Q119 / 150 · MHT CET 2026 (17 April Shift 2) · Physics
Two cylinders A and B fitted with pistons contain equal amount of an ideal diatomic gas at 303 K. The piston of cylinder A is free to move and that of cylinder B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in cylinder B is 63 K then the rise in temperature of the gas in A is
- A$35$ K
- B$40$ K
- Correct answerC$45$ K
- D$50$ K
Solution
For cylinder A, the piston is free to move, so the process is isobaric. The heat supplied is $Q_A = n C_P \Delta T_A$.
For cylinder B, the piston is fixed, so the process is isochoric. The heat supplied is $Q_B = n C_V \Delta T_B$.
Given that the same amount of heat is supplied to both cylinders, $Q_A = Q_B$.
$n C_P \Delta T_A = n C_V \Delta T_B$
$\Delta T_A = \dfrac{C_V}{C_P} \Delta T_B$
For an ideal diatomic gas, the molar heat capacity at constant volume is $C_V = \dfrac{5}{2} R$ and at constant pressure is $C_P = \dfrac{7}{2} R$.
$\dfrac{C_V}{C_P} = \dfrac{5}{7}$
Substituting $\Delta T_B = 63$ K:
$\Delta T_A = \dfrac{5}{7} \times 63 = 45$ K
Answer: $45$ K
For cylinder B, the piston is fixed, so the process is isochoric. The heat supplied is $Q_B = n C_V \Delta T_B$.
Given that the same amount of heat is supplied to both cylinders, $Q_A = Q_B$.
$n C_P \Delta T_A = n C_V \Delta T_B$
$\Delta T_A = \dfrac{C_V}{C_P} \Delta T_B$
For an ideal diatomic gas, the molar heat capacity at constant volume is $C_V = \dfrac{5}{2} R$ and at constant pressure is $C_P = \dfrac{7}{2} R$.
$\dfrac{C_V}{C_P} = \dfrac{5}{7}$
Substituting $\Delta T_B = 63$ K:
$\Delta T_A = \dfrac{5}{7} \times 63 = 45$ K
Answer: $45$ K
Hence, the correct answer is option: 'C'.