If 5 moles of an ideal gas expands from 10 L to a volume of 100 L at 300 K under isothermal and reversible condition then work, w, is -x J. The value of x is __________.
(Given R = 8.314 J K⁻¹ mol⁻¹)

Solution

Chemistry ma’am · 4 steps

Isothermal and reversible means one formula: w is minus nRT times the log of the volume ratio.

1. Pick the formula. In a reversible expansion the outside pressure stays matched to the gas pressure at every instant, so you integrate -p dV with p = nRT/V. At constant T this gives a logarithm of the volume ratio.

   w = -2.303 nRT log V₂/V₁

2. Put in numbers. nRT = 5 × 8.314 × 300 = 12471 J. The volume ratio is 100/10 = 10, and log 10 = 1.
3. Multiply. w = -2.303 × 12471 × 1 = -28720.7 J. The sign is negative because the gas does work on the surroundings as it expands.
4. So. w ≈ -28721 J, so x = 28721.

The common slip: 12471. Using log₁₀ but dropping the factor 2.303 gives 12471, not 28721. The 2.303 converts log₁₀ into the natural log.

The idea to keep. For isothermal reversible work, use w = -2.303 nRT log(V₂/V₁). A tenfold volume change makes the log equal to 1, so the answer is just 2.303 nRT.

Answer from the NTA final answer key. Solution by PrepWiser.