Equal masses in grams of H_2, N_2, Cl_2, and | vedclass

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(A) The work done in an isothermal reversible expansion is given by the formula: $W = -2.303 nRT \log_{10} (V_2/V_1)$.Since $R$, $T$, and the change i

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$H_2$

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(A) The work done in an isothermal reversible expansion is given by the formula: $W = -2.303 nRT \log_{10} (V_2/V_1)$.Since $R$, $T$, and the change in volume $(\Delta V = 10 \ dm^3)$ are constant, the work done $W$ is directly proportional to the number of moles $n$ $(W \propto n)$.The number of moles $n$ is given by $n = 	ext{mass} / 	ext{molar mass}$.Since the mass is equal for all gases, $n \propto 1/M$, where $M$ is the molar mass.Therefore, $W \propto 1/M$.To maximize the work done, we need the gas with the smallest molar mass.The molar masses are: $H_2 = 2 \ g/mol$, $N_2 = 28 \ g/mol$, $Cl_2 = 71 \ g/mol$, and $O_2 = 32 \ g/mol$.Since $H_2$ has the lowest molar mass, the work done is maximum for $H_2$.

Equal masses in grams of $H_2$, $N_2$, $Cl_2$, and $O_2$ are enclosed in cylinders separately. If these gases expand isothermally and reversibly by $10 \ dm^3$ at $300 \ K$, the work done by the gas is maximum for:

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