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

Question

Quantrex Academy · Accepted Answer

Correct answer: 2

Solid line represents reversible isothermal process. 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

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

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 \times1.5+1 \times₁⁺\frac{2}{3} \times2.5\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

One mole of an ideal gas is taken from a to b along two paths...

Correct answer

Step-by-step solution

More from Thermodynamics (C)