JEE Chemistry 2023 Question with Solution

Given: Initial volume is $2 \, \text{liters}$ and the gas expands isothermally into vacuum to a final volume of $6 \, \text{liters}$.

Find: Change in internal energy, $\Delta U$.

For an ideal gas undergoing an isothermal process, internal energy depends only on temperature:

$$\Delta U = f(T)$$

Since the process is isothermal, the temperature remains constant. Therefore,

$$\Delta U = 0$$

Hence, the change in internal energy is $0 \, \text{J}$. The numerical answer is 0.

Given: The gas is ideal and the expansion is isothermal.

Find: Whether volume change affects $\Delta U$.

For an ideal gas, internal energy is a function only of temperature, not of volume. Even though the gas expands from $2 \, \text{liters}$ to $6 \, \text{liters}$, the temperature does not change in an isothermal process.

Therefore,

$$\Delta U = 0$$

So the required nearest integer is 0.