Question

One mole of an ideal gas expands isothermally and reversibly from $10 \mathrm{dm}^{3}$ to $20 \mathrm{dm}^{3}$ at $300 \mathrm{~K} . \Delta \mathrm{U}$, q and work done in the process respectively are : Given : $\mathrm{R}=8.3 \mathrm{JK}^{-1}$ and $\mathrm{mol}^{-1}$ In $10=2.3$ $\log 2=0.30$ $\log 3=0.48$

• A. $0,21.84 \mathrm{~kJ},-1.26 \mathrm{~kJ}$
• B. $0,-17.18 \mathrm{~kJ}, 1.718 \mathrm{~J}$
• C. $0,21.84 \mathrm{~kJ}, 21,84 \mathrm{~kJ}$
• D. $0,178 \mathrm{~kJ},-1.718 \mathrm{~kJ}$

Hint:

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

Answer 1

The Correct Option is D

1. Given: - Isothermal expansion from $10 \mathrm{dm}^{3}$ to $20 \mathrm{dm}^{3}$ at $300 \mathrm{~K}$. - $\mathrm{R} = 8.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$.
2. Calculate the work done (w): \[ w = -nRT \ln \frac{V_2}{V_1} \] \[ w = -8.3 \times 300 \times \ln \left( \frac{20}{10} \right) \] \[ w = -1.718 \mathrm{~kJ} \]
3. Calculate the heat transferred (q): \[ q = -w = 1.718 \mathrm{~kJ} \]
4. Calculate the change in internal energy ($\Delta U$): \[ \Delta U = 0 \quad (\text{since} \Delta T = 0) \] Therefore, the correct answer is (4) $0,178 \mathrm{~kJ},-1.718 \mathrm{~kJ}$.