A monoatomic ideal gas of two moles is taken through a cyclic process starting from A as shown in the figure. The volume ratio are $\frac{V_B}{V_A}=2$ and  $\frac{V_D}{V_A}=4$. If the temperature  $T_A$ at  A is  ${27}^{o}C$ then

Given,
Number of moles,  $n=2$
$C_V=\frac{3}{2}R$ and  $C_p=\frac{5}{2}R$ (monoatomic)

$T_A={27}^{o}C=300K$

Let  $V_A=V_o$ then  $V_B=2V_o$  and   $V_D=V_C=4V_o$
a) Process  $A\to B$
 $V\alpha T$     $\Rightarrow \frac{T_B}{T_A}=\frac{V_B}{V_A}$

$\therefore T_B=T_A\left(\frac{V_B}{V_A}\right)=\left(300\right)\left(2\right)=600K$

$\therefore T_B=600K$

b) Process  $A\to B$

$V\alpha T$

$\Rightarrow p= cons\tan t$

$\therefore Q_{AB}=n{C}_{p}dT=n{C}_p(T_B-T_A)=\left(2\right)\left(\frac{5}{2}R\right)(600-300)$

$Q_{AB}=1500R \left(absorbed\right)$

Process  $B\to C$
T = constant
$\therefore dU=0$

$\therefore Q_{BC}=W_{BC}=nR{T}_B\ln \left(\frac{V_C}{V_B}\right)=\left(2\right)\left(R\right)\left(600\right)\ln \left(\frac{4V_0}{2V_0}\right)$

$=\left(1200R\right)\ln \left(2\right)=\left(1200R\right)\left(0.693\right)$

or  $Q_{BC}=831.6R$ (absorbed)
Process  $C\to D$  V = constant
$\therefore Q_{CD}=n{C}_{V}dT=n{C}_V(T_D-T_C)=n\left(\frac{3}{2}R\right)(T_A-T_B)$

$(T_D=T_{A}and{T}_C=T_B)=\left(2\right)\left(\frac{3}{2}R\right)(300-600)$

$Q_{CD}=-900R$ (released)
Process $D\to A$  T = constant
$\Rightarrow \Delta U=0$

$\therefore Q_{DA}=W_{DA}=nR{T}_D\ln \left(\frac{V_A}{V_D}\right)$

$=\left(2\right)\left(R\right)\left(300\right)\ln \left(\frac{V_0}{4V_0}\right)=600R\ln \left(\frac{1}{4}\right)$

$Q_{DA}=-831.6R \left(released\right)$

(c) In the complete cycle  $\Delta U=0$
Therefore, from conservation of energy
$$\begin{gathered} W_{net}=Q_{AB}+Q_{BC}+Q_{CD}+Q_{DA} \\ {W}_{net}=1500R+8316R-900R-831.6R or W_{net}=W_{total}=600R \end{gathered}$$