Question

Figure shows two processes A and B on a system. Let ∆Q1 and ∆Q2 be the heat given to the system in processes A and B respectively. Then
(a) ∆Q1 > ∆Q2 (b) ∆Q1 = ∆Q2 (c) ∆Q1 < ∆Q2 (d) ∆Q1 ≤ ∆Q2.
Figure

Answer 1

∆Q1 > ∆Q2.

Explanation:

Both the processes A and B have common initial and final points. So they have the same change in internal energy, ∆U.

In the PV diagram, the area under the curve represents the work done on the system, ∆W.

Area under curve A > area under curve B

∆W1 > ∆W2.

∆Q1 = ∆U + ∆W1

∆Q2 = ∆U + ∆W2

But ∆W1 > ∆W2, so ∆Q1 > ∆Q2.

∆Q1 and ∆Q2 denote the heat given to the system in processes A and B respectively.

Answer 2

option (a) is correct

Explanation:

All the A and B processes have similar beginning and stopping points. Thus, in both cases, the increase in internal energy is the same. External energy is a function of the state which is not dependent on the direction followed.

The area under the curve is the work done on the system in the P-V diagram. Since area below curve An area below curve B,  

$\begin{aligned}&\Delta \mathrm{W}_{1}>\Delta \mathrm{W}_{2}\\&\Delta Q_{1}=\Delta U+\Delta W_{1}\\&\Delta Q_{2}=\Delta U+\Delta W_{2}\\&\text { But } \Delta W_{1}>\Delta W_{2}\\&\Delta Q_{1}>\Delta Q_{2}\end{aligned}$

Here, in processes A and B, respectively, $\Delta Q_{1} \text { and } \Delta Q_{2}$, denote the heat given to the device.