Question

Four identical samples of an ideal gas are taken along the paths $A,B,C$ and $D$ from state $1$ to state $2$ as shown in the $PV$ indicator diagram. Let $Q,W$ and $\Delta U$ represent the heat supplied, work done and change in internal energy of the gas respectively. Choose the correct option among the following :

• A. $Q_A-Q_D=W_A-W_D$
• B. $Q_B-W_B>Q_C-W_C$
• C. $W_A<W_B<W_C<W_D$
• D. $Q_A<Q_B<Q_C<Q_D$

Answer 1

The correct option is A $Q_A-Q_D=W_A-W_D$

We know that area under $P-V$ curve gives the work done during the process.
Work done by the gas in all four processes is positive and is in the order
$W_A>W_B>W_C>W_D.....\left(1\right)$
Hence, option (c) is wrong.
The change in internal energy $\Delta U$ is the same for all processes as initial and final states for all processes are the same.
Using first law of thermodynamics for each process $A,B,C$ and $D$, we get
$Q_A=\Delta U+W_A......\left(2\right)$
$Q_B=\Delta U+W_B......\left(3\right)$
$Q_C=\Delta U+W_C......\left(4\right)$
$Q_D=\Delta U+W_D......\left(5\right)$
Hence, from $\left(1\right)$ we can say that, $Q_A>Q_B>Q_C>Q_D$
So, option (d) is also wrong.

From equation $\left(3\right)$ and $\left(4\right)$,
$Q_B-Q_C=W_B-W_C$
$\Rightarrow Q_B-W_B=Q_C-W_C$
i.e option (b) is wrong.

From equation $\left(2\right)$ and $\left(5\right)$,
$Q_A-Q_D=W_A-W_D$
Thus, option (a) is the correct answer.