In a thermodynamic process the pressure of a fixed mass of a gas is changed in such a manner that the gas…
In a thermodynamic process the pressure of a fixed mass of a gas is changed in such a manner that the gas released $30 \mathrm{~J}$ of heat and $18 \mathrm{~J}$ of work was done on the gas. If the initial internal energy of the gas was $60 \mathrm{~J}$, the final internal energy will be
$32 \mathrm{~J}$
$48 \mathrm{~J}$
$72 \mathrm{~J}$
$96 \mathrm{~J}$
Solution
Given that, heat released, $\Delta Q=-30 \mathrm{~J}$
Work done on the gas, $\Delta W=-18 \mathrm{~J}$
Change in internal energy $=\Delta U$
Initial internal energy, $U_i=60 \mathrm{~J}$
Let final internal energy, $U_f=U$
By using first law of thermodynamics,
$
\Delta Q=\Delta W+\Delta U
$
Substituting the values, we get
$
\begin{aligned}
-30 & =-18+\left(U_f-U_i\right) \\
-30+18 & =U-60 \Rightarrow U=48 \mathrm{~J}
\end{aligned}
$
Hence, the final internal energy of gas is $48 \mathrm{~J}$