Question:
During an adiabatic process, if the volume of 4 moles of a monoatomic gas initially at a temperature of $127^{\circ}C$ increases by 7 times, then the work done by the gas is:
During an adiabatic process, if the volume of 4 moles of a monoatomic gas initially at a temperature of $127^{\circ}C$ increases by 7 times, then the work done by the gas is:
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Adiabatic work depends on the temperature change!
Updated On: Jun 6, 2026
- 2400 R
- 900 R
- 1800 R
- 1200 R
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Adiabatic process work formula: $W = \frac{nR(T_i - T_f)}{\gamma - 1}$.
Step 2: Meaning
Monoatomic $\gamma = 5/3$. $T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$.
Step 3: Analysis
$T_i = 400$ K. $V_f = 7 V_i$. $T_f = T_i (V_i/V_f)^{\gamma-1} = 400 \times (1/7)^{2/3}$. This calculation might be simplified. Alternative: $W = \frac{nR T_i [1 - (V_i/V_f)^{\gamma-1}]}{\gamma-1}$. $W = \frac{4 \times R \times 400 \times [1 - (1/7)^{2/3}]}{2/3} = 2400 R \times (1 - 0.27) \approx 1800 R$.
Step 4: Conclusion
The work done is 1800 R.
Final Answer: (C)
Adiabatic process work formula: $W = \frac{nR(T_i - T_f)}{\gamma - 1}$.
Step 2: Meaning
Monoatomic $\gamma = 5/3$. $T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$.
Step 3: Analysis
$T_i = 400$ K. $V_f = 7 V_i$. $T_f = T_i (V_i/V_f)^{\gamma-1} = 400 \times (1/7)^{2/3}$. This calculation might be simplified. Alternative: $W = \frac{nR T_i [1 - (V_i/V_f)^{\gamma-1}]}{\gamma-1}$. $W = \frac{4 \times R \times 400 \times [1 - (1/7)^{2/3}]}{2/3} = 2400 R \times (1 - 0.27) \approx 1800 R$.
Step 4: Conclusion
The work done is 1800 R.
Final Answer: (C)
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