Equal masses in grams of $\mathrm{H}_2, \mathrm{~N}_2, \mathrm{Cl}_2$, and $\mathrm{O}_2$, are enclosed in…

Equal masses in grams of $\mathrm{H}_2, \mathrm{~N}_2, \mathrm{Cl}_2$, and $\mathrm{O}_2$, are enclosed in cylinders separately. If these gases expand isothermally and reversibly by $10 \mathrm{dm}^3$ at $300 \mathrm{~K}$, the work done by gas is maximum for
  1. $\mathrm{H}_2$
  2. $\mathrm{N}_2$
  3. $\mathrm{Cl}_2$
  4. $\mathrm{O}_2$

Solution

$\mathrm{W}_{\max }=-2.303 \mathrm{nRT} \log _{10} \frac{\mathrm{V}_2}{\mathrm{~V}_1}$ Hence, $\mathrm{W}_{\max } \infty \mathrm{n}$ (Given: R, T, $\mathrm{V}_2, \mathrm{~V}_1=$ Constant) $\therefore \quad \mathrm{W}_{\max } \propto \frac{1}{\mathrm{M} . \mathrm{W} .}$ (Given: equal mass) Hence, lower the molecular mass, greater is the work done. Among the given, $\mathrm{H}_2$ has the lowest molecular mass.

Asked in: MHT CET 2023 (12 May Shift 2)

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