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Level 1

Q No: 1 Three flasks of equal volumes contain CH 4, CO2 and Cl2 gases respectively. They will contain equal number of molecules if (1) the mass of all the gases is same (2) the moles of all the gas is same but temperature is different (3) temperature and pressure of all the flasks are same (4) temperature, pressure and masses same in the flasks Correct Answer: 3 Status: unattempted [X] Avogadro's Law Otherwise think in terms of n = Avogadro's Law :→ "At same condition of temperature and pressure, equal volumes of different gases contains equal number of molecules." Q No: 2 At 0ºC, the density of a gaseous oxide at 2 bar is same as that of nitrogen at 5 bar. What is the molecular mass of the oxide ? (1) 70 (2) 210 (3) 35 (4) 52 Correct Answer: 1 Status: unattempted [X]

Solve for unknown M

Q No: 3 8.2 L of an ideal gas weight 9.0 gm at 300 K and 1 atm pressure. The molecular mass of gas is(1) 9 (2) 27 (3) 54 (4) 81 Correct Answer: 2 Status: unattempted [X] Ideal gas equation

Q No: 4 A gas 'A' having molecular weight 4 diffuses thrice as fast as the gas B. The molecular weight of gas B is(1) 36 (2) 12 (3) 18 (4) 24 Correct Answer: 1 Status: unattempted [X] rA = 3 r B Grahamís Law of effusion or diffusion

Q No: 5 Equal volumes of two gases A and B diffuse through a porous pot in 20 and 10 seconds respectively. If the molar mass of A be 80, find the molar mass of B. (1) 80 g mol–1 (2) 20 g mol–1 (3) 40 g mol–1 (4) 160 g mol–1

Correct Answer: 2

Status: unattempted [X]

t1 = t2 (same time interval)

Q No: 6 Which is not correct in terms of kinetic theory of gases(1) Gases are made up of small particles called molecules (2) The molecules are in random motion (3) When molecules collide, they lose energy (4) When the gas is heated, the molecules moves faster Correct Answer: 3 Status: unattempted Kinetic Theory of Gases's postulates Elastic collisions. According to K.T.G., collision of gaseous molecules is elastic collision. Q No: 7 The average speed of an ideal gas molecule at 27ºC is 0.3 m sec–1. Calculate average speed at 927ºC. (1) 0.15 m/sec (2) 1.2 m/sec (3) 2.4 m/sec (4) 0.6 m/sec Correct Answer: 4 Status: unattempted [X]

Q No: 8 Critical temperature of the gas is the temperature(1) Below which it cannot be liquified (2) Above which it cannot be liquified (3) At which it occupies 22.4 L of volume (4) At which one mole of it occupies volume of 22.4 L Correct Answer: 2 Status: unattempted Definition of Critical temp. Critical Temperature (TC) :→ The temperature above which liquification of gas cannot take place however high pressure is applied on it. NA Q No: 9 The compressibility factor for an ideal gas is : (1) 1.5 (2) 1.0 (3) 2.0 Correct Answer: 2 Status: unattempted For ideal gas PV = nRT

PV = nRT for ideal gas

Q No: 10

(4) ∞

One litre of an unknown gas weighs 1.25 gm at N.T.P. which of the following gas pertains to the above data (1) CO2 (2) NO2 (3) N2 (4) O2 Correct Answer: 3 Status: unattempted

[X] N.T.P. (22.4 L 1 mole gas) M.W. (weight of 22.4 L) Decide from options. Weight of 1 L gas at N.T.P. = 1.25 g. ∴ M.W. = weight of 22.4 L = 1.25 × 22.4 = 28 g. from option N2 (M.W. = 28) Q No: 11 V versus T curves at constant pressure P1 and P2 for an ideal gas are shown in fig. Which is correct -

(1) P1 > P2 (2) P1 < P2 (3) P1 = P2 (4) All Correct Answer: 2

Status: unattempted

Higher slope, lower pressure. Otherwise at same temperature Higher volume Lower pressure (from Boyle's Law)

Q No: 12 If a gas is expanded at constant temperature(1) Number of molecules of the gas decreases (2) The kinetic energy of the molecule decreases (3) The kinetic energy of the molecules remains the same (4) The kinetic energy of the molecules increases Correct Answer: 3 Status: unattempted

Kinetic Energy is function of temperature only for ideal gas

K.E. of a gas is function of temperature only. Q No: 13 A box of 1L capacity is divided into two equal compartments by a thin partition which are filled with 2g H2 and 16gm CH4 respectively. The pressure in each compartment is recorded as P atm. The total pressure when partition is removed will be (1) P (2) 2P (3) P/2 (4) P/4 Correct Answer: 1 Status: unattempted

Does by opening door between two rooms, pressure of each room becomes 2 atm? or remains 1 atm?

after removing partition pressure will be same in both sides. NA Q No: 14

Helium atom is twice times heavier than a hydrogen molecule. At 25ºC the average K.E. of helium atom is (1) Twice that of hydrogen (2) Same as that of hydrogen (3) Four times that of hydrogen (4) Half that of hydrogen Correct Answer: 2 Status: unattempted

K.E. of gas does not depends upon M.W. K.E. of gas function of temperature only

Q No: 15 Which of the following molecule has the lowest average speed at 273 K ? (1) CO (2) CH4 (3) CO2 (4) C2H6 Correct Answer: 3 Status: unattempted

Choose higher M.W. from options.

Q No: 16 A real gas is expected to exhibit maximum deviations from ideal gas laws at (1) Low T and High P (2) Low T and Low P (3) High T and High P (4) High T and Low P Correct Answer: 1 Status: unattempted

At high temperature force of attraction deceases i.e. 'a' becomes very small. At low pressure, molecular volumes becomes insignificant i.e. b

Z=

Q No: 21 At 0ºC and 1 atm pressure, volume occupied by one mole of a gas is greater than 22.4L. Compressibility factor (z) for the gas is (1) greater than 1 (2) less than 1 (3) equal to 1 (4) can be greater or less than on Correct Answer: 1 Status: unattempted

For ideal gas PVm = RT Zreal < Zideal Compare (Vm)real and (Vm)ideal. For ideal at 0ºC and 1 atm Vm = 22.4 L for real gas if Z > 1, Vm > 22.4 L. if Z < 1, Vm < 22.4 L Level 3

Q No: 1

A mixture of hydrogen and helium is prepared such that the number of wall collisions per unit time by molecules of each gas is the same. Which gas has the higher concentration? (1) He (2) H2 (3) Both (4) None of these Correct Answer: 1 Status: unattempted

Number of molecuels colliding at wall/time =

NA Q No: 2 For a gaseous reaction, 2A(g) → 3B(g) + C(g) Whose extent of dissociation depends on temperature is performed in a closed container, it is known that extent of dissociation of A is different in different temperature range. With in a temperature range it is constant. (Temperature range T0 – T1, T1 – T2, T2 – T∞). A plot of P v/s T is drawn under the given condition. Given : tan55 = 1.42, tan50 = 1.19, tan60 = 1.73

If (1) (2) (3) (4)

is the degree of dissociation of A then in the temperature range Ti → Ti + 1

Correct Answer: 1

Status: unattempted

V = constant P Vs T is straight line and passes from origin. tanθ = n (Number of moles)

Q No: 3 For a gaseous reaction, 2A(g) → 3B(g) + C(g) Whose extent of dissociation depends on temperature is performed in a closed container, it is known that extent of dissociation of A is different in different temperature range. With in a temperature range it is constant. (Temperature range T0 – T1, T1 – T2, T2 – T∞). A plot of P v/s T is drawn under the given condition. Given : tan55 = 1.42, tan50 = 1.19, tan60 = 1.73

If initially 1 mole of A is taken in a 0.0821 l container then [R = 0.0821 atm lit/k] (1) (2) (3) (4) Correct Answer: 1,3 Status: unattempted

tan 50 = 1.19 = 1 + α α = 0.19 tan 55 = 1 + α 1.42 = 1 + α ; α = 0.42. NA Q No: 4 Select the correct option(s) for an ideal gas (1) Most probable speed increases with increase in temperature (2) Fraction of particles moving with most probable speed increases with increase in temperature (3) Fraction of particles moving with most probable speed are more for Cl 2 than H2 under similar condition of T, P & V. (4) Most probable speed is more for Cl2 than H2 at same temperature Correct Answer: 1,3 Status: unattempted

Q No: 5 For two gases A and B, P v/s V isotherms are drawn at T K as shown. T A & TB are critical temperatures of A & B respectively

Which of following is true? (1) TA < T < TB (2) TA > T > TB (3) TA > TB > T (4) none of above Correct Answer: 1 Status: unattempted

[X] Experiment Temp > TC then gas cannot be liquified. TA > T < TB NA Q No: 6 For two gases A and B, P v/s V isotherms are drawn at T K as shown. T A & TB are critical temperatures of A & B respectively

The correct statement(s) is/are (I) Pressure correction term will be more negligible for gas B at T K. (II) The curve for gas 'B' will be of same shape as for gas A if T > TB (III) Gas 'A' will show same P v/s V curve as of gas 'B' if T > TA (1) III only (2) II and III (3) II only (4) All

Correct Answer: 3

Status: unattempted

If a = 0 then attraction is zero so gas can not be liquified. T > (TC)B so gas show that graph. TA is critical point of A and above TC gas can not be liquified. Q No: 7 On the recently discovered 10th planet it has been found that the gases follow the relationship PeV/2 = nCT where C is constant other notation are as usual (V in lit., P in atm and T in Kelvin). A curve is plotted between P and V at 500 K & 2 moles of gas as shown in figure

The value of constant C is (1) 0.01 (2) 0.001 (3) 0.005 (4) 0.002 Correct Answer: 2 Status: unattempted

Applying PeV/2 = nCT.

NA Q No: 8 On the recently discovered 10th planet it has been found that the gases follow the relationship PeV/2 = nCT where C is constant other notation are as usual (V in lit., P in atm and T in Kelvin). A curve is plotted between P and V at 500 K & 2 moles of gas as shown in figure

Find the slope of the curve plotted between P Vs T for closed container of volume 2 lit. having same moles of gas (1) (2) 2000 e (3) 500 e (4) Correct Answer: 4

Status: unattempted

NA NA Q No: 9 On the recently discovered 10th planet it has been found that the gases follow the relationship PeV/2 = nCT where C is constant other notation are as usual (V in lit., P in atm and T in Kelvin). A curve is plotted between P and V at 500 K & 2 moles of gas as shown in figure

If a closed container of volume 200 lit. of O2 gas (ideal gas) at 1 atm & 200 K is taken to planet. Find the pressure of oxygen gas at the planet at 821 K in same container (1) (2) (3) 1 atm (4) 2 atm

Correct Answer: 1

Status: unattempted

NA NA Q No: 10 The molar volume of He at 10.1325 MPa and 273 K is 0.011075 of its molar volume at 101.325 KPa at 273 K.Calculate the radius of helium atom. The gas is assumed to show real gas nature. Neglect the value of a for He. (1) r = 1.33 × 10–8 cm (2) r = 2.33 × 10–8 cm (3) r = 1.33 × 10–6 cm (4) r = 5.23 × 10–3 cm Correct Answer: 1 Status: unattempted

For He a = 0

NA Q No: 11 Sign of initial slope of compressibility factor (z) versus P curves is ________ if a gas is below its Boyle's temperature and ________ if it is above its Boyle's temperature. (1) +ive, +ive (2) –ive, +ive (3) –ive, –ive (4) +ive, –ive Correct Answer: 2 Status: unattempted

For He a = 0

NA

Q No: 12 A commercial cylinder contains 6.91 m3 of O2 at 15 M Pa and 21°C. the critical constants for O2 are TC = –126°C, PC = 50 bar. Determine the reduced pressure and reduced temperature for O2 under these conditions. (1) PR = 3, TR = 2 (2) PR = 2, TR = 3 (3) PR = 5, TR = 2 (4) PR = 3, TR = 5 Correct Answer: 1 Status: unattempted

[X]

Put in the same unit. NA Q No: 13 Calculate the number of moles of gas present in the container of volume 10 lit at 300 K. If the manometer containing glycerin shows 5m difference in level as shown in diagram. Given : dglycerin = 2.72 gm/ml, dmercury = 13.6 gm/ml.

(1) 0.94 mole (2) 0.93 mole (3) 0.84 mole (4) 0.90 mole Correct Answer: 1 Status: unattempted eight is determine for glycerin. h1d1 = h2d2 hHg = 1M = 100 cm

Q No: 14 A manometer attached to a flask contains NH 3 gas have no difference in mercury level initially as shown in diagram. After the sparking into the flask, it have difference of 19 cm in mercury level in two columns. Calculate % dissociation of ammonia.

(1) 21% Correct Answer: 4

P1M1 = P2M2 76 × 17 = 95 × M2 M2 = 13.6

Q No: 15

(2) 22% Status: unattempted

(3) 24%

(4) 25%

1.0 × 10–2 kg of hydrogen and 6.4 × 10–2 kg of oxygen are contained in a 10 × 10–3 m3 flask at 473 K. Calculate the total pressure of the mixture. If a spark ignities the mixture. What will be the final pressure. (1) Ptotal = 27.54 × 105 N/m2, Pfinal = 19.66 × 105 N/m2 (2) Ptotal = 26.54 × 105 N/m2, Pfinal = 19.66 × 105 N/m2 (3) Ptotal = 22.54 × 105 N/m2, Pfinal = 21.66 × 105 N/m2 (4) Ptotal = 25.62 × 105 N/m2, Pfinal = 18.34 × 105 N/m2 Correct Answer: 1 Status: unattempted

NA

Q No: 16 Calculate relative rate of effusion of SO2 to CH4 under given condition (i) Under similar condition of pressure & temperature (ii) Through a container containing SO2 and CH4 in 3 : 2 mass ratio (iii) If the mixture obtained by effusing out a mixture three effusing steps. (1) (i) ; (ii) ; (iii) (2) (i) ; (ii) ; (iii) (3) (i) ; (ii) ; (iii) (4) None of these Correct Answer: 3 Status: unattempted

for

Q No: 17 Find the number of diffusion steps required to separate the isotopic mixture initially containing some amount of H2 gas and 1 mol of D2 gas in a container of 3 lit capacity maintained at 24.6 atm & 27 °C to the final mass ratio (1) 1 (2) 2 (3) 3 Correct Answer: 4 Status: unattempted

X → Number of effusion steps

. (4) 4

Q No: 18 Calculate the fraction of N2 molecules at 101.325 kPa and 300 K whose speeds are in the range of ump – 0.005 ump to ump + 0.005 ump. (1) 8.303 × 10–3 (2) 7.303 × 10–3 (3) 8.501 × 10–3 (4) 5.323 × 10–3 Correct Answer: 1 Status: unattempted

U2 – U1 = Ump + 0.005Ump – Ump + 0.005Ump = 0.01 Ump f = 0.01

0.01 Ump

Q No: 19

One mole of an ideal gas is subjected to a process in which P (in atm) = V (in Litre). If the process is operating from a initial pressure 1 atm to final pressure 10 atm (no higher pressure achieved during the process) then what would be the maximum temperature obtained & at what instant will it occur in the process. (1) 10,000 K (2) 20,000 K (3) 15,000 K (4) 19,000 K Correct Answer: 1 Status: unattempted

Moles remains constant.

NA

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Q No: 1 Three flasks of equal volumes contain CH 4, CO2 and Cl2 gases respectively. They will contain equal number of molecules if (1) the mass of all the gases is same (2) the moles of all the gas is same but temperature is different (3) temperature and pressure of all the flasks are same (4) temperature, pressure and masses same in the flasks Correct Answer: 3 Status: unattempted [X] Avogadro's Law Otherwise think in terms of n = Avogadro's Law :→ "At same condition of temperature and pressure, equal volumes of different gases contains equal number of molecules." Q No: 2 At 0ºC, the density of a gaseous oxide at 2 bar is same as that of nitrogen at 5 bar. What is the molecular mass of the oxide ? (1) 70 (2) 210 (3) 35 (4) 52 Correct Answer: 1 Status: unattempted [X]

Solve for unknown M

Q No: 3 8.2 L of an ideal gas weight 9.0 gm at 300 K and 1 atm pressure. The molecular mass of gas is(1) 9 (2) 27 (3) 54 (4) 81 Correct Answer: 2 Status: unattempted [X] Ideal gas equation

Q No: 4 A gas 'A' having molecular weight 4 diffuses thrice as fast as the gas B. The molecular weight of gas B is(1) 36 (2) 12 (3) 18 (4) 24 Correct Answer: 1 Status: unattempted [X] rA = 3 r B Grahamís Law of effusion or diffusion

Q No: 5 Equal volumes of two gases A and B diffuse through a porous pot in 20 and 10 seconds respectively. If the molar mass of A be 80, find the molar mass of B. (1) 80 g mol–1 (2) 20 g mol–1 (3) 40 g mol–1 (4) 160 g mol–1

Correct Answer: 2

Status: unattempted [X]

t1 = t2 (same time interval)

Q No: 6 Which is not correct in terms of kinetic theory of gases(1) Gases are made up of small particles called molecules (2) The molecules are in random motion (3) When molecules collide, they lose energy (4) When the gas is heated, the molecules moves faster Correct Answer: 3 Status: unattempted Kinetic Theory of Gases's postulates Elastic collisions. According to K.T.G., collision of gaseous molecules is elastic collision. Q No: 7 The average speed of an ideal gas molecule at 27ºC is 0.3 m sec–1. Calculate average speed at 927ºC. (1) 0.15 m/sec (2) 1.2 m/sec (3) 2.4 m/sec (4) 0.6 m/sec Correct Answer: 4 Status: unattempted [X]

Q No: 8 Critical temperature of the gas is the temperature(1) Below which it cannot be liquified (2) Above which it cannot be liquified (3) At which it occupies 22.4 L of volume (4) At which one mole of it occupies volume of 22.4 L Correct Answer: 2 Status: unattempted Definition of Critical temp. Critical Temperature (TC) :→ The temperature above which liquification of gas cannot take place however high pressure is applied on it. NA Q No: 9 The compressibility factor for an ideal gas is : (1) 1.5 (2) 1.0 (3) 2.0 Correct Answer: 2 Status: unattempted For ideal gas PV = nRT

PV = nRT for ideal gas

Q No: 10

(4) ∞

One litre of an unknown gas weighs 1.25 gm at N.T.P. which of the following gas pertains to the above data (1) CO2 (2) NO2 (3) N2 (4) O2 Correct Answer: 3 Status: unattempted

[X] N.T.P. (22.4 L 1 mole gas) M.W. (weight of 22.4 L) Decide from options. Weight of 1 L gas at N.T.P. = 1.25 g. ∴ M.W. = weight of 22.4 L = 1.25 × 22.4 = 28 g. from option N2 (M.W. = 28) Q No: 11 V versus T curves at constant pressure P1 and P2 for an ideal gas are shown in fig. Which is correct -

(1) P1 > P2 (2) P1 < P2 (3) P1 = P2 (4) All Correct Answer: 2

Status: unattempted

Higher slope, lower pressure. Otherwise at same temperature Higher volume Lower pressure (from Boyle's Law)

Q No: 12 If a gas is expanded at constant temperature(1) Number of molecules of the gas decreases (2) The kinetic energy of the molecule decreases (3) The kinetic energy of the molecules remains the same (4) The kinetic energy of the molecules increases Correct Answer: 3 Status: unattempted

Kinetic Energy is function of temperature only for ideal gas

K.E. of a gas is function of temperature only. Q No: 13 A box of 1L capacity is divided into two equal compartments by a thin partition which are filled with 2g H2 and 16gm CH4 respectively. The pressure in each compartment is recorded as P atm. The total pressure when partition is removed will be (1) P (2) 2P (3) P/2 (4) P/4 Correct Answer: 1 Status: unattempted

Does by opening door between two rooms, pressure of each room becomes 2 atm? or remains 1 atm?

after removing partition pressure will be same in both sides. NA Q No: 14

Helium atom is twice times heavier than a hydrogen molecule. At 25ºC the average K.E. of helium atom is (1) Twice that of hydrogen (2) Same as that of hydrogen (3) Four times that of hydrogen (4) Half that of hydrogen Correct Answer: 2 Status: unattempted

K.E. of gas does not depends upon M.W. K.E. of gas function of temperature only

Q No: 15 Which of the following molecule has the lowest average speed at 273 K ? (1) CO (2) CH4 (3) CO2 (4) C2H6 Correct Answer: 3 Status: unattempted

Choose higher M.W. from options.

Q No: 16 A real gas is expected to exhibit maximum deviations from ideal gas laws at (1) Low T and High P (2) Low T and Low P (3) High T and High P (4) High T and Low P Correct Answer: 1 Status: unattempted

At high temperature force of attraction deceases i.e. 'a' becomes very small. At low pressure, molecular volumes becomes insignificant i.e. b

Z=

Q No: 21 At 0ºC and 1 atm pressure, volume occupied by one mole of a gas is greater than 22.4L. Compressibility factor (z) for the gas is (1) greater than 1 (2) less than 1 (3) equal to 1 (4) can be greater or less than on Correct Answer: 1 Status: unattempted

For ideal gas PVm = RT Zreal < Zideal Compare (Vm)real and (Vm)ideal. For ideal at 0ºC and 1 atm Vm = 22.4 L for real gas if Z > 1, Vm > 22.4 L. if Z < 1, Vm < 22.4 L Level 3

Q No: 1

A mixture of hydrogen and helium is prepared such that the number of wall collisions per unit time by molecules of each gas is the same. Which gas has the higher concentration? (1) He (2) H2 (3) Both (4) None of these Correct Answer: 1 Status: unattempted

Number of molecuels colliding at wall/time =

NA Q No: 2 For a gaseous reaction, 2A(g) → 3B(g) + C(g) Whose extent of dissociation depends on temperature is performed in a closed container, it is known that extent of dissociation of A is different in different temperature range. With in a temperature range it is constant. (Temperature range T0 – T1, T1 – T2, T2 – T∞). A plot of P v/s T is drawn under the given condition. Given : tan55 = 1.42, tan50 = 1.19, tan60 = 1.73

If (1) (2) (3) (4)

is the degree of dissociation of A then in the temperature range Ti → Ti + 1

Correct Answer: 1

Status: unattempted

V = constant P Vs T is straight line and passes from origin. tanθ = n (Number of moles)

Q No: 3 For a gaseous reaction, 2A(g) → 3B(g) + C(g) Whose extent of dissociation depends on temperature is performed in a closed container, it is known that extent of dissociation of A is different in different temperature range. With in a temperature range it is constant. (Temperature range T0 – T1, T1 – T2, T2 – T∞). A plot of P v/s T is drawn under the given condition. Given : tan55 = 1.42, tan50 = 1.19, tan60 = 1.73

If initially 1 mole of A is taken in a 0.0821 l container then [R = 0.0821 atm lit/k] (1) (2) (3) (4) Correct Answer: 1,3 Status: unattempted

tan 50 = 1.19 = 1 + α α = 0.19 tan 55 = 1 + α 1.42 = 1 + α ; α = 0.42. NA Q No: 4 Select the correct option(s) for an ideal gas (1) Most probable speed increases with increase in temperature (2) Fraction of particles moving with most probable speed increases with increase in temperature (3) Fraction of particles moving with most probable speed are more for Cl 2 than H2 under similar condition of T, P & V. (4) Most probable speed is more for Cl2 than H2 at same temperature Correct Answer: 1,3 Status: unattempted

Q No: 5 For two gases A and B, P v/s V isotherms are drawn at T K as shown. T A & TB are critical temperatures of A & B respectively

Which of following is true? (1) TA < T < TB (2) TA > T > TB (3) TA > TB > T (4) none of above Correct Answer: 1 Status: unattempted

[X] Experiment Temp > TC then gas cannot be liquified. TA > T < TB NA Q No: 6 For two gases A and B, P v/s V isotherms are drawn at T K as shown. T A & TB are critical temperatures of A & B respectively

The correct statement(s) is/are (I) Pressure correction term will be more negligible for gas B at T K. (II) The curve for gas 'B' will be of same shape as for gas A if T > TB (III) Gas 'A' will show same P v/s V curve as of gas 'B' if T > TA (1) III only (2) II and III (3) II only (4) All

Correct Answer: 3

Status: unattempted

If a = 0 then attraction is zero so gas can not be liquified. T > (TC)B so gas show that graph. TA is critical point of A and above TC gas can not be liquified. Q No: 7 On the recently discovered 10th planet it has been found that the gases follow the relationship PeV/2 = nCT where C is constant other notation are as usual (V in lit., P in atm and T in Kelvin). A curve is plotted between P and V at 500 K & 2 moles of gas as shown in figure

The value of constant C is (1) 0.01 (2) 0.001 (3) 0.005 (4) 0.002 Correct Answer: 2 Status: unattempted

Applying PeV/2 = nCT.

NA Q No: 8 On the recently discovered 10th planet it has been found that the gases follow the relationship PeV/2 = nCT where C is constant other notation are as usual (V in lit., P in atm and T in Kelvin). A curve is plotted between P and V at 500 K & 2 moles of gas as shown in figure

Find the slope of the curve plotted between P Vs T for closed container of volume 2 lit. having same moles of gas (1) (2) 2000 e (3) 500 e (4) Correct Answer: 4

Status: unattempted

NA NA Q No: 9 On the recently discovered 10th planet it has been found that the gases follow the relationship PeV/2 = nCT where C is constant other notation are as usual (V in lit., P in atm and T in Kelvin). A curve is plotted between P and V at 500 K & 2 moles of gas as shown in figure

If a closed container of volume 200 lit. of O2 gas (ideal gas) at 1 atm & 200 K is taken to planet. Find the pressure of oxygen gas at the planet at 821 K in same container (1) (2) (3) 1 atm (4) 2 atm

Correct Answer: 1

Status: unattempted

NA NA Q No: 10 The molar volume of He at 10.1325 MPa and 273 K is 0.011075 of its molar volume at 101.325 KPa at 273 K.Calculate the radius of helium atom. The gas is assumed to show real gas nature. Neglect the value of a for He. (1) r = 1.33 × 10–8 cm (2) r = 2.33 × 10–8 cm (3) r = 1.33 × 10–6 cm (4) r = 5.23 × 10–3 cm Correct Answer: 1 Status: unattempted

For He a = 0

NA Q No: 11 Sign of initial slope of compressibility factor (z) versus P curves is ________ if a gas is below its Boyle's temperature and ________ if it is above its Boyle's temperature. (1) +ive, +ive (2) –ive, +ive (3) –ive, –ive (4) +ive, –ive Correct Answer: 2 Status: unattempted

For He a = 0

NA

Q No: 12 A commercial cylinder contains 6.91 m3 of O2 at 15 M Pa and 21°C. the critical constants for O2 are TC = –126°C, PC = 50 bar. Determine the reduced pressure and reduced temperature for O2 under these conditions. (1) PR = 3, TR = 2 (2) PR = 2, TR = 3 (3) PR = 5, TR = 2 (4) PR = 3, TR = 5 Correct Answer: 1 Status: unattempted

[X]

Put in the same unit. NA Q No: 13 Calculate the number of moles of gas present in the container of volume 10 lit at 300 K. If the manometer containing glycerin shows 5m difference in level as shown in diagram. Given : dglycerin = 2.72 gm/ml, dmercury = 13.6 gm/ml.

(1) 0.94 mole (2) 0.93 mole (3) 0.84 mole (4) 0.90 mole Correct Answer: 1 Status: unattempted eight is determine for glycerin. h1d1 = h2d2 hHg = 1M = 100 cm

Q No: 14 A manometer attached to a flask contains NH 3 gas have no difference in mercury level initially as shown in diagram. After the sparking into the flask, it have difference of 19 cm in mercury level in two columns. Calculate % dissociation of ammonia.

(1) 21% Correct Answer: 4

P1M1 = P2M2 76 × 17 = 95 × M2 M2 = 13.6

Q No: 15

(2) 22% Status: unattempted

(3) 24%

(4) 25%

1.0 × 10–2 kg of hydrogen and 6.4 × 10–2 kg of oxygen are contained in a 10 × 10–3 m3 flask at 473 K. Calculate the total pressure of the mixture. If a spark ignities the mixture. What will be the final pressure. (1) Ptotal = 27.54 × 105 N/m2, Pfinal = 19.66 × 105 N/m2 (2) Ptotal = 26.54 × 105 N/m2, Pfinal = 19.66 × 105 N/m2 (3) Ptotal = 22.54 × 105 N/m2, Pfinal = 21.66 × 105 N/m2 (4) Ptotal = 25.62 × 105 N/m2, Pfinal = 18.34 × 105 N/m2 Correct Answer: 1 Status: unattempted

NA

Q No: 16 Calculate relative rate of effusion of SO2 to CH4 under given condition (i) Under similar condition of pressure & temperature (ii) Through a container containing SO2 and CH4 in 3 : 2 mass ratio (iii) If the mixture obtained by effusing out a mixture three effusing steps. (1) (i) ; (ii) ; (iii) (2) (i) ; (ii) ; (iii) (3) (i) ; (ii) ; (iii) (4) None of these Correct Answer: 3 Status: unattempted

for

Q No: 17 Find the number of diffusion steps required to separate the isotopic mixture initially containing some amount of H2 gas and 1 mol of D2 gas in a container of 3 lit capacity maintained at 24.6 atm & 27 °C to the final mass ratio (1) 1 (2) 2 (3) 3 Correct Answer: 4 Status: unattempted

X → Number of effusion steps

. (4) 4

Q No: 18 Calculate the fraction of N2 molecules at 101.325 kPa and 300 K whose speeds are in the range of ump – 0.005 ump to ump + 0.005 ump. (1) 8.303 × 10–3 (2) 7.303 × 10–3 (3) 8.501 × 10–3 (4) 5.323 × 10–3 Correct Answer: 1 Status: unattempted

U2 – U1 = Ump + 0.005Ump – Ump + 0.005Ump = 0.01 Ump f = 0.01

0.01 Ump

Q No: 19

One mole of an ideal gas is subjected to a process in which P (in atm) = V (in Litre). If the process is operating from a initial pressure 1 atm to final pressure 10 atm (no higher pressure achieved during the process) then what would be the maximum temperature obtained & at what instant will it occur in the process. (1) 10,000 K (2) 20,000 K (3) 15,000 K (4) 19,000 K Correct Answer: 1 Status: unattempted

Moles remains constant.

NA

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