JEE Physics 2025 Question with Solution

Given: Initial temperature, $T_1 = 0^\circ \text{C} = 273 \, \text{K}$; final volume, $V_2 = \frac{V_1}{4}$; ratio of specific heats, $\gamma = \frac{C_p}{C_v} = \frac{3}{2}$.

Find: Change in temperature, $\Delta T$, during the sudden compression.

For a sudden adiabatic compression,

$$T_2 V_2^{\gamma - 1} = T_1 V_1^{\gamma - 1}$$

Therefore,

$$\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1}$$

Substituting the given values,

$$\frac{T_2}{273} = \left(\frac{V_1}{V_1/4}\right)^{\frac{3}{2} - 1}$$ $$\frac{T_2}{273} = 4^{\frac{1}{2}} = 2$$

So,

$$T_2 = 2 \times 273 = 546 \, \text{K}$$

Now the temperature change is,

$$\Delta T = T_2 - T_1 = 546 - 273 = 273 \, \text{K}$$

Therefore, the temperature increases by $273 \, \text{K}$.

Given: $T_1 = 273 \, \text{K}$, $V_2 = \frac{V_1}{4}$, and $\gamma = \frac{3}{2}$.

Find: The value of $\Delta T$.

For an adiabatic process,

$$T V^{\gamma - 1} = \text{constant}$$

Hence,

$$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$$

Rearranging,

$$T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1}$$

Using $V_2 = \frac{V_1}{4}$,

$$\frac{V_1}{V_2} = \frac{V_1}{V_1/4} = 4$$

Also,

$$\gamma - 1 = \frac{3}{2} - 1 = \frac{1}{2}$$

So,

$$T_2 = 273 \times 4^{1/2} = 273 \times 2 = 546 \, \text{K}$$

Now,

$$\Delta T = 546 - 273 = 273 \, \text{K}$$

Therefore, the required change in temperature is $273 \, \text{K}$.