習題六
14. Two formulas (other than the first law of thermodynamics) will be of use to us. It is
straightforward to show, from Eq. 19-11, that for any process that is depicted as a straight
line on the pV diagram, the work is
 pi  p f 
Wstraight  
 V
 2 
which includes, as special cases, W = pV for constant-pressure processes and W = 0 for
constant-volume processes. Further, Eq. 19-44 with Eq. 19-51 gives
 f 
 f 
Eint  n   RT    pV
2
2
where we have used the ideal gas law in the last step. We emphasize that, in order to obtain
work and energy in joules, pressure should be in pascals (N/m2) and volume should be in
cubic meters. The degrees of freedom for a diatomic gas is f = 5.
(a) The internal energy change is
5
5
 pcVc  paVa     0.50 103 Pa  4.0 m3   5.0 103 Pa  2.0 m3  
2
2
4
 2.0 10 J.
Eint c  Eint a 
(b) The work done during the process represented by the diagonal path is
 5.5 103 Pa 
 p  pc 
3
Wdiag   a
V

V



 c a 
  2.0 m 
 2 

2

which yields Wdiag = 5.5 ×103 J. Consequently, the first law of thermodynamics gives
Qdiag  Eint  Wdiag  (2.0 104  5.5 103 ) J  14.5 103 J  15 kJ.
(c) The fact that Eint only depends on the initial and final states, and not on the details of
the “path” between them, means we can write Eint  Eint c  Eint a  2.0 104 J for the
indirect path, too. In this case, the work done consists of that done during the constant
pressure part (the horizontal line in the graph) plus that done during the constant volume
part (the vertical line):
Windirect  5.0 103 Pa  2.0m3   0  1.0 104 J.
Now, the first law of thermodynamics leads to
Qindirect  Eint  Windirect  (2.0 104  1.0 104 ) J  1.0 10 4 J  10 kJ.
41. When the valve is closed the number of moles of the gas in container A is nA = pAVA/RTA
and that in container B is nB = 4pBVA/RTB. The total number of moles in both containers is
then
n  nA  nB 
VA  pA 4 pB 


  const.
R  TA TB 
此題活門打開後，一部分氣體由右室進入左室，因此粒子數改變為 nA, nB 。左右兩室
溫度維持原來的溫度，After the valve is opened, the pressure in container A is pA =
RnATA/VA and that in container B is pB = RnBTB/4VA. Equating pA and pB, we obtain
RnATA/VA = RnBTB/4VA, or nB = (4TA/TB)nA. Thus,

4T 
V p
4 pB 
n  nA  nB  nA 1  A   nA  nB  A  A 
.
TB 
R  TA
TB 

We solve the above equation for nA:
V ( p A TA + 4 pB TB )
.
(1+ 4TA TB )
nA nA¢n =
R

Substituting this expression for nA Ainto pVA = nARTA, we obtain the final pressure:
p 
55. (a)
nA RTA
p  4 pBTA / TB
 A
 3.2  105 Pa.
VA
1  4TA / TB
At point a, we know enough information to compute n:


 2500 Pa  1.0 m3
pV
n

1.1mol.
RT 8.31 J/mol  K   280 K 
(b) We can use the answer to part (a) with the new values of pressure and volume, and solve
the ideal gas law for the new temperature, or we could set up the gas law in terms of ratios
(note: na = nb and cancels out):
pbVb Tb

paVa Ta

 7.5kPa   3.0 m3 
Tb   280 K  
 2.5 103 K .

3 
2.5kPa
1.0
m



(c) As in the previous part, we choose to approach this using the gas law in ratio form:
pcVc Tc

paVa Ta
 2.5kPa   3.0 m3 
 Tc   280 K  
 8.4 102 K .

3 
 2.5kPa   1.0 m 
(d) The net work done as the gas progresses through the cycle is related to  “area” inside
that triangle (with area = 12 (base)(height) ), where we choose the plus sign because the
volume change at the largest pressure is an increase. Thus,
Qnet  Wnet 

1
2.0 m3
2
  5.0  10
3

Pa  5.0  103 J.
(e) The net energy added to the gas (as heat) is equal to the net work:
5.0  103 J.