The work done when 1 mole of a gas expands reversibly and isothermally from pressure of 5 atm to 1 atm at 300 K is
(Given, log 5 = 0.6989 and R = 8.314 J K^–1mol^–1)

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150 J

+4014.6 J

-4014.6 J

zero J

answer is C.

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Detailed Solution

$\begin{array}{l} W = - 2.303 nRTlog (\frac{P_{1}}{P_{2}}) \\ W = - 2.303 \times 1 mol \times 8.314 \frac{J}{Kmol} \times 300 K \times \log \frac{5 atm}{1 Atm} \end{array}$
W = –2.303 × 8.314 × 300 × 1 × log 5

W = –2.303 × 8.314 × 300 × 0.6989

W = –4014.58 J

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