Chemistry · Thermodynamics & Thermochemistry

JEE Main 2024 — 31 January, Shift 2 — Question 79

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 $-x \mathrm{~J}$ . The value of $x$ is $\qquad$ . (Given $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )

Answer: 28721

Numerical answer — enter this value.

It is isothermal reversible expansion, so work done negative

$\mathrm{W}=-2.303 \mathrm{nRT} \log \left(\frac{\mathrm{~V}_{2}}{\mathrm{~V}_{1}}\right)$

$=-2.303 \times 5 \times 8.314 \times 300 \log \left(\frac{100}{10}\right)$

$=-28720.713 \mathrm{~J}$

$\equiv-28721 \mathrm{~J}$

Answer key and solution verified before publishing.

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