Question

Consider the cyclic process $ABCA$, shown in Fig. performed on a sample of $2.0$ mole of an ideal gas. $A$ total of $1200 \,J$ of heat is withdrawn from the sample in the process. Find the work done (in joule) by the gas during the part $BC$.

Answer 1

In the cyclic process $\Delta U = 0$
From first law of thermodynamics for the cyclic process $Q= \Delta U +W$
$\therefore W = Q -A \Delta U = -1200 -0$
$ = -1200 J$

From $C$ to $A, \Delta V= 0 \therefore W_{CA} = 0$
For the whole cycle $W_{AB} + W_{BC} + W_{CA} = W = -1200$
As $W_{CA} =0, \therefore W_{AB}+ W_{BC} = - 1200J$....(i)
Work done from $A$ to $B$:
In the process $V\propto T$, so pressure remains constant. We know that $PV= nRT$
or $P\Delta V= nR\Delta T$
$\therefore W_{AB} = P\Delta V=nR\Delta T =2 \times 8.31\times \left(500 -300\right)$
$ = 3324J$
Substituting this value in equation (i), we get
$3324 +W_{BC} = -1200$
$\therefore W_{BC} = - 4524J$