Work Done In Isothermal Expansion

Question

Two moles of an ideal gas expand isothermally at 300 K from a volume V to a volume 2V in a slow reversible manner, so what is the work done by the gas? Take R = 8.3 J per mol kelvin.

RankGuru · Accepted Answer

As derived in NCERT Class 11, Chapter 12 (Thermodynamics), the work done by an ideal gas in a reversible isothermal expansion is $W = nRT \ln\!\left(\dfrac{V_f}{V_i}ight)$, where $n$ is the number of moles and the temperature is constant. Here $n = 2$, $T = 300$ K, and $\dfrac{V_f}{V_i} = \dfrac{2V}{V} = 2$, so $\ln 2 \approx 0.693$. Substituting: $W = 2 	imes 8.3 	imes 300 	imes 0.693 = 4980 	imes 0.693 \approx 3453$ J. Since temperature is constant, $\Delta \cup = 0$ and all this work is supplied by heat absorbed from the surroundings. The option 4980 J is wrong because it omits the $\ln 2$ factor, using $nRT$ alone. The option 1726 J is wrong because it uses one mole instead of two. The option 6906 J is wrong because it doubles the correct value, perhaps using $\ln 4$ or a factor error. A plausibility check on magnitude confirms the answer: $nRT = 2	imes8.3	imes300 = 4980$ J, and multiplying by $\ln 2 < 1$ must give a value below 4980 J, which the result of 3453 J indeed satisfies, so the magnitude and the logarithmic dependence are both reasonable.

Work Done In Isothermal Expansion

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