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A monoatomic gas supplied the heat $Q$ very slowly keeping the pressure constant. The work done by the gas will be
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Verified Answer
The correct answer is:
$\frac{2}{3} Q$
For monoatomic gas at constant pressure
$\frac{\Delta U}{Q}=\frac{1}{3}$
or $\quad \Delta U=\frac{Q}{3}$
Now applying first law of thermodynamics
$\begin{aligned} W & =\Delta Q-\Delta U \\ & =Q-\frac{Q}{3}=\frac{2 Q}{3}\end{aligned}$
$\frac{\Delta U}{Q}=\frac{1}{3}$
or $\quad \Delta U=\frac{Q}{3}$
Now applying first law of thermodynamics
$\begin{aligned} W & =\Delta Q-\Delta U \\ & =Q-\frac{Q}{3}=\frac{2 Q}{3}\end{aligned}$
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