Q123 / 150 · MHT CET 2026 (16 April Shift 1) · Physics
An ideal gas at $27^\circ$C is compressed adiabatically to $\left(\dfrac{8}{27}\right)$ of its original volume. If $\gamma = 5/3$, then the rise in temperature of a gas is
- A$300$ K
- B$675$ K
- Correct answerC$375$ K
- D$27$ K
Solution
For an adiabatic process, the relation between temperature and volume is given by $T V^{\gamma - 1} = \text{constant}$.
$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$
Given $T_1 = 27^\circ\text{C} = 300\text{ K}$, $V_2 = \dfrac{8}{27} V_1$, and $\gamma = \dfrac{5}{3}$.
$T_2 = T_1 \left( \dfrac{V_1}{V_2} \right)^{\gamma - 1}$
$T_2 = 300 \left( \dfrac{27}{8} \right)^{\dfrac{5}{3} - 1}$
$T_2 = 300 \left( \dfrac{27}{8} \right)^{\dfrac{2}{3}}$
$T_2 = 300 \left( \dfrac{3}{2} \right)^2 = 300 \times \dfrac{9}{4} = 675\text{ K}$
The rise in temperature is $\Delta T = T_2 - T_1 = 675\text{ K} - 300\text{ K} = 375\text{ K}$.
Answer: $375$ K
$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$
Given $T_1 = 27^\circ\text{C} = 300\text{ K}$, $V_2 = \dfrac{8}{27} V_1$, and $\gamma = \dfrac{5}{3}$.
$T_2 = T_1 \left( \dfrac{V_1}{V_2} \right)^{\gamma - 1}$
$T_2 = 300 \left( \dfrac{27}{8} \right)^{\dfrac{5}{3} - 1}$
$T_2 = 300 \left( \dfrac{27}{8} \right)^{\dfrac{2}{3}}$
$T_2 = 300 \left( \dfrac{3}{2} \right)^2 = 300 \times \dfrac{9}{4} = 675\text{ K}$
The rise in temperature is $\Delta T = T_2 - T_1 = 675\text{ K} - 300\text{ K} = 375\text{ K}$.
Answer: $375$ K
Hence, the correct answer is option: 'C'.