Q124 / 150 · MHT CET 2026 (19 April Shift 1) · Physics
A gas at normal temperature is suddenly compressed to one-fourth of its original volume. If $\gamma = 1.5$, then the increase in the temperature of the gas in Kelvin is ($\gamma$ is the ratio of specific heats)
- Correct answerA$273$
- B$373$
- C$473$
- D$573$
Solution
Normal temperature is $T_1 = 273$ K.
Initial volume $V_1 = V$.
Final volume $V_2 = \dfrac{V}{4}$.
Since the gas is suddenly compressed, the process is adiabatic.
The relation between temperature and volume for an adiabatic process is $T V^{\gamma - 1} = \text{constant}$.
$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$
$T_2 = T_1 \left( \dfrac{V_1}{V_2} \right)^{\gamma - 1}$
Substituting the given values:
$T_2 = 273 \left( \dfrac{V}{V/4} \right)^{1.5 - 1}$
$T_2 = 273 (4)^{0.5}$
$T_2 = 273 \times 2 = 546$ K.
The increase in temperature is $\Delta T = T_2 - T_1$.
$\Delta T = 546 - 273 = 273$ K.
Initial volume $V_1 = V$.
Final volume $V_2 = \dfrac{V}{4}$.
Since the gas is suddenly compressed, the process is adiabatic.
The relation between temperature and volume for an adiabatic process is $T V^{\gamma - 1} = \text{constant}$.
$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$
$T_2 = T_1 \left( \dfrac{V_1}{V_2} \right)^{\gamma - 1}$
Substituting the given values:
$T_2 = 273 \left( \dfrac{V}{V/4} \right)^{1.5 - 1}$
$T_2 = 273 (4)^{0.5}$
$T_2 = 273 \times 2 = 546$ K.
The increase in temperature is $\Delta T = T_2 - T_1$.
$\Delta T = 546 - 273 = 273$ K.
Hence, the correct answer is option: 'A'.