1m3 of an ideal gas at 500 K and 1000 kPa expands reversibly to 5 times its initial volume in an insulated container. If the specific heat capacity (at constant pressure) of the gas is 21 J/mole.K, the final temperature will be
A. 35 K
B. 174 K
C. 274 K
D. 154 K
Answer: Option B
Solution (By Examveda Team)
Insulated cylinder $$ \Rightarrow $$ adiabatic system$$\eqalign{ & \Rightarrow P{V^\gamma } = {\text{CONSTANT}} \cr & \Rightarrow T{V^{\gamma - 1}} = {\text{CONSTANT}} \cr & \Rightarrow \frac{{{T_1}}}{{{T_2}}} = \frac{{{V_2}^{\gamma - 1}}}{{{V_1}}} \cr & \Rightarrow \frac{{500}}{{{T_2}}} = {\left( 5 \right)^{\gamma - 1}} \cr & {C_P} - {C_V} = R\left( {{\text{For ideal gas}}} \right) \cr & \Rightarrow \gamma = 1.65 \cr & {\text{So, }}{T_2} = 176.05\,K \cr} $$
This Question Belongs to Chemical Engineering >> Chemical Engineering Thermodynamics
1m3 of an ideal gas at 500 K and 1000 kPa expands reversibly to 5 times its initial volume in an insulated container. If the specific heat capacity (at constant pressure) of the gas is 21 J/mole.K, the final temperature will be
A. 35 K
B. 174 K
C. 274 K
D. 154 K
We are given
Initial volume
V
1
=
1
m
3
V
1
=1m
3
Final volume
V
2
=
5
m
3
V
2
=5m
3
Initial temperature
T
1
=
500
K
T
1
=500K
Pressure
P
1
=
1000
kPa
P
1
=1000kPa
Process: reversible and adiabatic expansion (since it's an insulated container)
Specific heat at constant pressure:
C
p
=
21
J/mol
cdotp
K
C
p
=21J/molcdotpK
Step 1: Use the adiabatic relation for ideal gases:
For a reversible adiabatic process:
T
2
=
T
1
(
V
1
V
2
)
γ
−
1
T
2
=T
1
(
V
2
V
1
)
γ−1
We need to find
γ
γ, the heat capacity ratio:
γ
=
C
p
C
v
,
and
C
v
=
C
p
−
R
γ=
C
v
C
p
,andC
v
=C
p
−R
For an ideal gas,
R
=
8.314
J/mol
cdotp
K
R=8.314J/molcdotpK, so:
C
v
=
21
−
8.314
=
12.686
J/mol
cdotp
K
C
v
=21−8.314=12.686J/molcdotpK
γ
=
21
12.686
≈
1.655
γ=
12.686
21
≈1.655
Step 2: Apply the temperature-volume relation:
T
2
=
500
(
1
5
)
1.655
−
1
=
500
⋅
5
−
0.655
T
2
=500(
5
1
)
1.655−1
=500⋅5
−0.655
Now calculate
5
−
0.655
5
−0.655
:
log
10
(
5
)
≈
0.6990
⇒
−
0.655
⋅
log
10
(
5
)
≈
−
0.458
log
10
(5)≈0.6990⇒−0.655⋅log
10
(5)≈−0.458
10
−
0.458
≈
0.349
10
−0.458
≈0.349
So,
T
2
≈
500
⋅
0.349
=
174.5
K
T
2
≈500⋅0.349=174.5K
✅ Final Answer is C