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A mono atomic gas is supplied the heat Q very slowly keeping the pressure constant. The work done by the gas will be

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$\frac{3 Q}{5}$

$\frac{2 Q}{5}$

$\frac{2 Q}{3}$

$\frac{Q}{5}$

answer is C.

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Detailed Solution

$Δ Q = Δ U + Δ W \Rightarrow Δ W = {(Δ Q)}_{P} - Δ U = {(Δ Q)}_{P} [1 - \frac{Δ U}{{(Δ Q)}_{P}}]$

$= {(Δ Q)}_{P} [1 - \frac{C_{V}}{C_{P}}] = Q \times [1 - \frac{3}{5}] = \frac{2}{5} Q$

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