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Q. $2 $ mole of an ideal gas undergo isothermal and reversible expansion from $2$ litre to $20$ litre at $127^{\circ} C$ the work done by the gas is

Solution:

$ \omega =-2.303\, n\, R T \log \frac{V_{2}}{v_{1}} $
$=-2.303 \times 2+8.31 \times 400 \times \log \frac{20}{2} $
$=-15.3\, KJ $