Question

The average degrees of freedom per molecules for a gas is 6. The gas performs 25 J of work when it expands at constant pressure. Find the heat absorbed by the gas.

Answer 1

Let the change in temperature of the gas be $\Delta T$ and $n$ be the number of moles of the gas.

Work of expansion $W=P(V_2-V_1)$

Using $PV=nRT$ $⟹W=nR(T_2-T_1)=nR\Delta T$

$\therefore$ $25=nR\Delta T$ .......(1)

Also degree of freedom of the gas $f=6$

Heat absorbed by the gas (at constant pressure) $Q=n{C}_p\Delta T$

$\therefore$ $Q=(\frac{f}{2}+1)nR\Delta T$ $=(\frac{6}{2}+1)nR\Delta T=4\times 25=100$ J