Question

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 \(\left(\gamma=\frac{C_p}{C_v}\right)\)

• A. \(\frac{\mathrm{dT}_{\mathrm{A}}}{2}\)
• B. \(\frac{\mathrm{dT}_{\mathrm{A}}}{\gamma}\)
• C. \(\gamma dT_{A}\)
• D. \(2\mathrm{dT}_{\mathrm{A}}\)

Hint:

The free piston gives a constant-pressure process and the fixed piston a constant-volume process.

Answer 1

The Correct Option is C

Step 1: Reasoning
At constant volume no work is done, so all the heat raises the internal energy. At constant pressure part of the heat goes into work, so the temperature rise is smaller.

Step 2: Formula
$dT_B=\dfrac{Q}{nC_v}$ and $dT_A=\dfrac{Q}{nC_p}$.

Step 3: Ratio
$\dfrac{dT_B}{dT_A}=\dfrac{C_p}{C_v}=\gamma$.

Final Answer:
Hence $dT_B=\gamma\,dT_A$. \[ \boxed{\text{(C) }\gamma\,dT_{A}} \]