Q125 / 150 · MHT CET 2025 (26 Apr Shift 2) · Physics

Two cylinders A and B fitted with piston contain equal amount of an ideal diatomic gas at temperature T K. The piston of cylinder A is free to move while that of cylinder B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise of temperature of the gas in A is $\mathrm{dT}_{\mathrm{A}}$, then the rise in temperature of the gas in B is $\left(\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}\right)$

  • A
    $\frac{\mathrm{dT}_{\mathrm{A}}}{2}$
  • B
    $\frac{\mathrm{dT}_{\mathrm{A}}}{\gamma}$
  • Correct answer
    C
    $\gamma \mathrm{dT}_{\mathrm{A}}$
  • D
    $2 \mathrm{dT}_{\mathrm{A}}$

Solution

Given: Two identical cylinders containing ideal diatomic gas at temperature $T$, one with free piston (constant pressure), the other with fixed piston (constant volume). Both receive equal heat $dQ$.

The temperature rise in cylinder A (isobaric process) satisfies $dQ = n C_p dT_A$.

For cylinder B (isochoric process), $dQ = n C_v dT_B$.

Equating and simplifying: $n C_p dT_A = n C_v dT_B$ yields $dT_B = \frac{C_p}{C_v} dT_A = \gamma dT_A$.

Result: The temperature rise in cylinder B is $\gamma$ times that in cylinder A.

Hence, the correct answer is option: 'C'.

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