(C) for Helium, the Atomic Mass Is

5 轾4 3 PV 1.013 10 Pa3 p ( 0.150 m ) P21.7 (a) PV= Nk T : N = =臌 = 3.54 1023 atoms B -23 kB T (1.38 10 J K)( 293 K)

3 3 (b) K= k T = 1.38� 10-23( 293) J 6.07 10 - 21 J 2B 2 ( )

m = 6.64 10-27 kg molecule

12 3 3kB T mv= kB T : \v = = 1.35 km s 2 2 rms m

Nmv 2 P21.10 (a) PV= nRT = 3

Nmv 2 The total translational kinetic energy is = E : 2 trans

3 3 E= PV =3.00创 1.013 105 5.00 � 10- 3 2.28 kJ trans 2 2 ( )( )

mv2 3k T 3 RT 3( 8.314) ( 300) =B = = = 6.21 10-21 J (b) 23 2 2 2N A 2( 6.02 10 ) nRT P21.14 The piston moves to keep pressure constant. Since V = , then P

nRD T DV = for a constant pressure process. P

Q Q2 Q Q= nCP D T = n( C V + R) D T so DT = = = n( CV + R) n(5 R 2 + R) 7nR

nR骣2 Q 2 Q 2 QV and DV =琪 = = P桫7 nR 7 P 7 nRT

3 2 (4.40 10 J)( 5.00 L) DV = = 2.52 L 7 (1.00 mol) ( 8.314 J mol K)( 300 K) Thus, Vf= V i + D V =5.00 L + 2.52 L = 7.52 L

Q= nC D T + nC D T P21.20 ( P)isobaric ( V )isovolumetric

In the isobaric process, V doubles so T must double, to 2Ti .

In the isovolumetric process, P triples so T changes from 2Ti to 6Ti .

骣7 骣 5 Q= n琪 R(2 T - T) + n 琪 R( 6 T - 2 T) = 13.5 nRT = 13.5 PV 桫2i i 桫 2 i i i

P21.21 In the isovolumetric process A B , W = 0 and Q= nCV D T = 500 J

骣3R 2( 500 J) 500 J=n琪 ( T - T) or T = T + 桫2B A B A 3nR 2( 500 J) T =300 K + = 340 K B 3( 1.00 mol) ( 8.314 J mol K)

In the isobaric process B C , 5nR Q= nC D T =( T - T ) = -500 J . P2 C B Thus,

2( 500 J) 1 000 J (a) T= T - =340 K - = 316 K C B 5nR 5( 1.00 mol) ( 8.314 J mol K)

(b) The work done on the gas during the isobaric process is

WBC= - P B D V = - nR( T C - T B ) = - (1.00 mol) ( 8.314 J mol� K)( 316 K 340 J)

or WBC = +200 J

The work done on the gas in the isovolumetric process is zero, so in total

Won gas = +200 J . P21.29 (a) See the diagram at the right. P B 3 Pi g g (b) PB VB = P C VC Adiabatic

g g 3PVii = PV i C 1g 5 7 VC=(3) V i =( 3) V i = 2.19 V i C P i A VC =2.19( 4.00 L) = 8.77 L V(L) Vi = 4 L VC

TB=3 T i = 3( 300 K) = 900 K

(d) After one whole cycle, TA= T i = 300 K .

骣5 (e) In AB, Q= nC D V = n琪 R(3 T - T) = ( 5.00) nRT AB V桫2 i i i

QBC = 0 as this process is adiabatic

PC V C= nRT C = P i(2.19 V i) = ( 2.19) nRT i

so TC= 2.19 T i

骣7 Q= nC D T = n琪 R( T -2.19 T) =( - 4.17) nRT CA P桫2 i i i

For the whole cycle,

QABCA= Q AB + Q BC + Q CA =(5.00 - 4.17) nRT i = ( 0.829) nRT i DE =0 = Q + W ( int )ABCA ABCA ABCA

WABCA= - Q ABCA = -(0.829) nRT i = - ( 0.829) PV i i 5- 3 3 WABCA = -(0.829) ( 1.013创 10 Pa) ( 4.00 10 m) = - 336 J P21.31 (a) The work done on the gas is

Vb Wab = - PdV .

For the isothermal process,

V b 骣1 W= - nRT琪 dV ab a 桫V Va

骣Vb 骣 V a FIG. P21.31 Wab= - nRT a ln琪 = nRT ln 琪 . 桫Va 桫 V b

Thus,

Wab = 5.00 mol( 8.314 J mol K)( 293 K) ln( 10.0)

Wab = 28.0 kJ .

(b) For the adiabatic process, we must first find the final temperature, Tb . Since air consists primarily of diatomic molecules, we shall use

5R 5( 8.314) g air = 1.40 and C = = = 20.8 J mol K . V , air 2 2

Then, for the adiabatic preocess

g -1 骣Va 0.400 Tb= T a 琪 =293 K( 10.0) = 736 K . 桫Vb

Thus, the work done on the gas during the adiabatic process is

W- Q + D E = -0 + nC D T = nC T - T ab( int )ab( V) ab V( b a )

or Wab =5.00 mol( 20.8 J mol� K)( 736 = 293) K 46.0 kJ .

骣Va Thus, Pb= P a 琪 =1.00 atm( 10.0) = 10.0 atm . 桫Vb

g g For the adiabatic process, we have Pb Vb = P a V a .

g 骣Va 1.40 Thus, Pb= P a 琪 =1.00 atm( 10.0) = 25.1 atm . 桫Vb

P21.33 The heat capacity at constant volume is nCV . An ideal gas of diatomic molecules has three degrees of freedom for translation in the x, y, and z directions. If we take the y axis along the axis of a molecule, then outside forces cannot excite rotation about this axis, since they have no lever arms. Collisions will set the molecule spinning only about the x and z axes.

(a) If the molecules do not vibrate, they have five degrees of freedom. Random collisions put equal

1 amounts of energy k T into all five kinds of motion. The average energy of one molecule is 2 B

5 k T . The internal energy of the two-mole sample is 2 B

骣5 骣 5 骣 5 N琪 k T= nN 琪 k T = n 琪 R T = nC T . 桫2B A 桫 2 B 桫 2 V

5 The molar heat capacity is C= R and the sample’s heat capacity is V 2

骣5 骣 5 nCV = n琪 R = 2 mol 琪 ( 8.314 J mol K) 桫2 桫 2

nCV = 41.6 J K

For the heat capacity at constant pressure we have 骣5 7 骣 7 nCP= n( C V + R) = n琪 R + R = nR = 2 mol 琪 ( 8.314 J mol K) 桫2 2 桫 2

nCP = 58.2 J K (b) In vibration with the center of mass fixed, both atoms are always moving in opposite directions with equal speeds. Vibration adds two more degrees of freedom for two more terms in the molecular energy, for kinetic and for elastic potential energy. We have

骣7 nC= n琪 R = 58.2 J K V 桫2

骣9 and nC= n琪 R = 74.8 J K P 桫2

P21.36 (a) The ratio of the number at higher energy to the number at lower energy is e-DE kB T where DE is the energy difference. Here,

-19 骣1.60 10 J -18 DE =(10.2 eV) 琪 = 1.63 10 J 桫 1 eV

and at 0°C,

-23 - 21 kB T = (1.38� 10 J K)( 273 K) 3.77 10 J .

Since this is much less than the excitation energy, nearly all the atoms will be in the ground state and the number excited is

-18 25骣-1.63 10 J 25- 433 2.70� 10 exp琪 2.70 10 e . ( ) 桫3.77 10-21 J ( )

This number is much less than one, so almost all of the time no atom is excited .

(b) At 10 000°C,

-23 - 19 kB T = (1.38� 10 J K) 10 273 K 1.42 10 J .

The number excited is

-18 25骣-1.63 10 J 25- 11.5 20 2.70� 10 exp琪 � 2.70 10e 2.70 10 . ( ) 桫1.42 10-19 J ( )

n v 1 P21.37 (a) v =i i =�1( 2) + 2( 3) + 3( 5) + 4( 7) + 3( 9) � 2( 12) 6.80 m s av N 15 臌

n v 2 (b) v 2=i i = 54.9 m 2 s 2 ( )av N

2 so vrms = v =54.9 = 7.41 m s ( )av

骣N PVN A P21.45 (a) PV= 琪 RT and N = so that 桫N A RT

(1.00创 10-10)( 133) ( 1.00) ( 6.02 10 23 ) N = = 3.21 1012 molecules (8.314) ( 300)

1V 1.00 m 3 ℓ = = = (b) 21 2 2 1 212- 10 2 1 2 nV p d2 N p d 2 (3.21创 10 molecules)p ( 3.00 10 m) ( 2)

ℓ = 779 km

v (c) f = = 6.42 10-4 s - 1 ℓ

P21.51 (a) Pf = 100 kPa T f = 400 K

nRT f 2.00 mol( 8.314 J mol K)( 400 K) 3 V f = =3 =0.066 5 m = 66.5 L Pf 100 10 Pa

DEint =(3.50) nR D T = 3.50( 2.00 mol) ( 8.314 J mol� K)( 100 K) 5.82 kJ

W= - P D V = - nR D T = -(2.00 mol) ( 8.314 J mol� K)( 100 K) - 1.66 kJ Q= D Eint - W =5.82 kJ + 1.66 kJ = 7.48 kJ

(b) T f = 400 K

nRTi 2.00 mol( 8.314 J mol K)( 300 K) 3 Vf= V i = =3 =0.049 9 m = 49.9 L Pi 100 10 Pa

骣Tf 骣400 K P= P =100 kPa = 133 kPa W= - PdV = 0 V = constant f i 琪桫琪 since 桫Ti 300 K

DEint = 5.82 kJ as in part (a) Q= D Eint - W =5.82 kJ - 0 = 5.82 kJ

骣P 骣100 kPa V= V i =49.9 L = 41.6 L DE =(3.50) nR D T = 0 f i 琪桫琪 int since 桫Pf 120 kPa

T = constant

V f V 骣 dV 骣 f Pi W= -蝌 PdV = - nRTi = - nRT iln琪 = - nRT i ln 琪 V桫 Vi桫 P f Vi 骣100 kPa W = -(2.00 mol) ( 8.314 J mol� K)( 300 K) ln琪 + 909 J 桫120 kPa

Q= D Eint - W =0 - 910 J = - 909 J

CP CV + R 3.50R+ R 4.50 9 (d) Pf = 120 kPa g = = = = = CV C V 3.50 R 3.50 7

1 g 7 9 g g 骣P 骣100 kPa P V= PV : so V= V i =49.9 L = 43.3 L ff i i f i 琪桫琪桫Pf 120 kPa 骣Pf V f 骣120 kPa 骣 43.3 L T= T =300 K = 312 K f i 琪桫琪桫琪桫PVi i 100 kPa 49.9 L

DEint =(3.50) nR D T = 3.50( 2.00 mol) ( 8.314 J mol� K)( 12.4 K) 722 J Q = 0 ( adiabatic process)

W= - Q + D Eint =0 + 722 J = + 722 J

*P21.54 (a) W= nCV( T f - T i )

3 -2 500 J = 1 mol 8.314 J mol� KT 500 K 2 ( f )

T f = 300 K

g g (b) PVii = P f V f

g g 骣nRT 骣nRTi f g1- g g 1 - g Pi琪 = P f 琪 Ti P i = Tf P f 桫 Pi桫 P f

g( g -1) g g -1 g( g -1) T T ( ) Ti f 骣 f = Pf= P i 琪 Pi P f 桫Ti

(5 3)( 3 2) 5 2 骣Tf 骣300 P= P =3.60 atm = 1.00 atm f i 琪桫琪桫Ti 500

5- 3 3 PV (1.013创 10 Pa) ( 5.00 10 m ) P21.67 (a) n = = = 0.203 mol RT (8.314 J mol K)( 300 K)

骣P 骣3.00 T= T B =300 K = 900 K (b) B A 琪桫琪桫PA 1.00

TC= T B = 900 K 骣T 骣900 V= V C =5.00 L = 15.0 L C A 琪桫琪桫TA 300 FIG. P21.67

3 3 (c) Eint, A= nRT A = (0.203 mol) ( 8.314 J mol� K)( 300 K) 760 J 2 2 3 3 Eint, B= E int, C = nRT B = (0.203 mol) ( 8.314 J mol� K)( 900 K) 2.28 kJ 2 2

(d) P (atm) V(L) T(K) Eint (kJ) A 1.00 5.00 300 0.760 B 3.00 5.00 900 2.28 C 1.00 15.00 900 2.28

(e) For the process AB, lock the piston in place and put the cylinder into an oven at 900 K. For BC, keep the sample in the oven while gradually letting the gas expand to lift a load on the piston as far as it can. For CA, carry the cylinder back into the room at 300 K and let the gas cool without touching the piston.

(f) For AB: W = 0 DEint = E int, B - E int, A =(2.28 - 0.760) kJ = 1.52 kJ

Q= D Eint - W = 1.52 kJ

骣VC For BC: DEint = 0 , W= - nRTB ln 琪桫VB

W = -(0.203 mol) ( 8.314 J mol� K)( 900 K) ln( 3.00) - 1.67 kJ

Q= D Eint - W = 1.67 kJ

For CA: DEint = E int, A - E int, C =(0.760 - 2.28) kJ = - 1.52 kJ

W= - P D V = - nR D T = -(0.203 mol) ( 8.314 J mol� K)( = 600 K) 1.01 kJ

Q= D Eint - W = -1.52 kJ - 1.01 kJ = - 2.53 kJ

(g) We add the amounts of energy for each process to find them for the whole cycle. QABCA = +1.52 kJ + 1.67 kJ - 2.53 kJ = 0.656 kJ

WABCA =0 - 1.67 kJ + 1.01 kJ = - 0.656 kJ DE = +1.52 kJ + 0 - 1.52 kJ = 0 ( int )ABCA