JEE Main 2025 Apr 3 Shift 1, Physics Q12: Thermodynamic Processes

JEE Main2025Apr 3, Shift 1Physics

A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of 800 cm $^{3}$ and temperature 27°C. The change in temperature when the gas is adiabatically compressed to 200 cm $^{3}$ is :

(Take $γ = 1.5$ : $γ$ is the ratio of specific heats at constant pressure and at constant volume)

A
327 K
B
600 K
C
522 K
D
300 K

Solution

Adiabatic process: $T V^{γ - 1} = const .$

With $γ - 1 = 0.5$ , $T_{1} = 300$ K, $V_{1} = 800$ , $V_{2} = 200$ :

$T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} = 300 (\frac{800}{200})^{0.5} = 3004 = 600 K .$

The change in temperature: $Δ T = 600 - 300 = 300 K .$ The question asks for the change, so the answer is 300 K.

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JEE Main 2025 Apr 3 Shift 1, Physics Q12: Thermodynamic Processes

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