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(C) For an ideal gas,the work done in a process is given by $\Delta W = \int P \, dV$. Using the ideal gas equation $PV = nRT$,we have $V = \frac{nRT}

Vedclass · Accepted Answer

$RT \ln 2$

Vedclass · Answer

(C) For an ideal gas,the work done in a process is given by $\Delta W = \int P \, dV$. Using the ideal gas equation $PV = nRT$,we have $V = \frac{nRT}{P}$. Thus,$dV = \frac{nR}{P} dT$ (at constant pressure) or $dV = -\frac{nRT}{P^2} dP$ (at constant temperature).$1$. Path $AB$ (Constant pressure $P = 2P_0$): $\Delta W_{AB} = P \Delta V = nR \Delta T = 1 \cdot R \cdot (2T - T) = RT$.$2$. Path $BC$ (Constant temperature $T = 2T_0$): $\Delta W_{BC} = nRT \ln \left( \frac{V_f}{V_i} \right) = nRT \ln \left( \frac{P_i}{P_f} \right) = 1 \cdot R \cdot (2T) \ln \left( \frac{2P}{P} \right) = 2RT \ln 2$.$3$. Path $CD$ (Constant pressure $P = P_0$): $\Delta W_{CD} = P \Delta V = nR \Delta T = 1 \cdot R \cdot (T - 2T) = -RT$.$4$. Path $DA$ (Constant temperature $T = T_0$): $\Delta W_{DA} = nRT \ln \left( \frac{V_f}{V_i} \right) = nRT \ln \left( \frac{P_i}{P_f} \right) = 1 \cdot R \cdot (T) \ln \left( \frac{P}{2P} \right) = RT \ln \left( \frac{1}{2} \right) = -RT \ln 2$.Net work done $\Delta W = \Delta W_{AB} + \Delta W_{BC} + \Delta W_{CD} + \Delta W_{DA} = RT + 2RT \ln 2 - RT - RT \ln 2 = RT \ln 2$.

List-$I$	List-$II$
$A$. Zeroth law of thermodynamics	$I$. Direction of flow of heat
$B$. First law of thermodynamics	$II$. Work done is zero
$C$. Free expansion of a gas	$III$. Thermal equilibrium
$D$. Second law of thermodynamics	$IV$. Law of conservation of energy

One mole of an ideal gas having initial volume $V$,pressure $2P$ and temperature $T$ undergoes a cyclic process $ABCDA$ as shown below. The net work done in the complete cycle is:

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