Question

Figure shows two processes A and B on a system. Let $\Delta Q_1$ and $\Delta Q_2$ be the heat given to the system in processes A and B respectively. Then

• A. $\Delta Q_1>\Delta Q_2$
• B. $\Delta Q_1=\Delta Q_2$
• C. $\Delta Q_2<\Delta Q_2$
• D. $\Delta Q_1<=\Delta Q_2$

Answer 1

The correct option is A $\Delta Q_1>\Delta Q_2$

Both the processes $A$ and $B$ have common initial and final points. So, change in internal energy, $\Delta U$ is same in both cases. Internal energy is the state function that does not depend on the path forward.

In the $P-V$ diagram, the area under the curve represents the work done on the system, $\Delta W$. Since area under curve $A>$ area under curve $B$

$\Delta W_1>\Delta W_2$

Now,

$\Delta Q_1=\Delta U+\Delta W_1$

$\Delta Q_2=\Delta U+\Delta W_2$

But

$\Delta W_1>\Delta W_2$

$⟹\Delta Q_1>\Delta Q_2$

Here, $\Delta Q_1$ and $\Delta Q_2$ denotes the heat given to the system in processes $A$ and $B$, receptively.