Chemistry - Thermodynamics - I Question with Solution | TestHub

Solid graph represents isothermal path,

so work done in solid graph

${\mathrm{w}}_{\mathrm{s}}=-nRT\ln \left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)=-{\mathrm{P}}_1{\mathrm{V}}_1\ln \frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$

$=-2\ln \left(\frac{5.5}{0.5}\right)=-2\ln 11\mathrm{L}-\mathrm{atm}$

$\simeq -4.8\mathrm{L}\cdot$ atm ${\mathrm{w}}_{\mathrm{d}}={\mathrm{w}}_1+{\mathrm{w}}_2+{\mathrm{w}}_3=-\left[\begin{array}{lll}4& 1.5+1& 1+0.5^{\ast }\end{array}2.5\right]\mathrm{L}-\mathrm{atm}=-8.5\mathrm{L}-\mathrm{atm}$ $\mathrm{ratio}\left(\frac{{\mathrm{W}}_{\mathrm{d}}}{{\mathrm{W}}_{\mathrm{S}}}\right)=\frac{-8.5}{-4.8}\simeq 2$

From graph, the value 0.5 don't seem appropriate but we can not choose any number between 0.5 and 1 arbitarily. Hence we decided to take the number 0.5 . Incidently, if you choose any number between 0.5 to 1 , the answer will be the same.