Question:
Two cylinders A and B contain equal amounts of diatomic gas at T K. Piston A is free to move (isobaric), B is fixed (isochoric). If rise in temp of A is $dT_{A}$ for same heat $Q$, rise in B is}
Two cylinders A and B contain equal amounts of diatomic gas at T K. Piston A is free to move (isobaric), B is fixed (isochoric). If rise in temp of A is $dT_{A}$ for same heat $Q$, rise in B is}
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For the same heat, the temperature rise is higher in isochoric processes because no work is done.
Updated On: Jun 19, 2026
- $dT_{A}/2$
- $dT_{A}/\gamma$
- $\gamma dT_{A}$
- $2dT_{A}$
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The Correct Option is C
Solution and Explanation
Step 1: Formula
$Q = nC_p dT_A$ (for A) and $Q = nC_v dT_B$ (for B).
Step 2: Analysis
Since $Q$ is the same: $nC_p dT_A = nC_v dT_B$.
Step 3: Calculation
$dT_B = (\frac{C_p}{C_v}) dT_A$.
Given $\gamma = C_p/C_v$, we get $dT_B = \gamma dT_A$.
Step 4: Conclusion
Hence, the rise in temperature in cylinder B is $\gamma dT_{A}$. Final Answer: (C)
$Q = nC_p dT_A$ (for A) and $Q = nC_v dT_B$ (for B).
Step 2: Analysis
Since $Q$ is the same: $nC_p dT_A = nC_v dT_B$.
Step 3: Calculation
$dT_B = (\frac{C_p}{C_v}) dT_A$.
Given $\gamma = C_p/C_v$, we get $dT_B = \gamma dT_A$.
Step 4: Conclusion
Hence, the rise in temperature in cylinder B is $\gamma dT_{A}$. Final Answer: (C)
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