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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 $\frac{W_d}{W_s}$ is
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Verified Answer
The correct answer is:
2
Solid line represents reversible isothermal process.
So, work $W_s=-4 \times 0.5 \ln \left(\frac{5.5}{0.5}\right)=-2 \ln 11 \mathrm{~L}-\mathrm{atm}$
Dotted line represents irreversible process
So, work
$$
W_d=-\left\{4 \times 1.5+1 \times 1+\frac{2}{3} \times 2.5\right\}=-\left\{6+1+\frac{5}{3}\right\} \mathrm{L}-\mathrm{atm}=-\frac{26}{3} \mathrm{~L}-\mathrm{atm}
$$
So $\quad \frac{W_d}{W_s}=\frac{26}{3 \times 2 \ln 11} \approx 2$
Energetics
Conceptual
III
So, work $W_s=-4 \times 0.5 \ln \left(\frac{5.5}{0.5}\right)=-2 \ln 11 \mathrm{~L}-\mathrm{atm}$
Dotted line represents irreversible process
So, work
$$
W_d=-\left\{4 \times 1.5+1 \times 1+\frac{2}{3} \times 2.5\right\}=-\left\{6+1+\frac{5}{3}\right\} \mathrm{L}-\mathrm{atm}=-\frac{26}{3} \mathrm{~L}-\mathrm{atm}
$$
So $\quad \frac{W_d}{W_s}=\frac{26}{3 \times 2 \ln 11} \approx 2$
Energetics
Conceptual
III
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