At `27^@C` two moles of an ideal monatomic gas occupy a volume V. The gas expands adiabatically to a volume `2V`. Calculate
(a) final temperature of the gas
(b) change in its internal energy and
[ `R=8.31J//mol-K`]
At `27^@C` two moles of an ideal monatomic gas occupy a volume V. The gas expands adiabatically to a volume `2V`. Calculate
(a) final temperature of the gas
(b) change in its internal energy and
[ `R=8.31J//mol-K`]
(a) final temperature of the gas
(b) change in its internal energy and
[ `R=8.31J//mol-K`]
A
`(a)195K" "(b)2.7kJ`
B
`(a)189K" "(b)2.7kJ`
C
`(a)195K" "(b)-2.7kJ`
D
`(a)189K" "(b)-2.7kJ`
Text Solution
AI Generated Solution
The correct Answer is:
To solve the problem, we will break it down into two parts: (a) calculating the final temperature of the gas after adiabatic expansion, and (b) calculating the change in internal energy.
### Given Data:
- Initial temperature \( T_1 = 27^\circ C = 300 \, K \)
- Initial volume \( V_1 = V \)
- Final volume \( V_2 = 2V \)
- Number of moles \( n = 2 \)
- Universal gas constant \( R = 8.31 \, J/(mol \cdot K) \)
- For a monatomic ideal gas, \( \gamma = \frac{5}{3} \approx 1.67 \)
### Part (a): Final Temperature of the Gas
1. **Use the adiabatic relation for temperature and volume:**
\[
\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1}
\]
2. **Substituting the values:**
\[
\frac{T_2}{300} = \left(\frac{V}{2V}\right)^{1.67 - 1}
\]
\[
\frac{T_2}{300} = \left(\frac{1}{2}\right)^{0.67}
\]
3. **Calculating \( \left(\frac{1}{2}\right)^{0.67} \):**
\[
\left(\frac{1}{2}\right)^{0.67} \approx 0.6156
\]
4. **Now, calculate \( T_2 \):**
\[
T_2 = 300 \times 0.6156 \approx 184.68 \, K \approx 185 \, K
\]
### Part (b): Change in Internal Energy
1. **Internal energy \( U \) of an ideal gas is given by:**
\[
U = n C_v T
\]
Where \( C_v = \frac{R}{\gamma - 1} \).
2. **Calculate \( C_v \):**
\[
C_v = \frac{8.31}{1.67 - 1} = \frac{8.31}{0.67} \approx 12.4 \, J/(mol \cdot K)
\]
3. **Calculate initial internal energy \( U_1 \):**
\[
U_1 = n C_v T_1 = 2 \times 12.4 \times 300 = 7440 \, J
\]
4. **Calculate final internal energy \( U_2 \):**
\[
U_2 = n C_v T_2 = 2 \times 12.4 \times 185 \approx 4584 \, J
\]
5. **Change in internal energy \( \Delta U \):**
\[
\Delta U = U_2 - U_1 = 4584 - 7440 = -2856 \, J \approx -2.86 \, kJ
\]
### Final Answers:
- (a) Final temperature \( T_2 \approx 185 \, K \)
- (b) Change in internal energy \( \Delta U \approx -2.86 \, kJ \)
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