Question

Consider two processes on a system as shown in 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.

Options

• ∆W₁ > ∆W₂

• ∆W₁ = ∆W₂

• ∆W₁ < ∆W₂

• Nothing can be said about the relation between ∆W₁ and ∆W₂

Answer 1

∆W₁ < ∆W₂

 

Work done by the system, ∆W = P ∆ V

here,

P = Pressure in the process

∆V = Change in volume during the process

Let V_i and V_f​  be the volumes in the initial states and final states for processes A and B, respectively. Then,

\[\Delta W_1  =  P_1 \Delta V_1 \]

\[\Delta W_2  =  P_2 \Delta V_2 \]

But  \[\Delta V_2  = \Delta V_1 ,.............\left[ \left( V_{f_1} - V_{i_1} \right) = \left( V_{f_2} - V_{i_2} \right) \right]\]

\[ \Rightarrow \frac{\Delta W_1}{\Delta W_2} = \frac{P_1}{P_2}\]

\[ \Rightarrow \Delta W_1  < \Delta W_2..........\left[ \because P_2 > P_1 \right]\]