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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
- -900 J
- +2700 J
- -900 J
- +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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