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(D) For a reversible isothermal expansion, the work done is given by the formula: $W = -nRT \ln \frac{P_1}{P_2} = -2.303 nRT \log \frac{P_1}{P_2}$.Giv

Vedclass · Accepted Answer

$-3.82 	imes 10^{-2} 	imes T$

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(D) For a reversible isothermal expansion, the work done is given by the formula: $W = -nRT \ln \frac{P_1}{P_2} = -2.303 nRT \log \frac{P_1}{P_2}$.Given: $n = 2 \ mol$, $P_1 = 10 \ atm$, $P_2 = 1 \ atm$, $R = 8.3 \ J \ K^{-1} \ mol^{-1}$.Substituting the values: $W = -2.303 	imes 2 	imes 8.3 	imes T 	imes \log \frac{10}{1}$.Since $\log 10 = 1$, we get $W = -2.303 	imes 2 	imes 8.3 	imes T 	imes 1 = -38.2298 	imes T \ J$.To convert to $kJ$, divide by $1000$: $W = -38.2298 	imes 10^{-3} 	imes T \ kJ \approx -3.82 	imes 10^{-2} 	imes T \ kJ$.

At $T(K)$, $2 \ mol$ of an ideal gas is allowed to expand reversibly and isothermally from a pressure of $10 \ atm$ to $1 \ atm$. The work done (in $kJ$) is $(R=8.3 \ J \ K^{-1} \ mol^{-1})$.

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