Question

Calculate the work done by an external agent in compressing 1 mol of oxygen from a volume of 24 L and 1.32 atm pressure to 16 L at the same temperature.
1. 1.1 × 10³ J
2. 1.79 × 10³ J
3. 2.29 × 10³ J
4. 2.5 × 10³ J

Answer 1

Correct Answer - Option 1 : 1.1 × 10³ J

As the temperature remains constant, so this is isothermal process.

Work done in an isothermal process is given as W = -P₁ V₁ ln V₂/ V₁

W = - (1.32 × 10⁵ × 24 × 10^-3) ln (16 /22.4 L)

= 1.06 × 10³ J