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(B) For a reversible isothermal compression of an ideal gas,the work done is given by the formula: $W = -2.303 \, nRT \log \frac{P_1}{P_2}$ Given: $n

Vedclass · Accepted Answer

$3.458$

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(B) For a reversible isothermal compression of an ideal gas,the work done is given by the formula: $W = -2.303 \, nRT \log \frac{P_1}{P_2}$ Given: $n = 1 \, mol$,$T = 27 + 273 = 300 \, K$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$,$P_1 = 1 \, bar$,$P_2 = 4 \, bar$. Substituting the values: $W = -2.303 	imes 1 	imes 8.314 	imes 300 	imes \log(\frac{1}{4})$ $W = -2.303 	imes 8.314 	imes 300 	imes (-0.602)$ $W = +3.458 \, kJ$.

$1 \, mol$ of an ideal gas is compressed reversibly at a constant temperature of $27 \, ^oC$ from $1 \, bar$ to $4 \, bar$. Calculate the work done in $kJ$.

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