One mole of an ideal gas is taken from $a$ to $b$ along two paths denoted by the solid and the dashe

Question

One mole of an ideal gas is taken from $a$ to $b$ along two paths denoted by the solid and the dashed lines as shown in the graph below.\ If the work done along the solid line path is $w_{s}$ and that along the dotted line path is $w_{d}$, then the integer closest to the ratio $w_{d} / w_{s}$ is\

Professor PC Thomas Classes · Answer

(2): Solid line path work done $\left(w_{s}ight)$ is isothermal because $P V$ is constant (Boyle's law) and dashed line (horizontal) path work done $w_{d}$ is isobaric. Work done in vertical line is zero as $$ \Delta V=0 $$ Total work done on solid line path $\left(w_{s}ight)$ $=2.303 n R T \log \frac{V_{2}}{V_{1}}=2.303 P V \log \frac{V_{2}}{V_{1}}$ $=2.303 	imes 4 	imes 0.5 \log \frac{5.5}{0.5}=4.8 \mathrm{~L} \mathrm{~atm}$\ Total work done on dash line path $\left(w_{d}ight)$ $$ =P \Delta V $$ $=4 	imes(2-0.5)+1(3-2)+0.5(5.5-3)$\ $=6+1+1.25=8.25$\ So, $\frac{w_{d}}{w_{s}}=\frac{8.25}{4.8} \approx 2$\