Question

If one mole of an ideal gas at $\left(P_{1}, V_{1}\right)$ is allowed to expand reversibly and isothermally (A to $B$ ) its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value $( B \rightarrow C ) .$ Then it is restored to its initial state by a reversible adiabatic compression (C to A). The net workdone by the gas is equal to :

• A. $RT \left(\ln 2-\frac{1}{2(\gamma-1)}\right)$
• B. $-\frac{ RT }{2(\gamma-1)}$
• C. 0
• D. RT $\ln 2$

Answer 1

Answer: (A)

$A - B =$ isothermal process
$W _{ AB }= P _{1} V _{1} \ln \left[\frac{2 V _{1}}{ V _{1}}\right]= P _{1} V _{1} \ln (2)$
$B - C \rightarrow$ Isochoric process
$W _{ BC }=0$
$C - A \rightarrow$ Adiabatic process
$W _{ CA }=\frac{ P _{1} V _{1}-\frac{ P _{1}}{4} \times 2 V _{1}}{1-\gamma}$
$=\frac{ P _{1} V _{1}\left[1-\frac{1}{2}\right]}{1-\gamma}=\frac{ P _{1} V _{1}}{2(1-\gamma)}$
$W _{net} = W _{ AB }+ W _{ BC }+ W _{ CA }$
$\left\{ P _{1} V _{1}= RT \right\}$
$= P _{1} V _{1} \ln (2)+0+\frac{ P _{1} V _{1}}{2(1-\gamma)}$
$W _{ net } = RT \left[\ln (2)-\frac{1}{2(\gamma-1)}\right]$