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  • If one mole of an ideal gas at ( P1, V1) is allowed to expand reversibly and isothermally (A to B ), its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its

If one mole of an ideal gas at ( P1, V1) is allowed to expand reversibly and isothermally (A to B ), its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value (B\rightarrow C). Then it is restored to its initial state by a reversible adiabatic compression (C to A). The net work done by the gas is equal to :
 
Option: 1  -RT/(2(γ -1))

Option: 2  0 
Option: 3  RTln2
Option: 4  RT(ln2-1/2(Y-1))

Answers (1)

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\\ \mathrm{AB} \rightarrow \text{Isothermal process}\\ \mathrm{W}_{\mathrm{AB}}=\mathrm{nRT} \mathrm{In} 2=\mathrm{RT} \ln 2\\ \\ \\ \mathrm{BC} \rightarrow \text{Isochoric process}\\ \text{Work done in Isochoric process-}\\ \Delta W=P \Delta V \ \ \ \text { as } \Delta V=0 \\ \text { So } \Delta W=0 \\ \mathrm{W}_{\mathrm{BC}}=0 \\ \\ \\ \mathrm{CA} \rightarrow \text{Adiabatic process}\\ \text{Work done in adiabatic process -} \\ W=\int_{V_{i}}^{V_{f}} P d V=\frac{\left[P_{i} V_{i}-P_{f} V_{f}\right]}{(\gamma-1)}=\frac{n R\left(T_{i}-T_{f}\right)}{(\gamma-1)} \\ \\\\ \mathrm{W}_{\mathrm{CA}}=\frac{\mathrm{P}_{1} \mathrm{~V}_{1}-\(\mathrm{P}_{1} / 4) \times 2 \mathrm{~V}_{1}} {(1-\gamma)} =\frac{\mathrm{P}_{1} \mathrm{~V}_{1}} { 2(1-\gamma)} =\frac{\mathrm{RT}} { 2(1-\gamma)}\\ \\ \\ \mathrm{W}_{\mathrm{ABCA}}=\mathrm{RT} \ln 2+\frac{\mathrm{RT}} { 2(1-\gamma)} =\mathrm{RT}[\ln 2-\frac{1}{ 2(\gamma-1)}]

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