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Q.
A monatomic gas supplied the heat $Q$ very slowly keeping the pressure constant. The work done by the gas will be:
BHUBHU 2003
Solution:
Apply first law of thermodynamic In isobaric process
$Q=n C_{P} \Delta T$
Also, $d U=n C_{V} \Delta T$
$\therefore \frac{Q}{d U}=\frac{n C_{P} \Delta T}{n C_{V} \Delta T}=\frac{C_{P}}{C_{V}}=\gamma$
For monoatomic gas, $\gamma=\frac{5}{3}$
$\therefore \frac{Q}{d U}=\frac{5}{3} $
$\Rightarrow d U=\frac{3}{5} Q$
Now, form first law of thermodynamics
$Q=d U+W$
$\Rightarrow W=Q-d U $
$=Q-\frac{3}{5} Q $
$\therefore W=\frac{2}{5} Q$