Question

The average degree of freedom per molecule of a gas is $6$. The gas performs $25\text{J}$ work, while expanding at constant pressure. The heat absorbed by the gas is (in $\text{J})$

Answer 1

For constant pressure process,
Work done $\left(W\right)=nR\Delta T$
$\therefore \Delta T=\left(\frac{25}{nR}\right)$
Now, for above process $f=6$,
So, $C_P=(1+\frac{f}{2})R=4R$
$\therefore$ Heat absorbed $=n{C}_P\Delta T=n\times 4R\times \frac{25}{nR}$
$=100\text{J}$