91Ó°ÊÓ

The initial temperature is $20°c$, pressure at point $1$and at point $4$is $1atm$, the pressure at point $2$and at point $3$is $3atm$, the volume at point $1$and at point $2$is $50c{m}^3$and the volume at point $3$and at point $4$is $100c{m}^3$.

a. Consider the $pV$diagram of the engine given below -

From the Figure 1, the process $1$to $2$and $3$to $4$are constant volume process that isochoric process.

Conclusion :

Therefore, thelocalid="1648474263213" $pV$diagram of the heat engine is shown in Figure$1$.

The work done is the area of graph.

$$\begin{gathered} W=Areaofrec\tan gle \\ =\left(1→2\right)×\left(2→3\right) \end{gathered}$$

Substitute $\left(3atm-1atm\right)$for $\left(1→2\right)$and $\left(100c{m}^3-50c{m}^3\right)$for $\left(2→3\right)$in the above equation to find $W$.

$$\begin{gathered} W=\left(3atm-1atm\right)×\left(100c{m}^3-50c{m}^3\right) \\ =2atm\left(\frac{1.01×{10}^{5}Pa}{1atm}\right)×50c{m}^3\left(\frac{{10}^{-6}m}{1cm}\right) \\ =10.1J \end{gathered}$$

Conclusion :

Therefore, the work done per cycle is$10.1J$.

The ideal gas equation at point $1$is written as,

$$\begin{gathered} p_1{V}_1=nR{T}_1 \\ n=\frac{p_1{V}_1}{R{T}_1} \end{gathered}$$

$.P_1$is the process of gas at point $1$.

$.V_1$is the volume of gas at point $1$,

$.n$is the mole number of gas.

$.R$is the universal gas constant.

$.T_1$is the initial temperature of gas.

The ideal gas equation at point $2$is written as,

$$\begin{gathered} p_2{V}_2=nR{T}_2 \\ {T}_2=\frac{p_2{V}_2}{nR} \end{gathered}$$

$.P_2$is the pressure of gas at point $2$.

localid="1649580945108" $T_2$is the finial temperature of gas at point $2.$

Substitute $\frac{p_1{V}_1}{R{T}_1}$for $n$in the above equation.

localid="1649581059313" $$\begin{gathered} T_2=\frac{p_2{V}_2}{\frac{p_1{V}_1}{R{T}_1}R} \\ =\frac{p_2{V}_2{T}_1}{p_1{V}_1} \end{gathered}$$

Substitute $1atm$for localid="1649582671364" $P_1,3atm$for localid="1649582651192" $P_2,50c{m}^3$for localid="1649583173662" $V_{1}100c{m}^{3}for{V}_2$and $20°C$for $T_1$in above equation to find $T_2$.

$$\begin{gathered} T_2=\frac{\left(3atm\right)\left(100c{m}^3\right)\left(20°C+273\right)K}{\left(1atm\right)\left(50c{m}^3\right)} \\ =1758K \end{gathered}$$

Heat energy is given by,

$Q_H=n{C}_p∆T$ where $C_p=\frac{7}{2}R$for diatomic gas.

role="math" localid="1649583442545" $$\begin{gathered} Q_H=n{C}_p∆T=\frac{p_2{v}_2}{R{T}_2}\frac{7}{2}R(T_2-T_1) \\ =\left(\frac{3×{10}^5×100×{10}^{-6}}{1758}\right)\left(\frac{7}{2}\right)(1758-293) \\ =80J \end{gathered}$$

The thermal efficiency is given by

$\mathrm{η}=\frac{W}{Q_H}=\frac{10.1}{80}=0.129≃13\%$