When 5 moles of He gas expand isothermally and reversibly at 300 K from 10 litre to 20 litre, the magnitude of the maximum work obtained is __________ J. [nearest integer] (Given : R = 8.3 J K⁻¹ mol⁻¹ and log 2 = 0.3010)

Solution

Chemistry ma’am · 4 steps

Reversible and isothermal means maximum work: use nRT ln(V₂/V₁) and turn ln into 2.303 log.

1. Pick the formula. For a reversible isothermal expansion of an ideal gas, the work done by the gas is the maximum possible. Its magnitude depends only on n, T and the volume ratio.

   |Wₘₐₓ| = 2.303 nRT log V₂/V₁

2. Put in numbers. nRT = 5 × 8.3 × 300 = 12450 J. The volume ratio is 20/10 = 2, so log 2 = 0.3010.
3. Multiply out. |Wₘₐₓ| = 2.303 × 12450 × 0.3010 = 28672.35 × 0.3010 = 8630.4 J.
4. So. The magnitude of the maximum work is about 8630 J. The sign is negative by the IUPAC convention, because the gas does the work.

The common slip: 3747. Using log 2 = 0.3010 in place of ln 2 without the factor 2.303 gives about 3747 J, not 8630.

The idea to keep. For a reversible isothermal ideal-gas process, |W| = 2.303 nRT log(V₂/V₁). If the data give log values, never forget the 2.303 that converts ln to log.

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