JEE Chemistry 2025 Question with Solution

Given: One mole of an ideal gas expands isothermally and reversibly from $10 \, \mathrm{dm}^{3}$ to $20 \, \mathrm{dm}^{3}$ at $300 \, \mathrm{K}$.

Find: $\Delta U$, $q$ and work done.

For an isothermal process involving an ideal gas, the change in internal energy is zero.

For reversible isothermal expansion, the work done is

Substituting the given values,

Using $\ln 2 \approx 0.693$,

For an isothermal process of an ideal gas,

So,

The listed option writes the heat value as $178 \, \mathrm{kJ}$, but the solution working gives approximately $1.718 \, \mathrm{kJ}$. The intended matching option is D.

Therefore, the correct option is D.

Given: Isothermal expansion from $10 \, \mathrm{dm}^{3}$ to $20 \, \mathrm{dm}^{3}$ at $300 \, \mathrm{K}$, with $n=1$ and $R=8.3 \, \mathrm{J K}^{-1} \, \mathrm{mol}^{-1}$.

Find: $\Delta U$, $q$ and $w$.

1. Since temperature remains constant for an ideal gas,

2. Use the reversible isothermal work expression,

3. Put $V_2=20 \, \mathrm{dm}^{3}$ and $V_1=10 \, \mathrm{dm}^{3}$,

4. Evaluate,

5. For isothermal expansion of an ideal gas,

Hence,

Therefore, the correct option is D.