10.3 Solutions
443
(b) W =
dW = −
P dV
(1)
PV = nRT 1 = constant
or
P =
nRT 1
V
(T = constant for isothermal process)
(2)
Substituting (2) in (1)
W = −nRT
V 2
V 1
dV
V
= −nRT ln
V 2
V 1
(c) At the beginning of the cycle, P 1 = 1.01 × 10 5 Pa, V 1 = 1 m 3 , T 1 =
273.15 K.
At the end of stage 1, P 2 = 5.05 × 10 4 Pa, V 2 = 2 m 3 , T 2 = 273.15 K.
At the end of stage 2, P 3 = 1.01 × 10 5 Pa, V 3 = 2 m 3 , T 3 = T 2 P 3 /P 2 =
546.3 K.
(i) n =
P 1 V 1
RT 1
=
1.01 × 10 5 × 1.0
8.31 × 273.15
= 44.5 mol
(ii) Stage1, W 1 = −nRT ln (V 2 /V 1 )
= −44.5 × 8.31 × 273.15 ln(2/1) = −70046 J
Stage2, W 2 = P 2 V = P 3 (V 3 − V 2 ) = P(V 2 − V 2 ) = 0
Stage3, W 3 = P 3 (V 1 − V 3 ) = 1.01 × 10
5
(1 − 2)
= −1.015 × 10
5 J
(iii) U = 0, overall.
10.45 (a) and (b)
P 1 V 1
T 1
=
P 2 V 2
T 2
∴
T 1 =
P 1
P 2
V 1
V 2
T 2 =
P 1
3P 1
V 1
V 1
1083 = 361 K
n =
P 1 V 1
RT 1
=
(10 5 )(0.06)
(8.31)(361)
= 2 mol
P 3 V 3
T 3
=
P 2 V 2
T 2
∴ T 3 =
P 3
P 2
V 3
V 2
T 2 =
P 2
P 2
(V 2 /3)
V 2
1083 = 361 K
T 4 = T 3 = 361 K
(∵ process 3 → 4 isothermal process)
P 4 V 4
T 4
=
P 3 V 3
T 3