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Two moles of an ideal gas undergoes isothermal reversible compression from $20 {dm}^{3}$ to $5 {dm}^{3}$ at 300 K. The work involved in this process is (Given: log 2 = 0.301)

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$- 3.46 kJ$

$3.46 kJ$

$6.92 kJ$

$- 6.92 kJ$

answer is B.

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Detailed Solution

$w = - n R T \ln \frac{V_{2}}{V_{1}} = - (2.0 mol) (8.314 {JK}^{- 1} {mol}^{- 1}) (300 K) (2.303) \log (\frac{5 {dm}^{3}}{20 {dm}^{3}}) = 6916.6 J = 6.92 kJ$

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