One mole of an ideal gas is taken from a to b | vedclass

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(C) The work done in a $P-V$ graph is the area under the curve.For the dotted path $(w_d)$: The path consists of three rectangular steps.Step $1$: $P

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$2$

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(C) The work done in a $P-V$ graph is the area under the curve.For the dotted path $(w_d)$: The path consists of three rectangular steps.Step $1$: $P = 4 	ext{ atm}$,$\Delta V = (2.0 - 0.5) 	ext{ L} = 1.5 	ext{ L}$. Work = $4 	imes 1.5 = 6 	ext{ L-atm}$.Step $2$: $P = 1 	ext{ atm}$,$\Delta V = (3.0 - 2.0) 	ext{ L} = 1.0 	ext{ L}$. Work = $1 	imes 1.0 = 1 	ext{ L-atm}$.Step $3$: $P = 0.6 	ext{ atm}$ (approx),$\Delta V = (5.5 - 3.0) 	ext{ L} = 2.5 	ext{ L}$. Work = $0.6 	imes 2.5 = 1.5 	ext{ L-atm}$.Total $w_d = 6 + 1 + 1.5 = 8.5 	ext{ L-atm}$.For the solid path $(w_s)$: The path is an isothermal process. The equation is $PV = k$. At point $a$,$P=4, V=0.5$,so $k = 2 	ext{ L-atm}$.$w_s = \int_{V_a}^{V_b} P \, dV = \int_{0.5}^{5.5} \frac{k}{V} \, dV = k \ln\left(\frac{V_b}{V_a}ight) = 2 	imes 2.303 \log_{10}\left(\frac{5.5}{0.5}ight) = 4.606 \log_{10}(11) \approx 4.606 	imes 1.0414 \approx 4.797 	ext{ L-atm}$.Ratio $w_d / w_s = 8.5 / 4.797 \approx 1.77$.The integer closest to $1.77$ is $2$.

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

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