Solution_8_recitation - Solution to the problems in...

Solution to the problems in recitation 8
Problem 1
Known: Ideal gas in two compartments of a closed, rigid container.
Find: p
2B
, T
2B
and S
2B
-S
1B
;
p
2A
, T
2A
and S
2A
-S
1A
;
Q
12
(kJ);
Change of S for the universe.
Schematic and Given Data:
A
B
2
A
B
1
C
v
= 0.75 kJ/kg.K, C
p
/C
v
= 1.4, R = 0.300 kJ/kg.K
p
1A
= p
1B
= 180 kPa, T
1A
= T
1B
= 300 K, V
1A
= V
1B
= 0.5 m
3
, V2A = 0.75 m
3
, V2B = 0.25 m
3
Assumptions: 1. gas in each compartment is a closed system
2. part B undergoes a quasi-static process
3. Heat transfer occurs at reservoir temperature
4. Constant specific heat
5. No change in KE and PE
Analysis:
a) For gas in compartment B, it undergoes an isentropic (adiabatic and reversible) process. C
p
/C
v
= 1.4
p
1B
(V
1B
)
1.4
= p
2B
(V
2B
)
1.4
p
2B
= p
1B
(V
1B
/V
2B
)
1.4
= 180 kPa * (0.5 m
3
/0.25 m
3
)
1.4
p
2B
= 475 kPa
m
B
= (p
1B
V
1B
)/(RT
1B
) = 1 kg
From ideal gas law,
T
2B
/T
1B
= (p
2B
V
2B
)/(p
1B
V
1B
)
T
2B
= T
1B
(p
2B
V
2B
)/(p
1B
V
1B
) = 300 K (475 kPa * 0.25 m
3
)/(180 kPa * 0.5 m
3
) = 395.83 K
Entropy change should be zero for part B, since there is no heat transfer, and there is no internal irreversibility
for part B. according to entropy change for ideal gas
S
2B
-S
1B
= m
B
[C
v
ln(T
2B
/T
1B
) + R ln(V
2B
/V
1B
)]
= 1 (kg) [0.75 (kJ/kg.K) ln(395.83/300) + 0.3 (kJ/kg.K) ln(0.25/0.5)]