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An ideal gas expands isothermally from 10⁻³ m³ to 10⁻² m³ at 300 K against a constant pressure of 10⁵ Nm⁻². The work done on the gas is

  1. -900 J
  2. +2700 J
  3. -900 J
  4. +900 J

Correct answer: -900 J

Solution

In an isothermal process, the work done on the gas is calculated as W = -PΔV. Here, P = 10⁵ Nm⁻² and ΔV = (10⁻² - 10⁻³) m³ = 9 × 10⁻³ m³. Substituting, W = -(10⁵)(9 × 10⁻³) = -900 J.

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