A gas is expanded from an initial state to a final state along a path that consists of (a) an isothermal…
A gas is expanded from an initial state to a final state along a path that consists of (a) an isothermal expansion doing $40 \mathrm{~J}$ work, (b) an adiabatic expansion doing $\mathrm{W}$ work, (c) an isothermal expansion doing $30 \mathrm{~J}$ work. If the total change in the internal energy of the gas is $-20 \mathrm{~J}$, the work done by the gas during the adiabatic expansion $\mathrm{W}=$
$50 \mathrm{~J}$
$90 \mathrm{~J}$
$70 \mathrm{~J}$
$20 \mathrm{~J}$
Solution
Isothermal Expansion,
In this process temperature remains constant.
$\therefore \Delta \mathrm{V}=0$
Using 1st law of thermodynamics
$\Delta \mathrm{Q}_1=\Delta \mathrm{V}_1+\Delta \mathrm{W}_1$
$\Delta \mathrm{Q}_1=\Delta \mathrm{W}_1=40 \mathrm{~J}$ ... (1)
Adiabatic expansion,
In this process heat remains constant,
$\Delta Q_2=0$
$\Delta \mathrm{U}_2=-\mathrm{W}$ ... (2)
Isothermal expansion,
$\begin{aligned} & \Delta \mathrm{W}_3=30 \mathrm{~J} \\ & \Delta \mathrm{U}_3=0\end{aligned}$
$\Delta \mathrm{Q}_3=\Delta \mathrm{W}_3=30 \mathrm{~J}$ ...(3)
Change in internal energy, $\mathrm{V}=-20 \mathrm{~J}$
It can be written as
$\begin{aligned} & \mathrm{U}_1+\mathrm{U}_2+\mathrm{U}_3=-20 \mathrm{~J} \\ & 0+\mathrm{U}_2+0=-20 \quad \mathrm{U}_2=-20 \mathrm{~J}\end{aligned}$
Now, using equation (2),
$\begin{aligned} & \mathrm{W}=-\mathrm{V}_2=20 \mathrm{~J} \\ & \mathrm{~W}=20 \mathrm{~J}\end{aligned}$