Work Done In Expansion

Question

One mole of an ideal gas expands isothermally and reversibly at 300 K from a volume of 10 L to 50 L. What is the work done by the gas? (R = 8.314 J/mol·K)

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For a reversible isothermal expansion of an ideal gas, the work done by the gas is given by $w = -nRT\ln(V_f/V_i)$. This formula arises from integrating $p\,dV$ along a reversible path where $p = nRT/V$. Here $n = 1$ mol, $R = 8.314$ J/mol·K, $T = 300$ K, $V_i = 10$ L, $V_f = 50$ L. $w = -(1)(8.314)(300)\ln(50/10) = -2494.2 \times \ln(5) = -2494.2 \times 1.6094 = -4014$ J. The negative sign indicates the gas does work on the surroundings (energy leaves the system as work). Option -3988 J results from using $\log_{10}$ instead of natural logarithm without the necessary conversion factor. Option +4014 J has the wrong sign — work done BY the gas in expansion should be negative for the system. Option -2490 J omits the logarithm and uses a simple ratio incorrectly. This is a standard NCERT reversible work calculation. Plausibility check: for expansion from 10 L to 50 L at moderate temperature, a work magnitude near 4 kJ is physically reasonable for one mole of ideal gas.

Work Done In Expansion

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