91影视

$\begin{array}{r}{V}_2=3V_1\\ P_2=2P_1\end{array}$

Formula used:

Efficiency of the engine can be written as:

$e=\frac{W}{Q_h}$

Where,

$W$is the work done.

$Q_h$is the total heat absorbed.

Work done can be calculated as area under the curve.

$W=\left(V_2-V_1\right)\left(P_2-P_1\right)$

Plugging in the given values in the equation,

$\begin{array}{r}W=\left(3V_1鈭�V_1\right)\left(2P_1鈭�P_1\right)\\ W=2V_1{P}_1\end{array}$

From first law of thermodynamics, heat absorbed during the process A can be calculated as:

$\begin{array}{r}{Q}_A=n{C}_V\mathrm{螖}T\\ Q_A=n\frac{5}{2}R\left(T_2鈭�T_1\right)\\ Q_A=\frac{5}{2}Rn\left(\frac{P_2{V}_1}{nR}鈭�\frac{P_1{V}_1}{nR}\right)\\ Q_A=\frac{5}{2}Rn\left(\frac{2P_1{V}_1}{nR}鈭�\frac{P_1{V}_1}{nR}\right)\\ Q_A=\frac{5}{2}{P}_1{V}_1\end{array}$

Heat absorbed during the process $\mathrm{B}$can be calculated as:

$\begin{array}{r}{Q}_B=n{C}_P\mathrm{螖}T\\ Q_B=n\frac{7}{2}R\left(T_3鈭�T_2\right)\\ Q_B=\frac{7}{2}Rn\left(\frac{P_2{V}_2}{nR}鈭�\frac{P_2{V}_1}{nR}\right)\\ Q_B=\frac{7}{2}Rn\left(\frac{2P_1脳3V_1}{nR}鈭�\frac{2P_1{V}_1}{nR}\right)\\ Q_B=\frac{7}{2}脳2P_1\left(3V_1鈭�V_1\right)\\ Q_B=\frac{7}{2}脳4P_1{V}_1\\ Q_B=\frac{28}{2}{P}_1{V}_1\end{array}$

Total heat absorbed during the complete cycle is

$\begin{array}{r}{Q}_B=n{C}_P\mathrm{螖}T\\ Q_B=n\frac{7}{2}R\left(T_3鈭�T_2\right)\\ Q_h=Q_A+Q_B\\ Q_h=\frac{5}{2}{P}_1{V}_1+\frac{28}{2}{P}_1{V}_1\\ Q_h=\frac{33}{2}{P}_1{V}_1\end{array}$

Efficiency of heat engine can be calculated as:

$\begin{array}{r}e=\frac{W}{Q_h}\\ e=\frac{2P_1{V}_1}{\frac{33}{2}{P}_1{V}_1}\\ e=0.12\\ e=12\mathrm{\%}\end{array}$

Thus, the efficiency of the given cycle is $12\%$.

Let us use the formula

$T_c$temperature of cold reservoir

$T_h$temperature of hot reservoir

Highest value of temperature can be calculated as:

Temperature doubles as the pressure double and get tripled when the volume triples.

$\begin{array}{r}{T}_1鈭�P_1{V}_1\\ T_2鈭�P_2{V}_2\\ T_2鈭�2P_1脳3V_1\\ T_2鈭�6P_1{V}_1\end{array}$

Now, efficiency of the ideal engine can be calculated as:

$\begin{array}{r}e=1鈭�\frac{T_c}{T_h}\\ e=1鈭�\frac{T_c}{6T_c}\\ e=0.83\\ e=83\mathrm{\%}\end{array}$

The efficiency of the ideal engine operating between the same temperature extremes is$e=83\%$.