Question

Consider two processes on a system as shown in figure.
Figure
The volumes in the initial states are the same in the two processes and the volumes in the final states are also the same. Let ∆W1 and ∆W2 be the work done by the system in the processes A and B respectively.
(a) ∆W1 > ∆W2.
(b) ∆W1 = ∆W2.
(c) ∆W1 < ∆W2.
(d) Nothing can be said about the relation between ∆W1 and ∆W2.

Answer 1

Answer:

Correct answer is C.

Explanation:

Work done by the system,

ΔW=PΔV

Here,

P = Pressure in the process

ΔV = Change in volume during the process

Answer 2

option (c) is correct option

Explanation:

We know that

$\\ \\\\\Delta \mathrm{W}=\mathrm{P} \Delta \mathrm{V} \\ \Delta \mathrm{W} is Work done by the system,$

P is the Pressure in the process

∆V is  Change in volume during the cycle

Let $\mathrm{v}_{\mathrm{i}} \text { and } \mathrm{v}_{\mathrm{f}}$ be the volumes for processes A and B respectively in the initial states and final states. Then

$\begin{aligned}&\Delta W_{1}=P_{1} \Delta V_{1}\\&\Delta W_{2}=P_{2} \Delta V_{2}\end{aligned}$

$\Delta V_{2}=\Delta V_{1}, \quad\left[\left(V_{f 1}-V_{i 1}\right)=\left(V_{f 2}-V_{i 2}\right)\right]$

$\begin{aligned}&\frac{\Delta W_{1}}{\Delta W_{2}}=\frac{P_{1}}{P_{2}}\\&\Delta W_{1}<\Delta W_{2} \quad\left[P_{2}>P_{1}\right]\end{aligned}$