Consider the figure provided.
1 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at 18° C. If the piston is moved to position B, keeping the temperature unchanged, then 'x' L atm work is done in this reversible process.
x= ________ L atm. (nearest integer)
[Given : Absolute temperature = ° C+273.15, R=0.08206 L atm mol ⁻¹ K⁻¹]

Solution

Chemistry ma’am · 4 steps

Read the final volume carefully: B sits 90 L above A, so the gas goes from 10 L to 100 L.

1. Read the volumes. At A the gas fills V₁ = 10 L. Moving the piston to B adds another 90 L, so V₂ = 10 + 90 = 100 L. The temperature is T = 18 + 273.15 = 291.15 K.
2. Pick the formula. The expansion is isothermal and reversible, so the external pressure always matches the gas pressure nRT/V. Integrating P dV from V₁ to V₂ gives the work.

   |w| = 2.303 nRT log V₂/V₁

3. Put in numbers. log 100/10 = log 10 = 1, and nRT = 1 × 0.08206 × 291.15 = 23.89 L atm. So |w| = 2.303 × 23.89 × 1 = 55.02 L atm.
4. So. The gas does about 55 L atm of work as it expands, so w = -55 L atm in the IUPAC sign convention. The magnitude is x = 55.

The common slip: 52. Taking the final volume as 90 L instead of 10 + 90 = 100 L gives 23.89 × ln 9 ≈ 52, not 55.

The idea to keep. For reversible isothermal work, use 2.303 nRT log V₂/V₁ with T in kelvin. Always check whether a figure gives the final volume or only the change in volume.

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