Question

2000 mmol of an ideal gas expanded isothermally and reversibly from 20 L to 30 L at 300 K, calculate the work done in the process (\( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \)).

Hint:

For isothermal reversible expansion, work depends on the volume ratio; negative sign indicates work done by the system.

Answer 1

For isothermal reversible expansion, work done \( W = -n R T \ln \left( \frac{V_f}{V_i} \right) \).
\( n = 2000 \, \text{mmol} = 2 \, \text{mol} \), \( T = 300 \, \text{K} \), \( V_i = 20 \, \text{L} \), \( V_f = 30 \, \text{L} \), \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \).
\[ W = -2 \cdot 8.314 \cdot 300 \cdot \ln \left( \frac{30}{20} \right) = -2 \cdot 8.314 \cdot 300 \cdot \ln (1.5). \] \[ \ln (1.5) \approx 0.405465, \quad W \approx -2 \cdot 8.314 \cdot 300 \cdot 0.405465 \approx -2022.77 \, \text{J}. \] Answer: Work done \( \approx -2022.77 \, \text{J} \).