HOMEWORK ASSIGNMENT 6
Problems Chapter 6
6-2 Calculate the entropy change for the following processes a)
0 C to 80 0
1 kg of water is heated reversibly by an electric heating coil from 20
C . Assume constant pressure
∆ S =
Z
T f
T i
= 778 dQ
T
= 4 .
18 ∗ 10 3
=
∗
Z ln
T f
T i
1
T
293 .
nc
353 .
15
15 p dT = nc p ln
T
T f i b) 1 kg of ice melts at 0 0 C and 1 atm.
c)
∆ S =
∆ Q
T
=
`
T
=
1 kg of steam condenses at 100 0
3 .
34 ∗ 10 5
273 .
15
= 1222
C and 1 atm.
∆ S =
∆ Q
T
=
`
T
=
2 .
26 ∗ 10 6
373 .
15
= − 6057
6-7 a) b)
Derive an expression for the entropy of an ideal gas
As a function of T and V
T dS = dU + P dV =
µ
∂U
∂T
¶
V dT + (
µ
∂U
∂V
¶
T
S = nc
V
Z
= nc
V
T dT
T dT +
+ nR
Z
P dV
V dV
V
= nc
V ln T + nR ln V
+ P ) dV
+ constant
As a function of T and P
S =
= nc nc
V
V nc
P ln T + nR ln V + constant ln T + nR ln nRT
P
+ constant ln T − nR ln P + constant
6-11 a)
Consider a van der Waals gas
Show that c
µ
∂u
¶
∂v
T v
=
= is a function of T only. From (6.26)
T
T
µ
µ
∂P
∂T
∂
∂T
¶ v
−
RT
−
P b
− v a
2
¶¶ v
−
µ v
RT
− b
= v a
2
µ v
− a v 2
¶
1

Integrating this gives and so b) u = − a v
+ f ( T ) c v
=
µ
∂u
¶
∂T v
=
∂f ( T )
∂T
.
The specific internal energy is u = − a v
= − a v
+ f ( T ) = − a v
Z
T
+ c v dT
+
Z
T
+ const
∂f ( T )
∂T dT + const c) s =
=
The specific entropy is
Z
Z du
T
+
Z
T c v dT
T
P
T dv =
Z
T c v dT
T
+
Z µ
+ RT ln ( v − b ) + const
1
T
µ
∂u
¶
∂v
T
+
R v − b
− a
T v 2
¶ dv
2