Question:
In an isobaric process of an ideal gas, the ratio of heat supplied and work done by the system $\left(\frac{\text{Q}}{\text{W}}\right)$ is $\left[\frac{\text{C}_{\text{P}}}{\text{C}_{\text{V}}} = \gamma\right]$.
In an isobaric process of an ideal gas, the ratio of heat supplied and work done by the system $\left(\frac{\text{Q}}{\text{W}}\right)$ is $\left[\frac{\text{C}_{\text{P}}}{\text{C}_{\text{V}}} = \gamma\right]$.
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$Q : \Delta U : W$ for isobaric process is $\gamma : 1 : (\gamma - 1)$.
Updated On: Apr 26, 2026
- 1
- $\gamma$
- $\frac{\gamma}{\gamma-1}$
- $\frac{\gamma-1}{\gamma}$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
For an isobaric process: $Q = nC_P \Delta T$ and $W = P\Delta V = nR\Delta T$.
Step 2: Analysis
Ratio $\frac{Q}{W} = \frac{nC_P \Delta T}{nR \Delta T} = \frac{C_P}{R}$.
Step 3: Calculation
We know $R = C_P - C_V$.
$\frac{Q}{W} = \frac{C_P}{C_P - C_V}$.
Divide numerator and denominator by $C_V$:
$\frac{Q}{W} = \frac{C_P/C_V}{(C_P/C_V) - 1} = \frac{\gamma}{\gamma - 1}$.
Step 4: Conclusion
Hence, the ratio is $\frac{\gamma}{\gamma-1}$.
Final Answer: (C)
For an isobaric process: $Q = nC_P \Delta T$ and $W = P\Delta V = nR\Delta T$.
Step 2: Analysis
Ratio $\frac{Q}{W} = \frac{nC_P \Delta T}{nR \Delta T} = \frac{C_P}{R}$.
Step 3: Calculation
We know $R = C_P - C_V$.
$\frac{Q}{W} = \frac{C_P}{C_P - C_V}$.
Divide numerator and denominator by $C_V$:
$\frac{Q}{W} = \frac{C_P/C_V}{(C_P/C_V) - 1} = \frac{\gamma}{\gamma - 1}$.
Step 4: Conclusion
Hence, the ratio is $\frac{\gamma}{\gamma-1}$.
Final Answer: (C)
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