightly opened to sun such that temperature of black body (T) and chamber (T0) remains constant : (2006, 3M)

T

Let v , vrms and vp respectively denote the mean speed, root mean square sepeed and most probable speed of the molecules in an ideal monoatomic gas at asbolute temperature T. The mass of a molecule is m. Then : (a) no molecule can have a speed greater than

6.

7.

P

3 2 mv p 4

A

3

B

During the melting of a slab of ice at 273 K at atmospheric pressure : (1998; 2M) (a) positive work is done by the ice-water system on the atmosphere (b) positive work is done on ice-water system by the atmosphere (c) the internal energy of the ice-water increase (d) the internal energy of the ice-water system decreases

151

2

D

1 0

C 1

2

3

(a) the process during the path A → B is isothermal (b) heat flows out of the gas during the path B→C →D (c) work done during the path Α → B → C is zero (d) positive work is done by the gas in the cycle ABCDA

of the heater is negligible. The gas in the left chmaber expands pushing the partition until the final pressure in both chambers become 243 P0 /32. Determine (I) the final temperature of the gas in each chamber and (ii) the work done by the gas in the right chamber. (1984; 8M)

SUBJECTIVE QUESTIONS 1.

A lead bullet just melts when stopped by an obstacle. Assuming that 25 per cent of the heat is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is 27°C. (1981; 3M) (Melting point of lead = 327°C, specific heat of lead = 0.03 cal/g°C, latent heat of fusion of lead = 6 cal/g, J = 4.2 J /cal).

2.

A cyclie process ABCA shown in the V-T diagram is performed with a constant mass of an ideal gas Show the same process on a P-V diagram. (1981; 4M) (In the figure, CA is parallel to the V-axis and BC is parallel to the T-axis). V V2

V1

C

T2

T

3.

Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are 105 N/m2 and 6 L respectively. The final volume of the gas is 2 L Molar specific heat of the gas at constant volume is 3R/2. (1982; 8M)

4.

A solid sphere of copper of radius R and a hollow sphere of the same material of inner raidus r and outer radius R are heated to the same temperature and allowed to cool in the same environment. Which of them starts cooling faster? (1982; 2M)

5.

6.

Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0°C and a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at 62°C. What is the new value of the pressure inside the bulbs? The volume of the connecting tube is negligible. (1985; 6M)

8.

An electric heater is used in a room of total wall area 137 m2 to maintain a temperature of + 20°C inside it, when the outside temperature is –10°C. The walls have three different layers materials. The innermost layer is of wood of thickness 2.5 cm, the middle layer is of cement of thickness 1.0 cm and the outermost layer is of brick of thickness 25.0 cm. Find the power of the electric heater. Assume that there is no heat loss through the floor and the ceiling. The thermal conductivities of wood, cement and brick are 0.125, 1.5 and 1.0 W/m/°C respectively. (1986; 8M)

9.

A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle 5 cm containing mercury and the two equal ends containing air at the same pressure P. When the tube is held at an angle of 60° with the vertical direction, the length of the air column above and below the mercury column are 46 cm and 44.5 cm respectively. Calculate the pressure P in centimeters of mercury. (The temperature of the systm is kept at 30°C). (1986; 6M)

B

A T1

7.

One gram mole of oxygen at 27°C and one atmospheric pressure is enclosed in a vessel. (i) Assuming the molecules to be moving with vrms , find the number of collisions per second which the molecules make with one square metre area of the vessel wall. (ii) The vessel is now thermally insulated and moved with a constant speed v0 . It is then suddenly stopped. the process results in a rise of the temperature of the gas by 1°C. Calculate the speed v0 . (1983; 8M)

10. An ideal gas has a specific heat at constant pressure

5R . The gas is kept in a closed vessel of 2 volume 0.0083m3 , at a temperature of 300 K and a The rectangular box shown in figure has a partition pressure of 16 × 106 N/m2 . An amount of 2.49 × 104 J which can slide without friction along the legth of the of heat energy is supplied to the gas. Calculate the final box. Initially each of the two chambers of the box has temperature and pressure of the gas. (1987; 7M) one mole of a monoatomic ideal gas (γ = 5/3) at a 11. Two moles of helium gas (γ = 5/3) are initially at pressure P0 , volume V0 and temperature T0 . The temperature 27°C and occupy a volume of 20 L. The chamber on the left is slowly heated by an electric gas is first expanded at constant pressure until the heater. The walls of the box and the partition are volume is doubled. Then it undergoes an adiabatic thermally insulated. Heat loss through the lead wires change until the temperature returns to its initial value. (1988; 6M) 152 CP =

P

(i) Sketch the process on a P-V diagram. (ii) What are the final volume and pressure of the gas? (iii) What is the work done by the gas? 12. An ideal monoatomic gas is confined in a cylinder by a spring loaded piston of cross-section 8.0×10–3 m2 . Initially the gas is at 300 K and occupies a volume of 2.4 × 10–3 m3 and the spring is in its relaxed (unstretched, uncompressed) state. The gas is heated by a small electric heater until the piston moves out slowly by 0.1 m. Calculate the final temperature of the gas and the heat supplied (in joules) by the heater. the force constant of the spring is 8000 N/m, and the atmospherics pressure 1.0 × 105 Nm–2 . The cylinder and the piston are thermally insulated. The piston is massless and there is no friction between the piston and the cylinder. Neglect heat loss through the lead wires of the heater. The heat capacity of the heater coil is negligible. Assume the spring to be massless. (1989; 8M) Open atmosphere Heater

Rigid support

A

2atm

1atm

D 300K

B

C T

400K

(a) The net change in the heat energy. (b) The net work done. (c) The net change in internal energy 16. A cylindrical block of length 0.4 m and area of cross section 0.04 m2 is placed coaxially on a thin metal disc of mass 9.4 kg and of the same cross-section. The upper face of the cylinder is maintained at a constant temperature of 400 K and the initial temperature of the disc is 300 K. If the thermal conductivity of the material of the cylinder is 10 W/mK and the specific heat capacity of the material of the disc is 600 J/kg-K, how long will it take for the temperature of the disc to increase to 350 K? Assume, for purposes of calculation, the thermal conductivity of the disc to be very high and the system to be thermally insulated except for the upper face of the cylinder. (1992; 8M) 17. One mole of a monoatomic ideal gas is taken through the cycle shown in figure. (1993; 4 + 4 + 2M) P

13. An ideal gas having initial pressure P, volume V and temperature T is allowed to expanded adiabatically until its volume becomes 5.66V while its temperature falls to T/2. (1990; 7M) (i) How many degrees of freedom do gas molecules have? (ii) Obtain the work done by the gas during the expansion as a function of the initial pressure P and volume V.

7 R) at pressure, 2 PA and temperature TA is isothermally expanded to twice its initial volume. Finally gas is compressed at constant volume to its original pressure PA . (1991; 4 + 4M) (a) Sketch P-V and P-T diagrams for the complete process. (b) Calculate the net work done by the gas, and net heat supplied to the gas during the complete process. 15. Two moles of helium gas undergo a cyclic process as shown in figure. Assuming the gas to be ideal, calculate the following quantities in this process. (1992; 8M)

A

D

B

C V

A → B : adiabatic expansion B → C : cooling at constant volume C → D : adiabatic compression D → A : heating at constant volume. The pressure and temperature at A, B etc. are denoted by PA, TA, PB, TB etc. respectively. Given that TA = 1000 K, PB = (2/3) PA and PC = (1/3)P A, calculate the following quantities. (a) The work done by the gas in the process A → B. (b) The heat lost by the gas in the process B →C. (c) The temperature TD. (Given : (2/3)2/5 = 0.85]

14. Three moles of an ideal gas (CP =

18. An ideal gas is taken through a cyclic thermodynamic process through four steps. he amounts of heat involved in these steps are Q1 = 5960J, Q2 = – 5585J, Q3 = – 2980 J and Q4 = 3645 J respectively. The

153

corresponding quantities of work involved and W1 = 2200 J, W2 = – 825 J, W3 = – 11000 and W4 respectively. (a) Find the value of W4 . (b) What is the efficiency of the cycle? (1994; 6M) 19. A closed container of volume 0.2 m3 contains a mixture of neon and argon gases, at a temperature of 27°C and pressure of 1 × 105 Nm–2 . The total mass of the mixture is 28 g. If the molar masses of neon and argon are 20 and 40g mol–1 respectively. Find the masses of the individual gases in the container assuming them to be ideal (Universal gas constant R = 8.314 J/mol-K). (1994; 6M)

24. The apparatus shown in figure consists of four glass columns connected by horizontal sections. The height of two central columns B and C are 49 cm each. The two outer columns A and D are open to the atmosphere. A and C are maintained at a temperature of 95°C while the columns B and D are maintained a 5°C. The height of the liquid in A and D measured from the base line are 52.8 cm and 51 cm respecively. Determine the coefficient of thermal expansion of the liquid. (1997; 5M)

D 5°C

A

20. A gaseous mixture enclosd in a vessel of volume V consists of one gram mole of gas A with γ ( = Cp /Cv = 5/3) and another gas B with γ = 7/5 at a certain temperature T. The gram molecular weights of the gases A and B are 4 and 32 respectively. The gases A and B do not react with each other and are assumed to be ideal. The gaeous mixture follows the equation PV19/13 =K in adiabatic process. (2002; 5M) (a) Find the number of gram moles of the gas B in the gaseous mixture. (b) Compute the speed of sound in the gaseous mixture at 300 K. (c) If T is raised by 1 K from 300 K, find the percentage change is the speed of sound in the gaseous mixture. (d) The mixture is compressed adiabatically to 1/5 of its initial volume V. Find the change in its adiabatic compressibility in terms of the given quantities. 21. At 27°C, two moles of an ideal monoatomic gas occupy a volume V. The gas expands adiabatically to a volume 2V Calculate : (1996; 5M) (a) the final temperature of the gas, (b) change in its internal energy (c) the work done by the gas during this process

95°C B 5°C

C 95°C

25. A double-pane window used for insulating a room thermally from outside consists of two glass sheets each of area 1m2 and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state, the room-glass inter face and the glass-outdoor interface are at constant temperature of 27°C and 0°C respectively. Calculate the rate of heat flow through the window pane. Also find the temperatures of other interfaces. Given the conductivities of glass and air as 0.8 and 0.08 Wm –1 K–1 respectively. (1997C; 5M) 26. A sample of 2 kg monoatomic helium (assumed ideal) is taken through the process ABC and another sample of 2 kg of the same gas is taken through the process ADC (see fig.) Given molecular mass of helium = 4. (1997C; 5M)

4

22. The temperature of 100 g of water is to be raised from 24°C to 90°C by adding steam to it. Calculate the mass of the steam required for this purpose. (1996; 2M) 23. One mole of a diatomic ideal gas (γ = 1.4) is taken through a cyclic process starting from point A. The process A → B is an adiabatic compression. B → C is isobaric expansion, C → D an adiabatic expansion and D → A is isochoric. The volume ratio are VA / VB = 16and VC/VD = 2 and the temperature at A is TA = 300 K. Calculate the temperature of the gas at the points B and D and find the efficiency of the cycle. (1997; 5M) 154

2

(10N/m) 10 5

B

A

10

C D

20

V

(i) What is the temperature of helium in each of the states A, B, C and D? (ii) Is there any way of telling afterwards which sample of helium went through the process ABC and which went through the process ADC? Write Yes or No.

(iii) How much is the heat involved the process ABC and ADC?

B

2P1

27. One mole of an ideal monoatomic gas is taken round the cyclic process ABCA as shown in figure. Calculate: (1998;8M)

C

P1

A

B

3P0

T1

P0

A

V0

T

(a) The work done on the gas in the process AB and (b) the heat absorbed or released by the gas in each of the processes. Give answers in terms of the gas constant R.

C

2V0

2T1

V

(a) the work done by the gas. (b) the heat rejected by the gas in the path CA and the heat absorbed by the gas in the path AB. (c) the nest heat absorbed by the gas in the path BC. (d) the maximum temperature attained by the gas during the cycle. 28. A solid body X of heat capacity C is kept in an atmosphere whose temperature is TA = 300 K. At time t = 0, the temperature of X is T0 = 400 K. It cools according to Newton's law of cooling. At time t1 its temperature is found to be 350 K. At the time (t1 ) the body X is connected to a large box Y at atmospheric temperature TA through a conducting rod of length L, cross sectional area A and thermal conductivity. K. The heat capacity of Y is so large that any variation in its temperature may be neglected. The cross sectional area A of the connecting rod is small compared to the surface area of X. Find the temperature of X at time t = 3t1 . (1998; 8M) 29. Two moles of an ideal monoatomic gas initially at pressure P1 and volume V1 undergo an adiabatic compression until its volume is V2. Then the gas is given heat Q at constant volume V2 . (1999; 10M) (a) Sketch the complete process on a P-V diagram. (b) Find the total work done by the gas, the total change in internal energy and the final temperature of the gas. [Give your answer in terms of P1, V1, V2, Q and R] 30. Two moles of an ideal monoatomic gas is taken through a cycle ABCA as shown in the P-T diagram. During the process AB, pressure and temperature of the gas vary such that PT = constant. If T = 300 K, calculate (2000; 10M)

31. An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227°C. The specific heat S of the container varies with temperature T according to the empirical relation S = A + BT, where A = 100 cal/ kg-K and B = 2 × 10–2 cal/kg-K2 . If the final temperature of the container is 27°C, determine the mass of the container. (Latent heat of fusion for water = 8 × 104 cal/ kg, specific heat of water = 103 cal/kg-K). (2001; 5M) 32. A monoatomic ideal gas of two moles is taken through a cyclic process starting from A as shown in the figure.

VB VD The volume ratios are V = 2 and V = 4. If the A A temperature TA at A is 27°C. VD

D

C

V VB VA

B A TA

T TB

Calculate : (2001; 10M) (a) the temperature of the gas at point B. (b) heat absorbed or released by the gas in each process, (c) the total work done by the gas during the complete cycle. Express your answer in terms of the gas constant R. 33. A 5 m long cylindrical steel wire with radius 2 × 10–3 m is suspended vertically from a rigid support and carries a bob of mass 100 kg at the other end. If the bob gets snapped, calculate the change in temperature of the wire ignoring losses. (For the steel wire : Young's modulus = 2.1 × 1011 Pa; Density -= 7860 kg/ m3 ; Specific heat = 420 J/kg-K) (2001; 5M)

155

34. A cubical box of side 1 m contains helium gas (atomic weight 4) at a pressure of 100 N/m2 . During an observaion time of 1 s, an atom travelling with the root mean square speed parallel to one of the edges of the cube, was found to make 500 hits with a particular wall, without any collision with other atoms. Take

38. A cube of coefficient of linear expansion αs is floating in a bath containing a liquid of coefficient of volume expansion γl . When the temperature is raised by ∆T, the depth upto which the cube is submerged in the liquid remains the same. Find the relation between αs and γl showing all the steps. (2004; 2M)

25 J/mol-K and k = 1.38×10–23 J/K. (2002; 5M) 3 (a) Evaluate the temperature of the gas. (b) Evalulate the average kinetic energy per atom. (c) Evalulate the total mass of helium gas in the box.

39. One end of a rod of length L and cross-sectional area A is kept in a furnace of temperature T1 . The other end of the rod is kept at a temperature T2 . The thermal conductivity of the material of the rod is K and emissivity of the rod is e. It is given that T2 = Ts + ∆T, where ∆T < < Ts , Ts being the temperature of the surroundings. If ∆T ∝ (T1 – Ts ), find the proportionally constant. Consider that heat is lost only by radiation at the end where the temperature of the rod is T2 . (2004; 4M)

R =

35. An insulated box containing a monoatomic gas of molar mass M moving with as speed v0 is suddenly stopped. Find the increment in gas temperature as a result of stopping the box. (2003; 2M)

Ts

36. The top of an insulated cylindrical container is covered by a disc having emissivity 0.6 and conductivity 0.167 Wm–1 K–1 and thickness 1 cm. The temperture is maintained by circulating oil as shown : (a) Find the radiation loss to the surroundings in J/m2 s if temperature of the upper surface of disc is 127°C and temperature of surroundings is 27°C. (b) Also find the temperature of the circulating oil. Neglect the heat loss due to convection.

Insulated Furnance T1

Oil in

17 × 10 –8 Wm–2 K–4 ] 3

(2003; 4M)

37. The piston cylinder arrangement shown contains a diatomic gas at temperature 300 K. The cross-sectional area of the cylinder is 1m2 . Initially the height of the piston above the base of the cylinder is 1 m. The temperature is now raised to 400 K at constant pressure. Find the new height of the piston above the base of the cylinder. If the piston is now brought back to its original height without any heat loss, find the new equilibrium temperature of the gas. You can leave the answer in fraction. (2004; 2M)

1m

L Insulated

40. A metal of mass 1 kg at constant atmospheric pressure and at initial temperature 20°C is given a heat of 20000J. Find the following : (2005; 6M) (a) change in temperature, (b) work done and (c) change in internal energy. (Given : Specific heat 400 J/kg/°C, coefficient of cubical expansion, γ = 9 × 10–5/°C, density ρ = 9000 kg/m3. atmospheric pressure = 105N/m2)

Oil out

[Given : σ =

T2

41. In an insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed. Find the final temperature of the mixture (in kelvin). (2006; 6M) Given, Lfusion = 80 cal/g = 336 J/g. Lvaporization = 540 cal/g = 2268 J/g. S ice = 2100 J/kg, K = 0.5 cal/gK and S water = 4200 J/kg, K = 1 cal/gK. 42. A metal rod AB of length 10x has its one end A in ice at 0°C and the other end B in water at 100°C. If a point P on the rod is maintained at 400°C, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is 540 cal/g and latent heat of melting of ice is 80 cal/g. If the point P is at a distance of λx from the ice end A, find the value of λ. [Neglect any heat loss to the surrounding.] 43. A thin rod of negligible mass and area of cross section 4 × 10–6 m2 , suspended vertically from one end, has a length of 0.5 m at 100°C. The rod is cooled to 0°C, but

156

prevented from contracting by attaching a mass at the lower end. Find (i) this mass and (ii) the energy stored in the rod, given for the rod. Young's modulus = 1011 N/m2 , Coefficient of linear expansion 10–5 k–1 and g = 10 m/s 2 . (1997C; 5M)

figure. The atmospheric pressure is P0. 1.

The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder will then be : P0 (a) P0 (b) 2 P0 Mg P0 Mg + – (c) (d) 2 πR2 2 πR2

2.

While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this conditions, the distance of the piston from the top is: (2007; 4M)

ASSERATION AND REASON

1.

This section contains, statement I (assertion) and statement II (reasons). Statement-I : The total translation kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and its volume. A fixed thermally (2007; 4M) Because : Statement II : The molecules of a gas collide with each other and the velocities of the molecules change due to the collision. (a) Statement-I is true, statement-II is true; statementII is a correct explanation for statement -I (b) Statement-I is true, statement-II is true; statementII is NOT a correct explanation for statement -I (c) Statement-I is true, statement-II is false (d) Statement-I is false, statement-II is true

 2P0 πR 2   (2L ) (a)  2   πR P0 + Mg 

3.

COMPREHENSION Passage A fixed thermally conducting cylinder has a radius R and height L0. The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the

2R

L

 P0 π R2 Mg   (2 L) (b)  2   πR P0 

 P0 πR 2    (2L )  P0 π R2 + Mg   πR 2 P – Mg    (2 L )  0  (c)  (d) 2   πR P0  The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the L0 water surface in the tank is at H the same level as the top of the cylinder as shown in the figure. The density of the water is ρ. In equlibrium, the height H of the water column in the cylinder satisfies : (2007; 4M) (a) ρg (L0 – H)2 + P0 (L0 – H) + L0P0 = 0 (b) ρg (L0 – H)2 – P0 (L0 – H) – L0P0 = 0 (c) ρg (L0 – H)2 + P0 (L0 – H) – L0P0 = 0 (d) ρg (L0 – H)2 – P0 (L0 – H) + L0P0 = 0

L0

Piston

ANSWERS FILL IN THE BLANKS 1. 2R

2. partly solid and partly liquid

3.

4. 0°C

2T

9. temperature remains constant 10. 0.628 11. 60°C 13. 46.68°C 14. 5.5

5.

5803

6.

12. 192°

TRUE/FALSE 1. F 4. T 6. T

2. F 4. F 7. F

3. 5.

F F 157

1.71 ρrc

7.

4πKTR2 P

8.

Pt M

MATCH THE COLUMN 1.

Match the following for the given process

(2006; 6M)

P J

30

M

20 10

K

L

10

(A) (B) (C) (D)

3 20 V ( m )

Column I Process J → K Process K → L Process L → M Process M → J

Column II (p) (q) (r) (s)

Q> W< W> Q<

0 0 0 0

2.

Column 1 gives some devices and Column II gives some processes on which the functioning of these devices depend. Match the devices in Column I with the processes in Column II. (2007; 6M) Column I Column II (A) Bimetallic strip (p) Radiation from a hot body. (B) Steam engine (q) Energy conversion (C) Incandescent lamp (r) Melting (D) Electric fuse (s) Thermal expansion of solids

3.

Column I contains a list of processes involvin expransion of an ideal gas. Match this with Column II describing the thermodynamic change during this process. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS (2008; 7M) Column I Column II (A) An inuslated container has two chambers (p) The temperature of the gas decreases separated by a valve. Chamber I contains an ideal gas and the Chamber II has vaacuum. The valve is opened.

(B)

An ideal monoatomic gas expands to twice its original volume such that its pressure p ∝

(C)

where V is the volume of the gas. Real monoatomic gas expands to twice its

,

orginal

constant (r)

The gas loses heat

, where V is its volume V 4/ 3 An ideal monoatomic gas expands such that its (s) pressure p and volume V follows the behaviour shown in the graph

The gas gains heat

such that its pressure p ∝ (D)

1 V2

(q) The temperature of the gas increases or remains

1

158

OBJECTIVE QUESTION (ONLY ONE OPTION) 1. 8. 15. 22. 29. 36. 42. 49.

(a) (b) (c) (d) (b) (c) (b) (c)

2. (a) 9. (b) 16. (d) 23.(d) 30. (a) 37. (c) 43. (a)

3. 10. 17. 24. 31. 37. 44.

(b) (d) (a) (b) (b) None (c)

4. (d) 11. (b) 18. (c) 25. (a) 32. (a) 38. (b) 45. (c)

5. (b) 12. (d) 19. (d) 26. (a) 33. (a) 39. (c) 46. (a)

6. 13. 20. 27. 34. 40. 47.

(c) (b) (d) (c) (a) (c)

7. 14. 21. 28. 35. 41. 48.

(d) (c) (d) (b) (c) (c) ()

OBJECTIVE QUESTIONS (MORE THAN ONE OPTION) 1. (a, b, c, d) 7. (b, d) 10. (b, d)

2. (a, b, d) 3. 8. (a, b, c, d) 9.

(a, b) (b, d)

4. (a, c)

5. (c, d)

6. (b, c)

SUBJECTIVE QUESTIONS 1. 12.96 m/s 2. 3. – 972 J 4. Hollow sphere 5. (i) 1.96 × 1027 /s (ii) 36 m/s 6. (i) T1 = 12.94 T0 . T2 = 2.25 T0 (ii) – 1.875 RT0 7. 83.75 cm of Hg 8. 9000 W 9. 75.4 cm of Hg 6 2 10. 675K, 3.6 × 10 N/m 11. (i) (ii) 113 L, 0.44 × 105 N/m2 (iii) 12459 J 12. 800 K, 720 J 13. (i) f = 5 (ii) W = – 1.23 PV 14. (b) 0.58 RTA . 0.58 RTA 15. (a) 1152 J (b) 1152 J (c) zero 16. 166.32 s 17. (a) 1869.75 J (b) – 5297.6 J (c) 500 K 18. (a) 765 J (b) 10.82% 19. Mass of neon = 4.074g, mass of argon = 23.926 g 20. (a) 2 mol (b) 401 m/s (c) 0.167% (d) – 8.27 × 10–5 V 21. (a) 189 K (b) – 2767 J (c) 2767 J 22. 12 g 23. TB = 909K, TD = 791.4 K, 61.4% 24. 6.7 × 10–5 /°C 25. 41.6 W, 26.48°C, 0.52°C 26. (i) TA =120.34 K, TB =240.68 K, TC=481.36 K, TD =240.68 K (ii) No (ii) QABC=3.25 × 106 J, QADC=2.75 × 106 J 27. (a) P0 V0

25 P0V0 5 P0V0 (b) – P0V0 , 3P0 V0 (c) (d) 8 R 2 2

29. (b) (i) Wtotal 30. 31. 32. 33. 34.

 V 2 / 3   V 3 3  1    1 –1 = PV 1 1 = – PV (ii) ∆Utotal 2 1 1  V    V2 2  2   

(a) 1200 R (b) QAR = – 2100R, QBC = 1500R, QCA = 831.6R 0.495 kg (a) 600 K (b) 1500 R, 831.6 R, – 900 R, – 831.6 (R) (c) 600 R 4.568 × 10–3 °C (a) 160 K (b) 3.312 × 10–21 J (c) 0.3 g

35. ∆T =

37.

–2 K 4 t j   28. 300 + 12.5 e CL 

Mv02 3R

36. 595 W/m2 (b) 162.6°C

4 m, 448.8 K 3

38. γl = 2α,

K 39. Proportional constant =

4eσLTs3 + K 159

2/3

  

 K  

 2/3 V  Q PV –1 + Q (iii) T = + 11 1 Final  3R 2 R  V2  

40. (a) 50°C (b) 0.05 J (c) 19999.95 J 41. 273 K 43. (i) 40 kg (ii) 0.1 J

MATCH THE COLUMN 1. A-s, B-q, C-p, r, D-q, s 2. A-s, B-q, C-p, q, D-q, r 3. A-q, B-p, r, C-p, s, D-q, s

ASSERATION AND REASON 1. (b)

COMPREHENSION 1. (a)

2. (d)

3.

(c)

SOLUTIONS FILL IN THE BLANKS 1.

3 2

For monoatomic gasCV = R and for diatomic gas CV =

5 R 2

⇒ CV = 2.

3.

σT 4 (surface area of sum) become (Surface area of sphere having radius = distance of the earth and sun)

n1CV1 + n2 CV 2 n1 + n2

= =

1×

=

3 5 R + 1× R 2 2 = 2R 2

4 It value is given 1400 ⇒ σT

O → A the material remains solid but A → B it starts melting at constant temperature. ⇒ at P the solid and its liquid both will co-exist.

⇒

= constant πRT Puting, P = , we have V

T = constant V

4.

6.

V

So, if V is doubled, T becomes

Q = mL = 100 × 80 = 8000 cal.

5.

2 1 / 4

  

= 5803 K

The energy radiated by copper sphere is (per second)

⇒

 dT  mc  –  = σAT4  dt 

or

∫0 dt = 

or

t =

But cooling water from 25°C to 0°C provides Q′ = ms∆θ ⇒

 1.59 ×1011   ×  7 × 108   

Q = σ (T 4 − O ) A = σT 4 A which is responsible for decrease in temp by dT

2 times.

Melting of 100 g of ice requires

or

1400 r 2  T= 2  σ R 

 1400 =  5. 67 ×10 − 8 

2

T∝

R2 = 1400 r2

1/ 4

VP2

or

σT 4 4πR 2 4πr 2

Q′ = 300 ×1× ( 25 − 0) = 7500 cal. Whole ice will not melt and the temperature remains at 0°C. (finally)

Heat radiated by the sun per sec. unit area is σT 4 (Treating it as a black body) This heat received by unit area of earth per sec will

=

160

 –mc  σA 

t

100

∫200 T

–4

dt

mc  1 1  –   3σA  (100)3 (200)3 

7mc 7mc × 10–6 × 10–6 = 3 × 8 σA 24σA

Substituting the value

is same

 4 3 −6  7 πρ r (c)× 10 3  t=  3 × 8 × 5.67 × 10 −8 × 4πr 2 = 1.71 ρrc 7.

⇒

Heat generated per second must be radiated out of the sphere ⇒ Power of source = Rate of heat flow P= H =

⇒

temperaturedifference T = 1 t thermalresistance K 4πR 2

t=

9.

L=

1.

t = 3.3 ×10 2 s =

⇒

t = 5 .5 min

330 min 60

V =

3RT 1 ⇒V ∝ at same temp. M M

∴ For gases of diff. moleculer mass, we have different rms speeds.

100 k 2 100 2 5 −1 = B = ⇒ =1+ = θ kA 3 θ 3 3

2.

Slope of V - T graph

dV nR 1 = ∝ dT P P

(Slope) P1 > (Slope)P2 ⇒ P1 < P2

3 × 100 = 60 ° 5

12. Equilibrium will be achieved only if P at both sides of movable partition is same. P1 = P2

n1R T n2 RT = V1 V2

⇒

TRUE/FALSE

θ−0 100 − θ 11. In series H is same ⇒ l / k A = l / k A B A

⇒

W

× 0. 2m 2 m2 = 280 Js –1 This heat will melt the ice in t sec ⇒ (280 Js –1 ) (t s) = mL = (280 × 10–3 ) (3.3 × 105 J kg –1 )

∆ U = ∆Q − ∆W = 0

⇒ θ=

8 8 (360 °) = ( 360 °) = 192 ° 8+ 7 15

= 1400

average KE of particles remains unchanged Temperature is unchanged.

10. Under isothermal conditions PV = constant ∴ (P1 ) (2V) = PV ⇒ Pi = P/2 Under adiabtic conditons PVT = constant where γ = 1.67 1.67 ∴ (Pa ) (2V) = PV1.67 ⇒ Pa = P/(2)1.67 Therefore, the ratio Pa /Pi = (2/21.67) = 0.628

⇒

α=

14. Heat received per second = Intensity × Area

Pt M

As the system is an ideal gas which expands against free space ⇒ no work is done by it or on it ⇒ ∆W = 0 , ∆Q = 0 ∴ ⇒ ⇒

⇒

 3.5 ×104   T2 = 273.16)  4 K  3× 10  = 318.68k or temperature at given pressure will be (318.68 – 273)C or 45.68°C

Heat lost = Heat required P.t = M.L. ∴

α = Angle V2 towards

or

But this is valid only for t < < R. (as it is a shell) 8.

⇒

13. The gas thermometer works at constant volume. Therefore, T∝P or T2 /T1 = P2 /P1 or T2 = T1 (P2 /P1 )

P

4πkR2T P

V2 V V M 32 8 = 1 or 2 = 1 = = M 1 M 2 V1 M 2 28 7

⇒

3.

n1V2 = n2V1 4.

mass m = Now number of moles = and m molecular mass M 161

3RT M So, vrms not only depends on T but also on M. vrms =

Slope of P–V graph for isothermal process = P and slope for adiabatic process = γ P as γ > 1 ⇒ adiabatic process has larger slope than that of

isothermal process. ⇒ A represents isothermal process and B represents adiabatic process.

(1. 1+ 0.02) × 103 × 1× (80 − 15) 540 = 0.135 kg.

T M When oxygen gas dissociates into atomic oxygen its atomic mass M will remain half. Temperature is doubled. So, from Eq. (1) vrms will become two times. or

6. 7.

vrms ∝

Cp – CV = R > 0

5.

⇒ CP > CV

=

2

( E / t )1 4πR12 T14 = × ( E / t ) 2 4πR22 T24

1.

2.

6.

4.

CV 2 =

5 7 R γ2 = 2 5

12 3 = = 1.5. 8 2

In this case these two coaxial cylindrical layers are parallel to each other.

H =

θ1 − θ 2  l     KA 

2R

3 × 8.31 × 300 M ⇒

=

(∆Q) P = nCP ∆T  (∆Q)V CV 1 5 = = = ⇒ (∆Q)V = nCV ∆T  (∆Q) P C P γ 7 5 5 × ( ∆Q ) P = × 70 = 50 cal. 7 7

3 5 CV1 = Rγ1 = 2 3

,

R

M = 2 × 10 −3 kg = 2g or the gas is H2 . ∴ (a)

( ∆Q ) V =

CV1 +CV2

n1 = n2 = 1

,

H = H1 + H 2 , and

3RT M Room temperature T = 300 K

∴

3.

γ1CV1 +γ2CV2

CP =γ CV

will be the net rate of flow of heat.

vrms =

1930 =

n1CV1 + n2CV2

and

⇒ Temperature difference is same and H = H1 + H 2

Work done in cyclic process = area surrounded by P-V graph of this process = (2P – P) × (2V – V) = PV

∴

n1γ1CV1 + n2γ 2CV2

=

4

1 4  =   ×  = 1 4 2  ⇒ They radiate same heat. OBJECTIVE QUESTIONS (ONLY ONE OPTION)

n1CV1 + n2CV2

5 3 7 5 × + × = 3 2 5 2, 3 5 + 2 2

= σT 4 E / t = σAT 4 ⇒

n1CP1 + n2CP2

γ mixture =

=

According to the Stefan's law Energy (per sec per unit area) radiated by a black body

⇒

cw ∆θ L

=

3RT M

vrms =

5.

ms = (mw + weq )

θ1 − θ 2 k eff π (2 R) 2 l θ1 − θ 2 θ − θ2 K1πR 2 + 1 K 2 π( 4R 2 − R 2 ) l l

⇒ K eff = 7.

K1 + 3 K 2 4

The desired fraction is :

∆U nCV ∆ T CV 1 f = ∆ Q = nC ∆ T = C = γ P P

Heat given by steam = Heat required by water and calorimeter

ms L = ( mw + weq ) cw ∆θ 162

or

f=

∴

(d)

5 7 (as γ = ) 7 5

8.

Average speed of molecules depends only on absolute temperature and molecular mass of gas (mass of molecule)

8kT πm But all three vessels are at same temperature.

1/3

=

1 2 /3  1  (3) =   3 3

Vav =

⇒ Vav ∝

11.

vrms ∝ T When temperature is increased from 120 K to 480 K (i.e. four times), the root mean square speed will i.e.,

1 and hence average speed of O 2

m

molecules in A and C vessels will be same i.e. v1 . 9.

become

The heat lost = heat gained by each portion of a conductor during steady state flow. C

B ( 2 T)

⇒

A (T)

l

⇒

⇒

or

∴

T − TC

T −T 2 = C ⇒ T − TC = 2TC − 2T l l 2

dQ dT = mc = eσA(T 4 − T04 ) dt dt

Tsun Tnorthstar

1 λm

= =

( λ m) north star ( λ m ) sun

=

350 = 0. 69 510

15. In this case V is same in both vessels as well as n = 1 is same

⇒

nR T1 P1 T = V = 1 ⇒ T2 = 2T1 gives P2 nR T T2 2 V P2 = 2P1

4 3  3m   πr ρ ⇒ r =  3  4πρ 

16. Power radiated ∝ (surface area) (T)4 . The radius is 2

⇒

T=

molecule is 3 kT which depends on temperature only. 2 Therefore, if temperature is same, translational kinetic energyh of O2 and N2 both will be equal.

eσ A (T 4 − T04 ) ( dT / dt) 1 3mc 1 = eσ (dT / dt) 2 A2 (T 4 − T04 ) mc m=

2 times

14. Average translational kinetic energy of an ideal gas

TC 3 = T 1+ 2

1/3

But

Ιf T becomes double, V rms becomes

13. From Wien's displacement law λmT = constant

10. The rate of heat lost by spheres decides the rate of temperature fall i.e.

⇒ vrms ∝ T for same gas

= 12.42 ´ 10–21 J

 T − TB  = KA C  lCB 

TC (1 + 2 ) = 3T ⇒

3RT M

i.e. v rms (at 600 K ) = 2 ( 484 ) = 684 m/s and KE (600)

For TB ( 2T ) > T A (T )

T −T KA A C  l AC

4 to 2 times i.e., 2v.

3   and KE becomes 2 times  KE = KT  2  

⇒ B loses heat to A ⇒ C provides heat to B by equal amount As K is same as well as A then heat flowing from

or

12. vrms =

l 2

l

⇒

3RT M

vrms =

( dT / dt) 1 A 1 r  1 m  = 1 =  1  =  1  (dT / dt) 2 3 A2 3  r2  3  m2 

2 /3

1 times. 4 Temperature is doubled, therefore, T4 becomes 16 halved, hence, surface area will becomes

163

is maximum

1 times. New power = (450)   (16) = 1800 W. 4

⇒ U 2 is maximum ⇒ U 2 > U1 is correct. Eλ

17. Average kinetic energy per molecules per degree of

1 kT . Since, both the gases are diatomic 2 and at same temperature (300 K), both will have the same number of rotational degree of freedom i.e. two. Therefore both the gases will have the same average rotational kineic energy per molecules freedom =

U3

U2

∆PA 1 2 18. ∆P = 1.5 = 3 B

T is kept same (isothermal)

⇒

⇒

nA 2 = nB 3

But n =

∴

As

m M

⇒

 CP   dT A dTB =   CV  = γ (dTA ) [γ = 14 (diatomic)] [dTA = 30 K] = (1.4) (30K) dTB = 42K

γRT M

V =

VN 2 VHe

=

Energy value corresponding to this wavelength

7 / 28 5 = 5/ 4 3

7×3×4 ( 3) = 5 × 5 × 28 5

Hence, U = U O2 + U Ar

5  3  = 2  RT  + 4  RT  = 11RT 2  2  23. During adiabatic expansion, we know TVγ–1 = constant or

T1V1γ –1

= T2V2γ –1

For a monoatomic gas, γ =

b 2.88 ×10 6 nm K = T 2880 K = 1000 nm

⇒

7 and Helium is monoatomic 5

f  U = n  RT  2  where f = degree of freedom. = 5 for O2 and 3 for Ar

λm = λm

U3

22. Inernal energy of n moles of an ideal gas at temperature T is given by

20. From Wien's lawλm T = b it is clear that the wavelength corresponding to maximum energy radition is

⇒

λ

5 . 3

mA n A 2 = = mB nB 3 for same gas.

19. A is free to move, therefore, heat will be supplied at constant pressure ∴ dQA = nC P dTA ...(1) B is held fixed, therefore, heat will be supplied at constant volume. ∴ dQB = nC V dTB ...(2) But dQA = dQB (given) ∴ nC P dTA = nC V dTB ∴

γ=

1499 1500

489 500

21. Nitrogen is diatomic γ =

RT RT nA −nA V 2V = 2 RT RT 3 nB −nB V 2V

∴

999 1000

U1

= 2 × 1 kT or kT 2 Thus, ratio will be 1 : 1.

∴

T1  V2  =  T2  V1 

γ –1

5 3

 AL2  =   AL1 

(5/3)–1

(A = Area of cross section of piston)

164

2/3

 L2  =   L1 

temperatures we have

24. Wien's displacement law for a perfectly black body is— λm T = constant = Wien's constant b Here, λm is the minimum wavelength corresponding to maximum intensity I.

v1 = v2

m2 m1

29. Let θ be the temperature of the junction (say B). Thermal resistance of all the three rods is equal. Rate of heat flow through AB + Rate of heat flow through CB = Rate of heat flow through BD 90°C A

1 T From the figure (λm )1 < (λm )3 < (λm )2 Therefore, T1 > T3 > T2 λm ∝

or

25. The Process of converting – 10°C ice into 100°C steam involves change of phase twice and during each change of phase the temp remains constant (when energy is supplied is used to do work against intermolecular forces) and also the rise of temperature should be there hence only Ist graph is found to be correct.

B

D

0

C 90°C

90° – θ 90° – θ θ –0 + = R R R Here R = Thermal resistance ∴ 3θ = 180° or θ = 60°C ∴

26. The corresponding P-V graphs (also called indicaor diagram) in three different processes will be as shown: P

0°C

30. For an ideal gas U ∝ T But ∆ U = ∆Q − ∆W ⇒ ∆U < 0 if ∆Q < 0 and

2

∆W = 0 i.e. U decreases and hence T decreases.

1 3

31. For P-V graphs, slope = γP V1

V2

5 7 γ He   > γ O2   As 3 5

P

or Ar

Area under the graph gives the work done by the gas. (Area)2 > (Area)1 > (Area)3 ∴ W2 > W1 > W3

∆V nRT ∆ V nR ⇒ = 27. δ = and V = V∆T P ∆T P

⇒ Slope of P-V graph of monoatomic gas is greater than that of diatomic gas. ⇒ 2 should be He and 1 should be O2 32. For cyclic process ∆U = 0

P nR 1 × = nRT P T S.T. = [Constant] → rectangular hyperbola. (c)

Hence δ = ∴ ∴

28. v =

⇒

γRT M v=

M = Nm  R KB = ⇒ R = K B N M m

or N 2

⇒

∆ W AB + ∆W BC + ∆WCA = ∆Q = 5. 0 J

⇒

( 2 − 1)10 + O + WCA = 5

⇒

WCA = − 5 J

33. During isothermal process the bulk modulus of a gas is B =

γk BT but for monoatomic gases at same m

V∆ P =P ∆V

Hence β = 165

∆V 1 = ⇒ βP = 1 V∆P P

The remaining heat Q = Q1 – Q2 = 0.8 × 105 cal will melt a mass m of the ice, where

Which is the equation of rectangular hyperbola ∴ (a) 34. According to Kirchoff’s law of radiation “All good absorbers are good emitters”. ⇒ Being best absorber, a black body is also best emitter. ⇒ Initially it appears darkest as it absorbs maximum but finally when it reaches to the temperature of furnace it emits maximum and appears brightest.

39. Temperature of liquid oxygen will first increase in the same phase. Then, phase change (liquid to gas) will take place. During which temperature will remain constant. After that temperature of oxygen in gaseous state will further increase. Hence, the correct option is (c).

 dT  35. Rate of cooling  –  ∝ emissivity (e)  dt  From the graph,  dT   dT  –  >  – dt   y  dt  x

∴ ex > ey

40. Slope of adiabatic process at a given state (P, V, T) is more than the slope of isothermal process. The corresponding P-V graph for the two processes as shown in figure.

Further emissivity (e) ∝ absorptive power (a) (good absorbers are goods emitters also) ∴ ax > ay Hence, the correc answer is (c). Note : Emissivity is a pure ratio (dimensionless) while the emissive power has a unit J/s or watt.

∆L 36. α = L ∆θ 0

P

∆La = L1α at ,

P3

∆Ls = L 2 α s t

P1

⇒

∆La = ∆L s

⇒

L1α a = L 2α s

⇒

L1 α s = L2 α a

⇒

1+

⇒

Q 0.8 × 105 = = 1 kg L 80 × 103 So, the temperature of the mixture will be 0°C, mass of water in it is 5 + 1 = 6 kg and mass of ice is 2 – 1 = 1 kg. m=

C

A

B

L2 α =1 + a L1 αs V1

L1 αs = L1 + L 2 α a + α s

37. AC → adiabatic process BC → isobaric process AB → isothermal process ⇒ In P-V diagrams slopeA–C > SlopeA–B and Slope BC should be zero. Also (A → C and A → B both should be the rectangular hyperbola) ⇒ Option (b) may appear correct but temperature at B is less then that at C shows that volume at B should be less than volume at C which is not so. ⇒ none is correct.

V2

V

In the graph, AB is isothermal and BC is adiabatic. WAB = positive (as volume is increasing) and WBC = negative (as volume is decreasing) plus, |WBC| > |WAB |, as area under P-V graph gives the work done. Hence, WAB + WBC = W < 0 From the graph itself, it is clear that P3 > P1 . Hence, the correct option is (c). Note : At point B, slope of adiabatic (process BC) is greater than the slope of isothermal (process AB). 41. In Ist case ‘Rods’ are in parallel q =

dm l dQ = dt L dt

KA(100 − 0) KA(100 − 0) 2KA(100 ) + = lL lL l Finally they are connected in series.

q1 =

38. Heat released by 5 kg of water when its temperature fall from 20°C to 0°C is Q1 = mc∆θ = (5) (103 ) (20 – 0) = 105 cal when 2 kg ice at – 20°C comes to a temperature of 0°C, it takes an energy Q2 = mc∆θ = (2) (500) (20) = 0.2 × 105 cal

⇒

166

q2 =

KA(100 − 0) 2lL

∴

dv 2nRT = dT k ∴ (b)

q1 4 = q2 1

42. Power = Heat per sec = σAT 4 → Stefan’s law and

49. Thermal stress Given

according to Wein’s law λ mT = constant ⇒

T∝

dV 2 = VdT T

⇒

σ =Yα∆θ σ1 = σ2

Y1 α 2 3 ∴ Y1 α1 ∆θ = Y2 α2 ∆θ or Y = α = 2 2 1

1 λ m Now these two are combined.

A γ2 Hence Power Q ∝ λ4 or Q ∝ 4 λm m

1

C

2

QA : QB : QC =

⇒

2

2

2

(2) ( 4) ( 6) : : (3) 4 ( 4) 4 (5) 4

OBJECTIVE QUESTIONS (MORE THAN ONE OPTION)

= 0.05 : 0.0625 : 0.0576 QB is maximum.

1.

(a) Internal energy of an ideal gas depends only on its temperature ⇒ ∆ U = nCV ∆T for all processes (b) ∆ U + ∆ W = ∆ Q = 0 for an adiabatic process ⇒

43. Heat required to raise the temperature of water from

∆U = − ∆W

27 °C to 77°C = 2 × 4.2 × 103 × 50 ( m s ∆ θ)

(c) ∆ U = nCV ∆T = 0 for T → constant

Net heat gained by water in t sec = (1000 – 160)t = 840 t ⇒ 840 t = 2 × 4.2 × 103 × 50 or t = 500 sec. = 8 min 20 sec.

(d) ∆Q = 0 ⇒ Q = constant for adiabatic process ⇒ All options are correct. 2.

44. Filament of bulb heats glass of this bulb by radiation. (Primarily)

(a) Work done = Area under P=V graph. A1 > A2

45. According to Wein’s law λ m .T = constant ⇒ ⇒

T3 > T2 > T1

A

A

As λ m3 < λ m2 < λ m1 A1

T3 corresponds sun T2 corresponds filament

B

46. Calorie is the amount of heat required to raise the temperature of water from 14.5°C to 15.5°C at 760 mm of Hg. 47. 0.6σAT 4 (emitted) + 0.4σAT 4 (reflected)

B

A2

Given process

T1 corresponds welding arc.

V

V Isothermal process

∴ Wgiven process > Wisothermal process (b) In the given process P-V equation will be of a straight line with negative slope and positive 9ntercept i.e., P = – αV + β (Here α and β are positive constant) ⇒ PV = –αV2 + βV ⇒ nRT = – αV2 + βV

(

)

1 – αV 2 + βV ...(1) nR This is an equation of parabola in T and V.

⇒

= 1. 00σAT 4 will be max. energy radiated ⇒ (c) 48. pT = k ⇒

P

P

(d)

k p= now PV = nRT gives T

⇒

nRT 2 V= k 167

T=

dT = 0 = β – 2αV dV V=

B 2α

d 2T

Now,

2

Vav =

= – 2α = – ve

dV i.e., T has some maximum value. Now, T ∝ PV and (PA)A = (PV)B ⇒ TA = TB We conclude that temperature are same at A and B and temperature has a maximum value. Therefore, in going from A to B, T will first increase to a maximum value and then decrease. 3.

and V rms = ⇒

8RT 8kT = πM πm 3RT = M

3kT m

Vmp < V av < V rms F(v)

As the bodies have same outer surface area and the emit same power 1/ 4

⇒

e AσT A4 A = eB σTB4 A

T A  eB  =  ⇒ TB  e A 

1/ 4

⇒

T A  81  =  TB  1 

= 3 or TB =

As it is evident from the graph that few molecules can have speeds to a very large value and some molecules can have speeds slightly greater than zero.

T A 5802 = = 1934 K 3 3

Now, λ mB − λ mA = 1 µm and λ mB TB = b = λ mA T A ⇒

λ mA = λ mB

‘The average KE =

TB 1 = λm TA 3 B

(a) Coefficient of volumetric expansion is Now PV = nRT ⇒ V = ⇒ ⇒

1 3 2 3 2 m Vmp = mV mp 2 2 4 ⇒ Option (c) and (d) are correct. 6.

There is a decrease in volume during melting of an ice slab at 273 K. Therefore, negative work is done by icewater system on the atmosphere or positive work is done on the ice water system by the atmosphere. Hence, option (b) is correct. Secondly heat is absorbed during melting (i.e dQ is positive) and as we have seen, work done by ice-water system is negative (dW is negative). Therefore, from first law of thermodynamics dU = dQ – dW change in internal energy of ice-water system, dU will be positive or internal energy will increase.

7.

Let l0 be the initial length of each strip before heating. Length after heating will be :

∆V =γ V∆T

nRT P

nR dV dT dV 1 dT or γ = = ⇒ = P V T VdT T γ depends upon T. dV =

3 K BT 2 (c) As the pressure increases, volume decreases for

1 1 3kT 2 mV rms = m 2 2 m

=

1 3 λ mB − λ mB = 1 µm gives λ mB = = 1. 5 µm or 3 2 ⇒ Option (a) and (b)

4.

(b) Average Translational KE =

1  an ideal gas at a given temperature V ∝  P  ⇒ Mean free path decreases. (d) As average translational RE for each component is

αB αC d

3 k BT hence it is same for all gases in a given mixture. 2

5.

Vmp =

2RT = M

V

V mp V ap V rms

θ

2kT m 168

R

lB = l0 (1 + αB ∆T) = (R + d)θ lC = l0 (1 + αC∆T) = Rθ

and

⇒ Option (c) is wrong. Positive WD ( D → A → B) is greater than negative

 1 + α B∆T  R+d  =  R  1+ α C ∆T 

∴

WD ( B → C → D ) Hence in the complete cycle net work done is positive. ⇒ Option (d) is correct.

d = 1 + (αB – αC) ∆T R [From binomial expansion] 1+

∴

∴

R=

or R ∝ 8.

9.

SUBJECTIVE QUESTIONS 1.

3 1 2 mv  = ms∆θ + mL 4  2 

1 ( α B – α C ) ∆T

(8s ∆θ+ 8L ) 3 Substituting the value, we have

1 1 and ∝ | α – α | ∆T B C

or

The chamber is opened ⇒ Black body will absorb more radiation and hence emit more energy (as T is constant) Also for T constant both should be equal ⇒ option (a), (b) and (d) are correct.

C P − CV = R for all gases

CV =

5 7 R and C P = R for diatomic gas 2 2

CV =

3 5 R and C P = R for monoatomic gas 2 2

⇒

= C P .CV =

(8× 0.03 × 4.2 × 300) + (8× 6× 4.2) 3 = 12.96m/s

2.

V = constant (Slope) ⇒ V ∝ T T ⇒ It is an isobaric process. (Expansion) B → C is an isochoric process C → A is an isothermal process. (Compression) ⇒ Using above conditions that P-V diagrams can be drawn as follows A→ B

P

7 = 1.4 for diatomic gas 5

5 = 1.67 for monoatomic gas 3

V

3.

35 = 8.75 for diatomic gas 4

During adiabatic process

 Pi Vi − Pf V f W =  γ −1 

15 = 3.75 for monoatomic gas 4 ⇒ Option (b) and (d) are correct. =

CV =

10. As the graph of an isothermal process is rectangular hyperbola (-ve slope decreasing) and process A → B is – ve slope increasing ⇒ Option (a) is wrong. ∆ Q = ∆U + ∆W shows that ∆Q < O as ∆ WBCD < 0

⇒

v=

v =

C P + CV = 6 for diatomic gas = 4 for monoatomic gas

C P / CV = γ =

75% heat of retained by bullet.

Q f < Qi i.e. heat flows out of this system.

⇒ Option (b) is correct Work done during the path A → B → C is area of semi circle (positive WD ( A → B ) > negative WD ( B → C )] ⇒ Net WD is positive

169

3 5 5 R ⇒ C P = R and γ = 2 2 3

V Pf = Pi i V  f ⇒

   

W=

γ

5/ 3   = 10 5  6  = 6.24 × 10 5 2     

10 5 × 6 × 10 −3 − 6.25 × 105 × 2 × 10 −3 5   − 1 3 

3 [ 6 − 12. 5] × 10 2 2 = 150 [ −6 .5 ] = – 972 J =

4.

dQ = eσAT 4 will be same for both as e A and T dt are same

Further

dT dQ dT 1 = ⇒ ∝ But mc dt dt dt m ⇒ For same outer radium and material, the mass of hollow sphere is less ⇒ its rate of cooling will be more ⇒ it cools down faster than the solid sphere 5.

or

T1 + T2 =

5/3

If n molecules collide per sec per unit area then change

1 dp = Pressure A dt

of gas becomes P = n( 2mv rms )

P 1. 01× 10 5 27 = = 1. 96 × 10 / sec. 32 2mv rms  483 .4 2  6.02 × 10 26 

7.

1 mgasv02 2

As the bulbs are connected by using a tube ⇒ if temperature difference is created thus to keep pressure same, some gas (say x mole) moves from hot bulb to cold bulb. ⇒ Initially we have 76 × V = nR 273 and finally ( n + x) R 273 = P′V = (n − x) R 335

5 ×R 2 ∆T × = v02 −3 1(32 × 10 ) 2 × 1×

or

∆ T = 1°C

v0 = ⇒

6.

5 2 × × 8.31 2 1000 32

v0 = 36 m/s

Applying

n+x =

or

x=

335 (n − x ) 273

62 n 608

P′ ( n + x) x 62 = =1+ =1+ 76 n n 608 P′ = 83 .75 cm of Hg

⇒ 8.

PV = constant for both the chambers. T

V1 =

⇒

Also,

P0V0 PV1 PV2 243P0 = = = (Here, P ) T0 T1 T2 32 ∴

1–5/3

3  = (1)  R  (T0 = 2.25T0 ) 2  = – 1.875 RT0

(b) When the gas filled vessel, moving with speed

nCV ∆T =

...(1)

 T0   243 P0    =   T  2  32P0  Solving this equation, we get T2 = 2.25 T0 From Eq. (1), T1 = 12.94 T0 (ii) Work done by the gas in right chamber (∆Q = 0, adiabatic process) ∆W = – ∆U = nC V (Ti – Tf )

i.e. + 2m vrms

v0 is suddenly stopped (adiabatic process) then entire KE is used as thermal energy to raise temp of gas

243 T 16 0

Applying the P-T equation of adiabatic process to the right chamber,

particletothewallis −[ − mvrms − ( mv rms )]

n=

 16   243  (T1 + T2 ) = T0  

or

(a) During each collision momentum delivered by the

in momentum per sec. per unit area is

 32   T2  V2 =     V0  243   T0  V1 + V2 = 2V0

and

Let R1 , R2 and R3 be the thermal resistance of wood, cement and brick. All the resistance are in series. Hence,

20°C

32  T1    V0 243  T0 

R1

R2

V

R = R1 + R2 + R3 170

R3

10°C

2.5 × 10 –2 1.0 × 10–2 25 × 10–2 l + + (as R = ) 0.125 ×137 1.5 × 137 1.0 × 137 KA = 0.33 × 10–2 °C/W ∴ Rate of heat transfer

2.49 × 10 4 ∆Q = 300 + T f = Ti + PV  5  n(C p − R )  R − R RT  2 

=

= 300 +

temperaturedifference dQ = thermalresistance dt

Pi 675 6 Hence Pf = T × T f = 1.6 × 10 × 300 i

30 =

W 0.33×10–2 = 9000 W ∴ Power of heater should be 9000 W 9.

x

P

P 5 cm (a)

5cm

P1

cm 44. 5

11. PV diagram is shown below AB process is isobaric ∴ V∝T P = constant i.e if V is doubled, T also becomes two times. TA = 300 K ∴ TB = 600 K VA = 20 L ∴ VB = 40 L

m 46 c

P2

30°

PA =

(b)

Now the tube is tilted at an angle of 30° with the horizontal lower end compresses to 44.5 and upper end expands to 46. But total length remains unchanged ⇒ We have ( 2 x + 5) cm = ( 44 .5 + 5 + 46 )cm

x = 45 .25cm If equilibrium (in tilted tube) after isothermal process P′V ′ = PV below lower surface of mercury P2 × A × 44.5 = P × A × x = P × A × 45 .25 Now, above the upper surface of mercury P1 × A × 46 = P × A × x = P × A × 45. 25

⇒

46 × 5 3

5/3

 600  or    300 

5/3–1

 2.49 × 105  =    Pc  

 2.49 × 105 =  Pc 

2/3

   

 300  Pc = (2.49 × 105)    600  Pc = 0.44 × 105 N/m2 Similarly, using PVr = constant we can find that Vc = 113 L The corresponding P-V graph is shown below 5

P (× 10)N/m

2

A

B

C

0.44

44. 5 46 × 5 × 44.5 P2 = = 75.4 cm of Hg 45.25 3 × 45 .25 P = 75 .4 cm of Hg P=

20

(iii)

10. As the volume of gas remains same ⇒

20 × 10 –3 = 2.49 ×105 N /m2 Process BC is adiabatic. So, applying TγP1–γ = constant

2.49

44.5 5 P2 − P2 = 46 2

⇒

(2)(8.31)(300) =

5/2

P2 × 44. 5 = P1 × 46 Also comparing pressures in cm of mercury in titled tube we get P2 = P1 + 5 sin 30

P2 =

nRT A VA

∴

⇒

 1. 5  5 P2   = ⇒  46  2

Pf = 3 .6 × 10 6

⇒

When tube is horizontal the lengths of air columns at both sides of mercury column are equal. x

2. 49 × 10 4 × 300 3 × = 675 K 6 −3 2 1. 6 × 10 × 8. 3× 10

∆W = 0 ⇒ ∆ Q = ∆U = nCV ∆T 171

40

113

V (L)

WAB = Area under P-V graph = (2.49 × 105 ) (40 – 20) × 10–3 = 4980 J WBC = – ∆U = nC V (TB – TC)

γ = 1.4 (diatomic)

3  = (2)  × 8.31 (600–300) = 7479J 2  Wnet = (4980 + 7479) J = 12459 J

∴

12. The piston will finally be in equilibrium ⇒

PG f A = Patm∆ + Kx

⇒

PG f = Patm +

f =

⇒

 Vi Using Pf = Pi  Vf

(ii)

W=

⇒

FGas

γ

(for adiabatic process)

TiV f  Kx  =  Patm +  Now T f = Ti PiVi PiVi  A  Pf V f

 V + 0.1xA  Kx  =  i 1 + Ti Vi   APatm 

1 ( Pt Vt − Pf V f ) γ −1

=

1 [ PV − 0.09ρ × 5.66V ] 1.4 − 1

=

10 × [1 − 0.51]PV 4

=

4.9 = 1. 225 = 1. 23 J 4

Fspring

 0.1× 8 × 10 = 1 + 2.4 × 10 −3 

   

we get Pf = 0.09 P

Kx A Fatm

−3

2 = 5 degree of freedom 1. 4 − 1

Pi → Patm 14. (i) The P-V and P-T diagram are shown below

 8000 × 0.1  1 +  8 × 10 − 3 × 1× 10 5 300 

P

P PA

A

PA

 1 = 1 + (1 + 1)300  3 PA 2

= 800 K

PA 2

B

C

A

C

B

Heat Supplied by heater ∆ Q = ∆U + ∆W + U spring VA

1 ∆ Q = nCV δT + P∆V + KX 2 2 PV TR

=

10 5 × 2.4 × 10 −3 3 × R[800 − 300 ] 300 × R 2

∴P ∝

VA =

nRT A 3RT A = PA PA

= nRT A log e

VB + nR(TC − T A ) + 0 VA

= 2.3. 3× 3RT A log10

T′ V  ( γ − 1) ln   = ln T V′ ln (T / T ′ ) ln 2 = = 0. 4 ln (V ′ / V ) ln(5.66)

T

∆UAB = 0 The total work done ⇒ W = WA→B + WB→C + WC→A

2 Number of degrees of freedom f = γ − 1

( γ − 1) =

TA

1 , V is doubled. Therefore, P will remain half. V

Further,

now T ′V ′ (γ −1) = TV γ−1 (for adiabatic process) ⇒

TA 2

(PV)C =

1 + 10 5[3.2 − 2.4] × 10 −3 + 8000 × ( 0.1) 2 2 = 600 + 80 + 40 = 720 J

13. (i)

V

( PV ) A TA ∴ TC = 2 2 (ii) Process A-B : T = constant

1 3  2  R  ∆T + P∆V + KX 2 2 

=

2VA

2V A T  + 3R A − T A  VA 2  

3 = 2.08RTA − RTA = 0.58RTA = Qnet = 0. 58RTA 2 15. (a) 172

A → B is an isobaric process

⇒ dQ = dU + PdV

P

⇒ ∆ QAB = nCV ∆T + RAT = n(CV + R) ∆T = nC p ∆T

A B

and hence for C → D we have ∆ QCD = −nC p∆ T

B → C and D → A are isothermal processes ⇒ ∆U = 0 for thus ⇒ ∆Q = ∆ W

 PB Hence ∆ QBC = nRTBC ln   PC

D

 2  = 2(8.31)400 ln  1 

V

1  and ∆ QDA = 2(8.31)( 300 ) ln   = −3455 J 2 ⇒

Now, work done in the process A-B will be WAB =

Net change in heat energy

∆ QT = ∆QAB + ∆QBC + ∆QCD + ∆QDA ∆ QT = 0 + 4607 − 3455 = 1152 J .

(b)

Net WD = ∆Q = 1152 J ( ∆U = 0 for cyclic process) U is state function ⇒ ∆U = 0 (for cyclic process)

8.31 (850 − 1000 ) 5 1− 3 or WAB = 1869.75J (b) B -C is an isochoric process (V = constant)

TB P = B TC PC

∴

16. The heat flowing from cylinder to disc will increase the temperature of disc.

mdisc Sdisc 350 k

∫

300 k

dθdisc dt

TC = 425K

3  Therefore, QBC = nC V ∆T = (1)  R  ( TC – TB ) 2 

t

∫ dt 0

⇒

m S L  400 − 300  0.4(600 ) 0.4 t = D D C ln ln 2 = KC A 10(( 0.04)  400 − 350 

⇒

t = 166 . 32 sec.

17. Given TA = 1009 K, PB =

3 =   (8.31)(425–850) 2 QBC = – 5297.6 J Therefore, heat lost in the process BC is 5297.6J. (c) C-D and A-B are adiabtic process. Therefore, γ

2 1 PA , PC = PA 3 3

PC1– γ TCγ

CP 5 Number of moles, n = 1, γ = C = 3 R (monoatomic) V (a) A B is an adiabatic process therefore

PA1– γ TAγ

=

 PC   (1/3) PA   TB =   850K TC =   PB   (2/3) PA 

∴

( 400 − θdisc Q = = Kcylinder A Lcylinder t

dθ D KC A = 400 − θD M D S D LC

PB1– γ TBγ

=

1–5/3 5/3

P  T  t –γ ⇒ C = D  PD  TC 

...(1)

γ

...(2)

Multiplying Eqs. (1) and (2), we get γ

 TB = TA    PB 

3 = (1000)   2

1– γ γ PD TD

 T  t –γ P PA1– γ TAγ = PB1– γ TBγ ⇒ A =  B  PB  TA 

1– γ PA  γ

∴

R (TB – TA ) 1– γ

=

⇒

(c)

C

–2/5

3 2 = (1000)   = (1000)   2 3 TB = (1000) (0.85) TB = 850 K

PC PA  T T 1– γ =  D B PD PB  TC TA  2/5

...(3)

Process B-C and D-A are isochoric. (V = constant) Therefore,

173

PC TC P T and A = A = PB TB PD TD

Multiplying these two equations, we get

PC PA TC TA = PD PB TBTD

Let the number of moles of gas B be n B = n The internal energy of the mixture = internal energy of gas A + internal energy of gas B.

...(4)

From Eqs. (3) and (4), we have

 T DT B  T T  C A

1

 TC TA   1− γ    =   TB TD   1− γ  γ

(n A + n B )

or

(425)(1000) TC TA TD = = K 850 TB

or

TD = 500K

γ mixture

R R T = n A γ − 1 T + nB γ − 1 T A B –1

Solving this, we get n = 2 Note : For γmixture we can directly use the formula

n n1 n + 2 γ mixutre –1 = γ1 –1 γ 2 –1 (b) Molecular weight of the mixture will be given by

W 4 = ( Q1 + Q2 + Q3 + Q4 ) − (W1 + W2 + W3 )

n M + nBM B M = A A nA + nB

= (5960 − 5585 − 2980 + 3645 ) − ( 2200 − 825 − 1100 ) = 765 J (b)

R

1+n 1 n = + (19/13)–1 (5/3)–1 (7/5)–1

18. (a) In a cyclic-process ∆U = 0 as U is a state function ⇒ ∆W = ∆Q ⇒

nRT γ –1

....(1) Since, the mixture obeys the law PV19/13 = constant (in adiabatic process) Therefore, γmixture = 19/13 (PV Y = constant) Substituting the values in Eq. (1) we have

TC TA =1 TDTB

⇒

U=

Therefore,

 TC TA   =   TB TD 

 TC T A   T T   D B

or

Since,

M = 22.67 Speed of sound in a gas is given by

WD × 100 % Efficiency = Q (absorbed)

γRT M Therefore, in the mixture of the gas v=

 2200 − 825 − 1100 + 765  =  × 100% 5960 + 3645   = 10.82%

v=

(19 / 13)(8.31)(300 )

22 .67 × 10 −3 v = 401m/s

19. If mixture contains m gm neon then it will have (28– m)gm argon gas.

(c)

v ∝

T v = KT1/2 ...(2)

M 28 − m ∴ Mixture has moles of neon and moles 20 40

or

of argon

⇒

dv 1 –1/2 = KT dT 2

⇒

 dT  dv = K   2 T 

M 28 − m  RT +   20 40  V

 ∴Net pressure P = 

m 28 − m  RT +   20 40  V

 5 i.e. 1 ×10 = 

⇒

1  7  8.31 × 300  1 =  m −  +  20 40   10  0.02  ⇒ m = 4.074 gm neon is present and 28 − m = 23 .936 gm argon is present in the mixture

dv K  dT  = . v v  2 T  K = v

⇒

20. (a) Number of moles of gas A are n A = 1 (given) 174

=

1 T

1  dT  T 2 T

(From Eq. 2)

 1  dT   =    2 T 

m/s

1  dT dv ×100 =  2 T v

⇒

⇒ ∆ W = O − (− 2767 ) ⇒

  × 100 

22. mst L + mst. S∆Qsteam = Heat Lost by steam

1 1  =  × 100 2  300  = 0.167 Therefore, percentage change in speed is 0.167%

mw S∆θwater = Heat gained by water ms [L + S (100 − 90 )] = 100 × [90 − 24 ]

⇒

1 (d) Compressibility = = β (say) Bulkmodulus

ms =

Adiabatic bulk modulus is given by

dP   Badi = γ P  B = – dV / V   ∴ Adiabatic compressibility will be given by 1 1 1 = and β 'adi = γP γ P ' γ P (5) γ

βadi =

[PVγ = constant] [∴

PVγ

= P' (V/5)r ⇒ P' = P

(5)γ]

23. The corresponding P-V diagram is as shown Given : TA = 300 K, n = 1, γ = 1.4, VA /VB = 16 and VC/ VB = 2 Let VB = V0 and PB = P0 Then, VC = 2V0 and VA = 16V0 Temperature at B : Proces A-B is adiabatic. Hence,

γ

–1   1  ∴ ∆ β = β 'adi – β adi = 1 –   γP   5   =

 

γ –1

19     1 13  = 1 –  5    19  (1 + 2)(8.31)(300)      13 

–V

19    γ = γ mixture = 13    ∆β = – 8.27 × 10–5 V

V  TD = TC  C   VD 

T2 = T1 (V1 /V2 ) γ−1 = 300 (V / 2V ) (5 / 3−1)

=

300 = 189 K ( 4)1/ 3

η=

3  ∆ U = nCV ∆T = n R  (T F − Ti ) 2 

1.4–1

 2 = (1818)   16 

Networkdoneinthecycle ×100 Heatabsorbedinthecycle

or η =

3 × 8.31(189 − 300 ) 2 ∆ U = −2767 J ∆ W = ∆Q − ∆U but process is adiabatic = 2×

(c)

γ –1

TD = 791.4 K Efficiency of cycle : Efficiency of cycle (in percentage) is defined s

T2189 K (b)

TAVAγ –1 = TBVBγ –1

 VA  or TB = TA    VB  = (300) (16)1.4–1 TB = 909 K Temperature at D : B → C is an isobaric process (P = constant) ∴ T ∝ V, VC = 2VB ∴ TC = 2TB = (2) (909) K TC = 1818 K Now, the process C-D is adiabatic. Therefore,

  1 γ  –V nRT  1 –     P − γ ( nA + nB ) RT   5    V   

21. (a)

100 × 1× 66 6600 = = 12 g 540 + 1× 10 550

=

WTotal × 100 Q+ ve

 Q  Q+ve – Q– ve × 100 = 1 – 1  ×100 Q+ve  Q2 

...(1)

where Q1 = Negative heat in the cycle (heat released) and Q2 = Positive heat in the cycle (heat absorbed) In the cycle

∆Q = 0

175

QAB = QCD = 0 (Adiabatic process)

25. As the heat flows, through series combination of these layers of glass air and glass

5  QDA = nC V ∆T = (1)  R  (T A – TD ) 2  (CV =

⇒

5 R for a diatomic gas) 2 5 × 8.31(300–791.4) 2 = – 10208.8J =

or

QDA

27 − θ1 θ − θ2 θ −O KG A = 1 K AA = 2 KG A lG lA lG

⇒

l A kG θ1 + θ2 = 27 °C and θ1 − θ 2 = θ 2 l k G A

⇒

51θ 2 = 27 ° i.e. θ 2 =

Also H =

27 = 0. 53°C 51

θ 2 K G A 27 0. 8 = × ×1 lG 51 0. 01

= 42. 35 J S–1 = 42.35 W 26. Number of gram moles of He,

A B

P0

m 2 ×103 = = 500 M 4 (i) VA = 10m3 : (i) PA = 5 × 104 N/m2 n=

C D

∴

A

V0

2V0

16V0

∴ or

where γ is coefficient of volumetric expansion. In equilibrium, there should be no net horizontal displacement of liquid in arm BC. ⇒

PB . A − PC A = 0 or PB = PC

⇒

P0 + hAρ 95 g − hB ρ 5 g = P0 + hDρ s g − hc ρ 95 g

ρ0 1 + ρ (95)

ρ0 1 + ρ(5)

101 .8 (1 + 5 γ ) = 100 (1 + 95 g γ )

ρ = 2 × 10 −4 °C − 1

PAVA (10)(5 × 104 ) = nR (500)(8.31)

(10)(10 ×104 ) TB = K (500)(8.31) TB = 240.68 K VC = 20m3 , PC = 10 × 104 N/m2

TC

( 20)(10 × 10 4 ) K (500 (8.31)

or TC = 481.36 K and VD = 20m3 , PD = 5 × 104 N/m2

( hA + hC )ρ 95 = (hB + hD )ρ 5

(52 .8 + 49)

TA =

or TA = 120.34K Similarly, VB = 10m3 , PB = 10 × 104 N/m2

V

m m ρ0 24. Density of liquid ρ b = V = V (1 + γt ) = (1 + ρt ) 0 0

= (51 + 49 )

⇒ θ1 = 50θ 2

and θ1 = 26 .47°C

 10208.8  η = 1–  ×100  26438.3  η = 61.4%

or

⇒

⇒

5 0. 8 = θ2 × 1 0.08

7  Q BC =  (8.3)(1818 − 909 ) J 2 or QBC = 26438.3 J Therefore, substituting Q1 = 10208.8 J and Q2 = 26438.3J in Eq. (1), we ge ∴

we get same Rate of flow H1 = H 2 = H 3

TD =

(20)(5 × 104 ) K (500)(8.31)

or TD = 240.68 K (ii) No, it is not possible to tell afterwards which sample went through the process ABC or ADC. But during the process if we note down the work done in both the processes, then the process which require more work goes through process ABC. (iii) In the process ABC

3  dU + nC V ∆T = n  R  (TC – TA ) 2  176

3 = (500)   8.31(481.36–120.34) J 2 ∆U = 2.25 × 106 J and ∆W = Area under BC = (20 – 10) (10) × 104 J = 106 J ∴ ∆QABC = ∆U +∆W = (2.25 × 106 + 106 )J ∆QA B C = 3.25 × 106 J In the process ADC, dU will be same (because it depends on initial and final temperatures only) dW = Area under AD = (20 – 10) (5 × 104 )J = 0.5 × 106 J ∴ ∆QADC = ∆U +∆W = (2.25 × 106 + 0.5 × 106 )J ∆QA B C = 2.75 × 106 J

dQCA =

5 5 ( P0 V0 – 2 P0 V0 ) = – P0 V0 2 2

5 PV 2 v 0 (ii) Heat absorbed in path AB : (process is isocharic) ∴ dQAB = CV dT = CV (Tf – Ti ) Therefore, heat rejected in the process CA is

 Pf V f PV  = CV  – i i R   R

27. (a) ABCA is a clockwise cyclic process. ∴ Work done by the ga.s

=

CV (P V – Pi Vi ) R f f

=

3 (P V – Pi Vi ) 2 f f

3 (3P0 V0 – P0 V0 ) 2 dQAB = 3P0 V0 ∴ Heat absorbed in the process AB is 3P0 V0 . (c) Let dQBC be the heat absorbed in the process BC : Total heat absorbed. dQ = dQCA + dQAB + dQBC =

3P0

P0

B

C

A

V0

 5  dQ =  – P0V0  + (3P0 V0 ) + dQBC 2   V

2V0

P0V0 dQ = dQBC + 2 Change in internal energy, dU = 0 ∴ dQ = dW

W = + Area of triangle ABC

1 (base) (height) 2

=

∴

1 = (2V0 – V 0 )(3P0 – P0 ) 2 W = P0 V0 (b) Number of moles n = 1 and gas is monoatomic, therefore CV =

3 5 R and CP = R 2 2

CV 3 C 5 = and P = R 2 R 2 (i) Heat rejected in path CA : (process is isobaric) ∴ dQCA = Cp dT = Cp (Tf – Ti ) ⇒

∴

(

CP P f V f − PiVi R

dQBC =

P0V0 2

∴ Heat absorbed in the process BC is +

P0V0 2

(d) Maximum temperature of the gas will be somewhere between B and C. Line BC is a straight line. Therefore, P-V equation for the process BC can be written as : P = – mV + c (y = mx + c)

2P0 Here, m = V and c = 5P0 0

 Pf V f PV  – i i = CP  R   R

=

P0V0 dQBC + = P0 V0 2

 2P  P = –  0 V + 5P0  V0 

)

Multiplying the equation by V,

 2P  2 PV = –  0  V + 5P0 V  V0 

Substituting the values

177

(PV =RT for n = 1)

2KAt1   = − 2 ln 2 + CL  

 2P0  2  V + 5P0 V RT =   V0  or

T=

For T to be maximum, ⇒

5P0 –

⇒

V=

1 2P0 2  V   5P0V – R V0 

2KAt1  T − 300  ln  2 × 4 = − CL  50  ...(1)

T2 − 300 × 4 = e −2KAt1 / CL 50

dT =0 dV

T2 = 300 +

4P0 .V = 0 V0

T2 = 300 + 12.5e −2 At1 / CL Kelvin. 29. (a)

5V0 4

P – V diagram is given as shown in figure. (b) (i)work is done only in the adiabatic process AB not in isochoric process BC

5V0 (on line BC), temperature of the gas is 4 maximum. From Eq. (1) this maximum temperature will be : i.e., at V =

Tmax

Tmax

P

C B

P2

3 1  5V  2 P  5V   = 5P0  0  – 0  0   R  4  V0  4   

=

P1

25 P0V0 8 R

A

28. Upto t ≤ t1 , temperature falls only according to the

V2

Newton’s law of cooling i.e. only through the radiation. radiation ⇒ − 350

⇒

∫

400

⇒

ln

dT = k (T − T A ) dt

⇒

t1

∫

dt = −k dt (T − 300 ) 0

(350 − 300 ) = − kt1 ⇒ ( 400 − 300 )

t1 =

AS P2 = P1

1 ln 2 k

dT kA = k (T − T A ) + (T − T A ) dt CL (Here K is thermal conductivity of the rod, A its area of cross section whereas L is its length and c-thermal capacity of solid body) ⇒

−

−

∫

350

⇒

ln

P1V1 − P2V 2 +0 5   − 1 3 

V1r V2r

  V1  r−1  3 = P1V1 1 −    2   V2  

as well as radiation.

T2

V

V1

WTotal = W AB + W BC =

But for t > t1 cooling takes place both by conduction

⇒

50 −2KAt1 / CL e Kelvin 4

(ii)

⇒ ∆ U AB = −∆ W =

∫

 V 3 P1V1  1 2 V  2

  

r −1

 − 1 

and ∆ U BC = ∆QBC − ∆WBC but ∆ WBC = 0

3t1

dt KA   = k +  dt T − TA  CL 

As ∆QAB = 0 (adiabatic process)

(isochoric) ⇒

t1

T2 − 300 KA   = − 2K + t1 350 − 300 CL  

∆ U BC = Q

⇒ ∆ U T = ∆U AB + ∆U BC = 178

 V 3 P1V1  1 2 V  2

  

r −1

 − 1 + Q 

In the process Pvx = constant

R (iii) 2 ( r − 1) (TC − T A ) = nCV ∆T = ∆U T =

 V 3 P1V1  1 2  V2

  

r −1

Molar heat capacity, C =

 − 1 + Q 

Here the process is P2 V = constant or PV1/2 = constant i.e.,

But PAV A = 2RT A (as n = 2) ⇒ T A = ⇒ TC = T A +

=

[( 5 / 3) − 1]  3 2R  V1   V2

P1V1 P1V1 + 2R 2R

TC =

 V P V  1  2 1 1  V 2  

P1V1  V1  2 R  V 2

  

  

2/3

2/3

+

−

  

PAV A 2R

r −1

P

  − 1 + Q    

P1V1 Q + 2 R 3R

Q 3R

V and

∴

WAB

=∫

VB

VA

VB

K

VA

V

=∫ PdV .

dV

= 2  ( PB2VB )VB – ( PA2V A)V A ](K = P 2V    = 2 [PB VB – PA VA ] = 2 [nRTB – nRTA ] = 2nR [T1 – 2T1 ] = (2) (2) (R) [300 – 600] = – 1200 R ∴Work done on the gas in the process AB is 1200 R. (b) Heat absorbed/released in different processes. Since, the gas is monoatomic

2P1

P1

C

2T1

+

 PC   QC→A =WC→A = nRTC ln   PA 

31. Heat lost by container = Heat gained by ice cube (melting and after it) 27 + 273

− mc

∫ SdT = m

ice

[L + SW ( 300 − 273 ) ]

227 +273 300

− mc

∫ ( A + BT )dT = 0.1[8 × 10

4

]

+ 10 3 × 27 = 10.7 × 10 3

500

300

⇒

 BT 2  − mc  AT + = 10700  2  500 

(

)

  2 × 10 2 − mc 100 ( 300 − 500 ) + 300 2 − 500 2  = 10700 2   mc =

A

T1

R

= (2) (R) (600) in (2) QC→A =831.6 r (absorbed)

5 3 5 R and CP = R and γ = 3 2 2 B

C =

 2P1  = nR(2T1 ) ln    P1 

= 2 K  VB – VA  = 2  KVB – KVA 

Therefore, CV =

1 2

5  5  = (2)  R  (TC – TB) = 2  R  (2T 1 – T1 ) 2  2  = (5R) (600 – 300) QB→C = 1500 R (absorbed) Process C → A : Process is isothermal. ∴ ∆U = 0

K

=

x=

R 5 1 –1 1– 3 2 ∴ C = 3.5 R ∴ QA→B = nC∆T =(2) (3.5R) (300 – 600) or QA→B =– 2100 R Process B→ C : Process is isobaric. ∴ QBC = nCp∆T ∴

30. (a) Number of moles n = 2, T1 = 300 K During the process A → B PT = constant or P2 V = constant = K (say) ∴

R R + γ –1 1– x

10700 200 × 100 + 800 × 200 × 10

mc = 0.495 kg

T

179

−2

=

107 216

32. Given : Number of moles n = 2

QCD = – 900 R (released) Processs D → A : T = constant

 VA   QDA =WDA = nRTD in   VD 

V 4V0

D

C VB = 2 and VA B VD = 4 VA

2V0 V0

 V0   =(2) (R) (300) in   4V0 

A TA

TB

T

3 5 R and CP = R (Monatomic) 2 2 TA = 27°C = 300 K Let VA = V0 and VB = 2V0 and VD = VC = 4V0 (a) Process A → B : CV =

TB VB V ∝ T⇒ T = T A A

1 =600 R in   4 QDA = – 831.6 R (released) (c) In the complete cycle : dU = 0 Therefore, from conservation of energy Wnet = QAB + QBC + QCD + QDA Wnet = 1500 R + 831.6R – 900 R – 831.6R or Wnet = Wtotal = 600R 33. When the bob gets snapped, the elastic potential energy stored in the wire is used to raise the temperature of this wire.

V  TB = TA  B  = (300) (2) = 600 K  VA 

∴

∴ TB = 600 K (b) Process A → B : V ∝ T⇒P = constant ∴ QAB = nC P dT = nC P (TB – TA )

∴

QBC

1 ∆θ = ms

 1 Mg / π r 2  Y  2

=

QAB Process B → C : T =constant ∴ dU = 0

   

ms ∆θ = ∆U =

=

5  = (2)  R  (600–300) 2  = 1500 R (absorbed)

 VC =WBC = nRTB ln  V  B

⇒

(

1  stress2  2  Y

  × Volume  

)2 × πr 2l  

 M 2 g2 1 2   2 4 πr l  2 ρ (π r l) s  2π r Y 

(100 ) 2 (10 ) 2 2

(

 22  7860 × 420 ×   × 2 × 10 −3  7 

)4 × 2.1× 1011

= 4. 568 × 10 −3°C 34. Between two collisions the particle goes to opposite wall and comes back and hence it covers a distance

 4V0   =(2)(R) (600) ln   V0 

2l ⇒ it strikes after every ( 2l / v rms )s time ⇒ times per second = 500

=(1200 R) ln (2) =(1200 R) (0.693) or QBC =831.6 R (absorbed) Processs C → D : V = constant ∴ QCD = nC V dT = nC V (TD – TC)

as l = 1m ⇒ v rms = 2 × 1 × 500 = 1000 M/S. ⇒

3 RT 10 6 × 4 × 10 −3 = = 1000 ⇒ T = M 3 × ( 25 / 3)

3  = n  R  (TA – TB ) 2 

⇒ T = 160 K (b) Average KE per atom

3  = (2)  R  (300–600) 2 

=

3 3 kT = (1.38 × 10 −23 ) (160) 2 2 = 3.312 × 10 −23 J

180

vms 2L

(c)

⇒ Density of liquid will decrease

Mass of helium gas in the box m = no. of moles × molecular mass =n×M

ρ0 V = V0 (1 + γ∆T ) and ρ = (1 + γ ∆T )

PV 100 × 1 ×M = × 4 = 0 .3 g 25 RT   = ( 160 )    3 

In this way during floating the upthrust balances weight ⇒ v inρ L g But depth upto which the cube is immersed is given same

35. Decrease in kinetic energy = increase in internal energy of the gas ∴ ∴ 36. (a)

 m  3  1 2 mv0 = nCV ∆T =    R  ∆T 2  M  2  ∆T =

Mv02 3R

Rate of heat loss due to radiations from upper

Q = eσA(T 4 − T 04 ) lid is t ⇒

[

]

Q 17 = 0.6 × × 10 −8 400 4 − 300 4 A t 3

= ( 595W / m 2 ) A (b) This heat radiated per second must pass through the lid via conduction KA ⇒

(θ i − 127 ) t

595 595 × 10 Qi − 127 = = K 0.167 θ = 127 + 35.6 θ = 162 .6°C

= 35 .6

4  400  T  m= m h 2 = h1  2  = (1.0)   3  300   T1 

[as

K ∆T 4 3 (T1 − T2 ) = T S × 4 = 4T S ∆T eσl TS

1.4–1

   

⇒

K [T1T S ∆T ] = 4T33∆T eσl

⇒

∆ T = K (T1 − T S ) / 4eσ[T S3 + K

[

K hence k = 4eσLT 3 + K S

[

0 .4

= 448.8K

∆T 0 ∆Q > 0 D → ( q)( s )

⇒ J

30

COMPREHENSION 20

M

10

L

K 10

20

1.

Since it is open from top, pressure will be P0 . ∴ (a)

2.

When piston is in equilibrium Force exerted by the gas inside in downward direction + weight of piston = force exerted by atmosphere.

3 V(m)

Process K → L → isobaric ⇒ V ∝ T but V ↑ ⇒ T ↑ ⇒ ∆ U > O and W > 0

PA + mg = P0 A

⇒

But PAL ′ = P0 A2L

∆Q > O

Process L → M ⇒ isochoric ⇒ W = O ⇒

∆ Q = ∆ U but P ↑ ⇒ T ↑ hence ∆U > 0

⇒

k ∆Q > 0

Process M → J ⇒

V ↓ hence W < 0

2P0 L L′

2P0 L mg = P0 + L′ A

But ( PV ) J < ( PV ) M

2.

P=

⇒

or

2 P0 L mg P0 + πR 2

⇒

nRT J < nRTM ⇒

⇒

∆U < O

⇒

∆Q < O

(a)

bimetallic strip works if one metal expands more

Pressure inside and outside the cylinder should be equal at same height of water

than other A → (s )

⇒

(b)

Þ

TJ < T M

L′ =

 p0 πR 2  L′ =  2L 2  ( mg + P0 πR 

⇒

∆ Q = ∆U + W

3.

P1 = P2

steam engine converts one form of energy into other B → (q)

(c)

In candescent lamp radiates when filament is heated C → (q)

(d)

P

electric fuse protects the circuit elements if it melts first at high currents d – (r) 1

3.

(a)

In case of free expansion if heat is neither added or removed then ∆W = 0 ⇒ ∆U = 0 ⇒ U constant

A → (q) and T constant (b)

PV 2 = constant Þ ⇒

⇒ (c)

(d)

nRTV = const.

1 1 or T ∝ T V V ↑ ⇒ T ↓ B → ( p)( r ) ∆U < 0 V∝

pV 4 / 3 = const. nRTV 1/ 3 = K V ↑ ⇒ T ↓ C → ( p)( s) Pressure and volume both are increasing

183

2

P0 + ρg ( L0 − H ) = P Initially air inside cylinder was at atmospheric pressure ⇒ P0 L0 A = P( L 0 − H ) A P0 L0 L0 − H

⇒

P=

⇒

P0 + ρg ( L0 − H ) =

⇒

P0 ( L 0 − H ) + ρg ( L 0 − H ) 2 − P0 L 0 = 0

P0 L0 (L0 − H )