Question 641: Thermodynamics & KTG

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

Accepted Answer

Step 1: Recall specific heats for diatomic ideal gas
For a diatomic gas (ignoring vibrational d.o.f. at moderate temperatures):
$ C_v = \frac{5}{2} R,  \quad C_p = C_v + R = \frac{7}{2} R $
$ \gamma = \frac{C_p}{C_v} = \frac{7/2}{5/2} = \frac{7}{5} $
Step 2: Heat supplied in each case
For cylinder A (isobaric):
$ Q = n C_p dT_A $
For cylinder B (isochoric) with temperature rise $dT_B$:
$ Q = n C_v dT_B $
Since $Q$ is the same in both cases:
$ n C_p dT_A = n C_v dT_B $
Step 3: Relating temperature rises
$ dT_B = \frac{C_p}{C_v} dT_A = \gamma \, dT_A $
Final Answer:
The rise in temperature of the gas in B is
$ \boxed{\gamma \, dT_A} $
Correct option: C

Answer

$\frac{\mathrm{dT}_{\mathrm{A}}}{2}$

Answer

$\frac{\mathrm{dT}_{\mathrm{A}}}{\gamma}$

Answer

$\quad \gamma d T_A$

Answer

$\quad 2 \mathrm{dT}_{\mathrm{A}}$

Thermodynamics & KTG

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