JEE MAIN - Physics (2025 - 3rd April Morning Shift - No. 1)
A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of $800 \mathrm{~cm}^3$ and temperature $27^{\circ} \mathrm{C}$. The change in temperature when the gas is adiabatically compressed to $200 \mathrm{~cm}^3$ is:
(Take $\gamma=1.5 ; \gamma$ is the ratio of specific heats at constant pressure and at constant volume)
Explanation
To determine the change in temperature when a gas is adiabatically compressed, we are given the following initial conditions:
Initial volume, $ V_1 = 800 \, \text{cm}^3 $
Final volume, $ V_2 = 200 \, \text{cm}^3 $
Initial temperature, $ T_1 = 27^\circ \text{C} = 300 \, \text{K} $
Adiabatic index (ratio of specific heats) $ \gamma = 1.5 $
In an adiabatic process, the relationship between temperature and volume is given by the equation:
$ TV^{\gamma-1} = \text{constant} $
Applying this to the initial and final states:
$ 300 \times (800)^{0.5} = T_2 \times (200)^{0.5} $
Solving for the final temperature $ T_2 $:
$ T_2 = 300 \times \left( \frac{800}{200} \right)^{0.5} = 300 \times (4)^{0.5} $
$ T_2 = 300 \times 2 = 600 \, \text{K} $
Thus, the change in temperature is:
$ \Delta T = T_2 - T_1 = 600 \, \text{K} - 300 \, \text{K} = 300 \, \text{K} $
Therefore, when the gas is adiabatically compressed, the temperature increases by 300 K.
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