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(D) For an isothermal reversible expansion process,the work done is given by the formula:$W = -2.303 \ nRT \log \frac{P_1}{P_2}$Given:$n = 1 \ mol$,$R

Vedclass · Accepted Answer

$1729$

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(D) For an isothermal reversible expansion process,the work done is given by the formula:$W = -2.303 \ nRT \log \frac{P_1}{P_2}$Given:$n = 1 \ mol$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$P_1 = 210 \ kPa$,$P_2 = 105 \ kPa$Substituting the values:$W = -2.303 	imes 1 	imes 8.314 	imes 300 	imes \log \frac{210}{105}$$W = -2.303 	imes 8.314 	imes 300 	imes \log 2$Using $\log 2 \approx 0.3010$:$W = -2.303 	imes 8.314 	imes 300 	imes 0.3010$$W \approx -1729 \ J$The magnitude of work done is $|W| = 1729 \ J$.

$1 \ mol$ of an ideal gas expands isothermally and reversibly by decreasing pressure from $210 \ kPa$ to $105 \ kPa$ at $300 \ K$. What is the work done (in $J$)? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

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