Selected Questions and Problems in Physics - Gladkovauestions and Problems in Physics - Gladkova

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SELECTED QUESTIONS AND PROBLEMS IN PHYSICS

P. A . fjjaflKOBa, H . II. KyTMJiOBCKaa

C6opHHK 3a«aH n BonpocoB no 2ox and /?2nit, we use B oyle’s law Plox^lox = P 2ox^2oxi P in it^ in lt = P 2nitP 2nit for each gas separately: P io x F io x P2ox=

' ^ 2ox P in itF m it

P2m==

V'tnit

2 X IV

P a X 3 X 10“ 3 m 3

n

v 105 P t

= ---------- 1 X 10-* m*----------= U . b X l U 3 x 1 0 * P a x 7 x 1 0 - 3 m3

0 t ^

= --------- 1 x 1 0 ^ m*------------ 2.1 X 1 0

Pa, Do

Pa.

Finally, we obtain p = 0.6 X 105 P a + 2.1 X 105 Pa = 2.7 X 106 Pa. Answer. The pressure of the gas m ixture in the cylinders is 270 kPa. Problem 14. The air in a balloon at a tem perature of 20°C and a pressure of 99.75 kPa has a volume of 2.5 1. W hen the balloon is immersed in w ater at a tem perature of 5°C, the air pressure in it increases to 2 X 106 Pa. W hat is the change in the volume of the air in the balloon?

29

§ 3. Equation of State for an Ideal Gas

Given: t x = 20°C, p 1 = 99.75 kPa, and Vx = 2.5 1 are the tem perature, pressure, and volume of the air before the balloon is immersed in w ater, t2 = 5°C and p 2 = 2 X 105 Pa are the tem perature and pressure of the air after the im m er­ sion of the balloon in water. Find: the change A V in the volume of the air in the bal­ loon. Solution. Before the balloon is immersed in w ater, the state of the air in it is characterized by the param eters />!, V1, and 7V After im m ersion, the param eters are p 2, K2, and 7V Let us w rite the param eters of the gas in states 1 and 2 separately, expressing them in SI units: state 1

state 2

p x = 9.975 X 104 Pa, Vx = 2.5 X 10-3 m3, Tx = 293 K,

p 2 = 2 X 105 Pa, V2 = ?

T 2 = 278 K.

As the air changes from state 1 to state 2, all three param ­ eters change. Consequently, we m ust use the equation of state to determ ine the final volume V 2: P\V\ _ T1

P2V2 T2

w hen ce

Pi Vi T2

T iP2 ’ 9.975 X 104 Pa x 2.5 X 10"3 m 3 X 278 K 293 K X 2 X 105 Pa

AV = 2.5 X 10-3 m3 -

= 1.2

X

10 3 m3.

1.2 x 1 0 '3 m3 = 1.3 X l O ”3 m 3.

Answer. The change in the volume of the a ir is 1.3 X 10-3 m 3 = 1.3 1. Problem 15. A cylinder having a volume of 0.6 m3 con­ tains oxygen at a tem perature of 27°C. A pressure gauge on

Ch. I. Fundam entals of M olecular P h ysics

30

the cylinder indicates 11.7 MPa of excess pressure.2 Reduce the volum e of the oxygen to normal conditions and d eter­ mine its mass. Given: V± = 0.6 m3 is the volume and T x = 27°C is the tem perature of the oxygen in the cylinder, p g = 11.7 MPa is the reading of the pressure gauge, T 0 = 273 K, p 0 = 1.013 X 106 Pa are the tem perature and pressure under nor­ m al conditions. From tables, we take the density p0 = I.43 kg/m 3 of oxygen under norm al conditions, the m olar mass of oxygen, M = 32 X 10~3 kg/mol, and the m olar gas constant R = 8.31 J/(m o l-K ). Find: the volume V0 of the oxygen under norm al condi­ tions and the mass m of the oxygen in the cylinder. Solution. From the reading of the pressure gauge, we de­ term ine the gas pressure in the cylinder: p x = p g + Po — I I .7 X 106 Pa + 0.1013 X 106 Pa - 11.8 x 106 Pa. R e ­ ducing the volume to norm al conditions means determ ining the volum e the gas would occupy at a tem perature of 273 K and a pressure of 1.013 X 105 Pa. We shall w rite the param eters of the oxygen in SI units for the tw o statas: state 1 p x = 11.8 x 106 Pa, V t = 0 .6 m 3,

T x = 300 K,

state 2 Po = 1.013 X 105 Pa, T0 = 273 K, Vo = ?

We solve the problem by using the equation of state for an ideal gas: PiV i Ti

PqVq T0

whence T/’ °“

P \V \T

T iPo

o ’

T/ 11.8 X 106 Pa X 0.6 m 3 X 273 K ao a 3 Vo= 300 K X 1.013 X 105-Pa-------= 6 3 ‘6 m

2 The excess pressure in d icated on a pressure gauge is the difference betw een the pressure of the gas in the cy lin d er and atm ospheric pres­ sure.

§ 3. Equation of State for an Ideal Gas

31

From V0 and p0, we can determ ine the mass of the oxygen: m = p0F 0, m = 1.43 kg/m3 X 63.6 m3 = 91 kg. Answer. The volume of the oxygen under norm al condi­ tions is 63.6 m3, the mass is approxim ately 91 kg. Remark. The problem can be solved by using the equation PjF, = ^ R T X, from which we first determ ine the mass ni = ^H 1 ! We can find the gas volume from the density of the oxygen under normal conditions: Vq =

m /p 0 .

Problem 16. A cylinder contains acetylene a t 27°C under a pressure of 4.05 M Pa. W hat w ill the pressure in the c y lin ­ der be after half the mass of the gas has been used up if the tem perature has fallen thereby to 12°C? Given: T x = 300 K and p x = 4.05 X 106 Pa are the in i­ tial tem perature and pressure of the gas in the cylinder, T2 = 285 K is the tem perature of the rem aining gas, m 2 = 0.5 m x is the mass of the consumed gas. Find: the pressure p 2 of the gas rem aining in the cylinder. Solution. In this problem, the tem perature, pressure, and mass of the gas change. Therefore, it is best to use the equa­ tions of state for the two cases in the form p .V ^ ^ -R T ,

and

R T 2.

Dividing the first equation by the second term wise and c a n ­ celling out m 1, M , R , and the unknown volum e Vj, we obtain

Pi _ p2

7*1 o.5r 2 ’

whence Pi Pi

0-5 P l T 2 Tx

’

0.5 X 4.05 X 106 1’a X 285 K 300 K

192.4 x 104 Pa

cz. 1.92 x 106 Pa. Answer. The pressure of the acetylene rem aining in the cylinder is approxim ately 1.92 MPa.

Ch. I. Fundam entals of M olecular P hysics

32

Problem 17. D eterm ine the density of hydrogen a t a tem ­ perature of 17°C and a pressure of 204 kPa. Given: T = 290 K and p = 2.04 x 105 Pa are the tem ­ perature and pressure for which the density of hydrogen has to be determ ined. From tables, we obtain the m olar mass of hydrogen, M = 2 X 10~3 kg/mol, and the m olar gas con­ sta n t R = 8 .3 1 J/(m ol-K ). F in d : the hydrogen density p. Solution. We w rite the equation of state p F = - ^ - R T . D iv id in g both sides by V , we obtain p = RT

m /V is the density. Then p = p M PM

P = RT

YX\f

Jt^. ^ 1

, where

whence

_ 2.04 X 105 Pa x 2 X 10~3 kg/m ol 8.31 J/ (m ol-K ) X 290 K

= 0.17 kg/m 3.

Answer. The density of hydrogen is 0.17 kg/m 3. Problem 18. P lo t the graphs for an isotherm al, isobaric, and isochoric process in p - V coordinates. Solution. An isotherm al process occurs a t a constant tem ­ perature T = const and obeys B oyle’s law. According to P,MPq Is o b a r

0.2

0.1

p XJ T 2 > T x. Consequently, the upper isotherm corresponds to the higher tem perature. (Questions and Problems 3.1. A vessel contains 10.2 1 of a gas under norm al condi­ tions. W hat volume w ill the gas occupy a t a tem perature of 40°C and a pressure of 1 M Pa? 3.2 . A gas occupies a volum e of 4 1 a t a tem perature of r>0°C and a pressure of 196 kP a. At w hat pressure w ill this gas occupy a volume of 16 1 if heated to 20°C? 3.3. At w hat tem perature do 4.0 m 3 of a gas produce a pressure of 150 kPa if the same mass of the gas under norm al conditions has a volume of 5 m 3? 3.4. A 45-1 cylinder contains oxygen a t a tem perature of M 'C and a pressure of 1.52 M Pa. W hat volume would the gas occupy under norm al conditions? 3.5. Compressed oxygen for w elding is stored in 20-1 cy l­ inders a t a pressure of 9.8 M Pa and a tem perature of 290 K . Mcduce the volum e of oxygen to norm al conditions. 3.6. A 6-1 cylinder contains 0.1 kg of a gas a t a tem per­ ature of 300 K and a pressure of 9.44 X 105 Pa. D eterm ine th e m olar mass and identify the gas. i o'kio

34

Ch. I. Fundam entals of Molecular P hysics

3 .7 . W hat am ount of gas (in moles) is contained in a 10-1 cylinder at a pressure of 0.29 MPa and a tem perature of 17°C? 3 .8. Determ ine the mass of carbon dioxide stored in a cylinder having a volume of 40 1 at a tem perature of 13°C and a pressure of 2.7 M Pa. 3.9 . Determ ine the am ount of gas (in moles) occupying a volum e of 25 1 a t a pressure of 1.4 X 105 Pa and a te m p e ra ­ ture of 300 K. 3.10. A m ountain-clim ber takes in 5 g of air w ith each breath under norm al conditions. W hat volume of air m ust he inhale in the m ountains where the atm ospheric pressure is lower than th a t at sealevel and is 79.8 kPa a t a tem perature of —13°C? 3.11. Under norm al conditions neon occupies a volume of 12.4 1. How m any tim es w ill the pressure of the gas increase if it is placed in a vessel having a volume of 5.6 1 a t a tem ­ perature of 318 K? 3.12. An ideal gas occupies a volume of 2 1 at a pressure of 1.33 kPa and a tem perature of 15°C. W hat w ill be the pressure if the tem perature is doubled, w hile the volume de­ creases by 0.25 of the in itia l value? 3 .1 3 . A 40-1 cylinder contains 64 g of oxygen under a pressure of 213 kPa. D eterm ine the tem perature of the gas. 3.14. 42 g of acetylene are kept in a 20-1 cylinder a t a tem perature of 17°C. D eterm ine the am ount of substance in the gas and its pressure. 3.15. A 40-1 cylinder contains 1.98 kg of carbon dioxide a t 0°G. W hen the tem perature is increased through 48 K, the cylinder explodes. At w hat pressure does the explosion occur? 3.16. C alculate the m olar mass of butane if 2 1 of the gas have a mass of 4.2 g a t a tem perature of 15°C and a pressure pf 87 kPa. C alculate the num ber of gas molecules in 1 m 3. 3.17. At the beginning of the compression stroke in a d ie­ sel engine, the tem perature of the air was 40°C and the pres­ sure was 78.4 kPa. As a result of compression, the volume was reduced by a factor of 15, and the pressure increased to 3.5 M Pa. D eterm ine the tem perature of the air a t the end of the compression stroke. 3.18. The volume of 265 g of a gas a t a tem perature of 273 K and a pressure of 5 MPa is 60 1. W hat gas is it?

§ 3. Equation of State for an Ideal Gas

35

3.19. D eterm ine the mass of air in a room of 6 X 5 X .'I in3 at a tem perature of 293 K and a pressure of 1.04 X K)ft Pa. The m olar mass M of air is 29 X 10“3 kg/mol. 3.20. W hat w ill the final tem perature of a gas m ixture in mi internal com bustion engine be if it occupies a volume of 40 dm 3 in the cylinder at a tem perature of 50°C under nor­ mal atm ospheric pressure and is compressed by the piston to n volume of 5 dm 3 and a pressure of 15.2 X 105 Pa? 3.21. A cylinder contains a gas at a tem perature of 7°C mid a pressure of 91.2 M Pa. W hat w ill be the pressure after 0,25 of the mass of the gas has flown out of the cylinder and the tem perature has risen to 27°C? 3.22. A closed vessel of 2-m3 volume contains 1 kg of nitrogen and 1.5 kg of oxygen. Determ ine the pressure of flm m ixture in the vessel if the tem perature of the m ixture In 17°C. 3.23. D eterm ine the density of oxygen a t a tem perature of 47°C and a pressure of 105 Pa. 3.24. The pressure of air 10 km above the surface of the Knrth is about 30.6 kPa and the tem perature is 230 K. D e­ term ine the density of the air, the num ber density of the molecules, and their root-m ean-square velocity a t this a lt i ­ tude. 3.25. 7 g of a gas contained in a cylinder a t 27°C produce n pressure of 4.9 X 104 P a. 4 g of hydrogen a t 60°C produce a pressure of 43.5 X 104 Pa in the same volum e. D eterm ine the molar mass of the unknown gas and identify it. 3.26. W hat w ill be the increase in the mass of air in a room if the atm ospheric pressure changes from 9.84 X 104 to 10.1 X 104 Pa and the air tem perature rem ains unchanged mid equal to 273 K? The size of the room is 4 X 5 X 2.5 m 3. 3.27. A vessel containing 10 1 of air under a pressure of 1 MPa is connected to a 4-1 em pty vessel. D eterm ine the final air pressure in the vessels, assuming th a t the process is Isotherm al. 3.28. A vessel contains a gas under a pressure of 5 X 106 Pa. W hat w ill the gas pressure be if 3/5 of the mass of the gas has flown out, the tem perature being m aintained constant? 3.29. P lo t the graphs of an isotherm al process in V -T mid p -T coordinates. 3.30. The bladder of a football having a volume of 2.5 1 ;i*

36

Ch. I. Fundam entals of Molecular P hysics

m ust be inflated to a pressure of 300 kP a. The pump takes in 0.14 1 of air under norm al atm ospheric pressure during one stroke. How m any strokes are required if the bladder was in itia lly em pty? 3.31. Given two equations of isotherm al processes for the same mass of a gas: p l V1 = 6 and p 2V 2 = 9. P lo t the graphs in p-V coordinates and answer the following ques­ tions: (1) in which units is the product p V expressed? (2) w hich of the isotherm s w ill be further from the coordinate axes and why? 3.32. D ry atm ospheric air consists of oxygen, nitrogen, and argon. D isregarding other components whose concen­ tratio n s are very sm all, determ ine the masses of these gases

T

F ig. 6

F ig. 7

in 1 m3 of atm ospheric air under normal conditions if their p artial pressures are 2.1 X 104 Pa for oxygen, 7.8 X 104 Pa for nitrogen, and 103 Pa for argon. 3.33. R epresent an isobaric process graphically in V - T , p-V, and p - T coordinates. 3.34. Figure 6 shows two isobars plotted for the same mass of a gas. W hich of the two isobaric processes occurs at a higher pressure and why? 3.35. A gas is heated isobarically from T x to T 2 (Fig. 7). W hat changes occur in the gas? 3.36. W hat volume w ill a gas occupy a t 348 K if its volume a t 35°C is 7.5 1? The process is isobaric. 3.37. A gas occupies a volume of 10 1 a t 27°C. To w hat tem perature should it be cooled isobarically to reduce its volume by 0.25 of the in itia l value? 3.38. Through how m any degrees K elvin should the tem ­ perature of a gas be increased in an isobaric process to in-

§ 4. The Change in the Internal Energy

37

crease its volume by a factor of 1.3 in com parison w ith the volume occupied by it at 0°C? 3.39. A certain mass of a gas is taken through a closed cycle 1-2-3-1 in which the state of the gas changes isobaricnlly-isochorically-isotherm ally. Represent these processes in V T and p -V coordinates. 3.40. A balloon of volume 4500 m 3 is filled w ith helium at 290 K . The mass of the envelope is 677 kg. D eterm ine the lifting force of the balloon at a tem perature of 27°C, assuming th a t the atm ospheric pressure rem ains unchanged iiiid is 102 kP a.

§ 4. THE CHANGE IN THE INTERNAL ENERGY DURING HEAT TRANSFER AND DUE TO MECHANICAL WORK

llasic Concepts and Formulas When several bodies having different tem peratures come into contact, heat transfer takes place, as a result of which the tem perature of the bodies equalizes, and therm al equilib­ rium sets in. The in tern al energy of the bodies being heated increases a t the expense of the energy given away by the bodies w ith higher tem peratures. I t has been established that the change in the internal energy of a body is propor­ tional to its mass and to the change in its tem perature: AU = Q = cm A T , where Q is the am ount of heat (a m ea­ sure of the change in the in tern al energy) expressed in joules (J), and c is the proportionality factor known as the specific heat of a substance. For exam ple, the specific heat of a lu ­ m inium is 920 J/(k g -K ), which means th a t the in te rn al energy of 1 kg of alum inium increases by 920 J when its tem perature is raised through 1 K . In order to determ ine a specific heat, the heat balance equation is employed. This equation is constructed by estab ­ lishing the processes in which the energy is released (during cooling or as a result of fuel combustion) and the processes in which energy is absorbed (as a result of heating). For exam ple, a body of mass m x and specific heat cx has a tem perature T x and is exchanging heat w ith a body of mass m2, which has a specific heat c2 and a tem perature T 2, It is known th a t T x > T 2. Then the first body gives away

38

Ch. I. Fundam entals of Molecular P hysics

the following am ount of heat: (?giv — c1m 1 (T x

0 ).

The second body receives as a result of the heat transfer the following am ount of heat: Qrec = C2m 2 (0 — T 2), where 0 is the final tem perature of the two bodies. Since energy is conserved, we can w rite @glv + C*rec = 0. This gives c1m 1 (7 \ — 0 ) = c2m 2 (0 — T 2). This is the heat balance equation for the system of two bodies. A body can be heated a t the expense of the energy released during fuel com bustion. The am ount of heat liberated is pro­ portional to the mass of the burnt fuel and depends on which fuel it is: Q = qmf, where q is the specific heat of combustion of the fuel in joules per kilogram (J/kg) and m t is the mass of the burnt fuel. N ot all the heat liberated during the com bustion of a fuel is usefully spent on heating. A fraction of the heat dissi­ pates. This is taken into account by introducing the efficiency of a heater: 7j — ^useful $spent (?urefui = cm

and Qspent = qmt, we have cm A T

n = -----------

The internal energy of a body or a system of bodies m ay change when a m echanical work is done. The m echanical energy of a body or a system of bodies can be transform ed com pletely into internal energy, i.e. be spent on heating: Q — AZ?k + A2?p, where AE k and Ai?p are the changes in the kinetic and po­ te n tia l energies of the body. The first law of therm odynam ics (the law of energy conser­ vation). The am ount of heat transferred to a closed system

§ 4. The Change in the Internal Energy

39

is spent on increasing its internal energy and on the m echan­ ical work done against external forces: Q = AC/ + A, where Q is the am ount of heat transferred to the system , \ U is the change in its internal energy, and A is the work done by the system. For isochoric processes, V = const, and hence A = p AF = 0, Q = A *7, i.e. the heat transferred to a gas is com pletely used to in ­ crease its internal energy. In isobaric processes, p = const, and hence Q = A t/ + 4 , i.e. the heat transferred to a gas is spent on increasing its internal energy and on the work done to expand the gas. For isotherm al processes, T = const, and hence the in ­ ternal energy does not change. Therefore, Q = A . The heat supplied to a gas during an isotherm al process is com pletely spent on the m echanical work done by the gas. A diabatic processes occur in the absence of any heat exchange w ith the surroundings, i.e. Q = 0. Therefore, the first law of therm odynam ics in this case will have the form Q = A U + A , or A = - A U, i.e. an adiabatically expanding gas does work at the expense of the change in its internal energy, being cooled thereby. Worked Problems Problem 20. In order to determ ine the specific heat of copper, a copper cylinder of mass 0.5 kg is heated to 100°C and then placed in an alum inium calorim eter of mass 40 g, containing 300 g of w ater at a tem perature of 15°C. As a result of heat exchange, the calorim eter stabilizes at tem ­ perature 26°C. W hat is the specific heat of copper in this experim ent? Compare the obtained result w ith the ta b u la t­ ed value and determ ine the absolute and the relative error. Given: m c = 0.5 kg is the mass of the copper cylinder, T = 373 K is the in itia l tem perature of the cylinder, ma = 0.04 kg is the mass of the calorim eter, m w = 0.3 kg is the

40____________Ch. I. F u n d a m e n tals of M olecular P h y sics

m ass of the w ater, 71, = 288 K is the in it ia l tem perature of th e w ater and ca lorim eter, and 0 = 299 K is the final te m ­ perature of th e w ater, ca lorim eter, and c y lin d er. From ta ­ b le s, w e tak e th e specific h ea t of alu m in iu m , ca — 880 J/(k g - K ), and the specific h e a t of w ater, cw = 4187 J /(k g -K ). T h e tab u la ted specific h ea t of copper is ctab — 380 J /(k g -K ). F ind: the specific heat of copper cc, the absolute error Ac

and the relative error Ac/ctab

the measurem ent.

S o lu tio n . A s a r e su lt of h ea t e xch ange, th e tem perature of a ll the b odies in the calo rim eter eq u a liz e s. T he heated c y l­ inder g iv e s off heat @g i v and is cooled from T to 0 :

Cgiv = ccm c (T — 0 ). T he ca lo rim eter and w ater receive h eat, and their t e m ­ perature rises from T l to 0 :