Question

The work done by 2 mole of an ideal gas at 300 K in a reversible isothermal expansion from 10 litres to 20 litres is

• A. – 3458 J
• B. + 3458 J
• C. – 1730 .4 J
• D. + 1730.4 J

Answer 1

The correct option is B + 3458 J

According to the available data :
$N=2mol,V_1=10L,V_2=20L,T=298K,R=8.134J{K}^{-1}mo{l}^{-1}\therefore \text{Amount of work done in reversible isothermal expansion}w=2.303nRTlog\frac{v_2}{v_1}=2.303\times \left(2mol\right)\times \left(8.314J{k}^{-1}mo{l}^{-1}\right)\left(300K\right)\times 0.3010=3457.9J(\because log\frac{20}{10}=0.3010)$

protip: use approximations: log 2 value and R as well!