Prove directly (by calculating the heat taken in and the heat expelled) that a Carnot engine using an ideal gas as the working substance has an efficiency of$\frac{1-t_c}{t_h}$

Carnot engine using an ideal gas as the working substance has the efficiency of $1-\frac{T_c}{T_h}$

The Carnot cycle begins with the isothermal expansion of 1 mol of gas, which changes its state from $(P_1,V_1,T_h)to(P_2,V_2,T_c)$.The heat absorbed $\left(Q_H\right)$by the gas from the source at constant temperature

$T_h$is given by:

$Q_h=W_1=R{T}_h{\log }_e\left(\frac{V_2}{V_1}\right)---\left(1\right)$

The Carnot cycle's second stage is the adiabatic expansion of 1 mol of gas taking its state from$\left(P_2,V_2,T_h\right)$to $\left(P_3,V_3,T_c\right)$The work done $\left(W_2\right)$by the gas is given by:

$W_2=\frac{R\left(T_h-T_c\right)}{\gamma -1}---\left(2\right)$

The Carnot cycle's third stage involves isothermal compression of 1 mol of gas taking its state from$\left(P_3,V_3,T_c\right)to\left(P_4,V_4,T_h\right)$by the gas to the sink at constant temperature $T_c$is given by

$Q_c=W_3=R{T}_c{\log }_e\left(\frac{V_3}{V_4}\right)---\left(3\right)$

The fourth stage of the Carnot cycle is adiabatic compression, which involves compressing 1 mol of gas to its original condition.

$\left(P_4,V_4,T_c\right)$to $\left(P_1,V_1,T_h\right)$The work done $\left(W_4\right)$on the gas is given by:

$W_4=\frac{R\left[T_h-T_c\right]}{\gamma -1}---\left(4\right)$

The efficiency of Carnot engine is given by

$\begin{array}{r}\eta =\frac{\text{output work}}{\text{heat supplied}}\\ \eta =\frac{Q_h-Q_c}{Q_h}=1-\frac{Q_c}{Q_h}---\left(5\right)\end{array}$

Substitute (1) and (3) in (5)

$\eta =1-\frac{R{T}_c{\log }_e\left(\frac{V_3}{V_4}\right)}{R{T}_h{\log }_e\left(\frac{V_2}{V_1}\right)}---\left(6\right)$

For an adiabatic expansion:

$T{V}^{\gamma -1}=\text{Constant}$

In the second stage, for an adiabatic expansion:

$\begin{array}{r}{T}_h{V}_2^{\gamma -1}=T_c{V}_3^{\gamma -1}\\ \frac{T_c}{T_h}=\frac{V_2^{r-1}}{{V_3}^{r-1}}={\left(\frac{V_2}{V_3}\right)}^{r-1}\\ \left(\frac{V_2}{V_3}\right)={\left(\frac{T_c}{T_h}\right)}^{\frac{1}{r-1}}------\left(7\right)\end{array}$

In the fourth stage, for an adiabatic compression:

$\begin{array}{r}{T}_c{V}_4^{\gamma -1}=T_h{V}_1^{\gamma -1}\\ \frac{T_2}{T_1}=\frac{V_1^{\gamma -1}}{V_4^{\gamma -1}}={\left(\frac{V_1}{V_4}\right)}^{\gamma -1}\\ \left(\frac{V_1}{V_4}\right)={\left(\frac{T_c}{T_h}\right)}^{\frac{1}{\gamma -1}}---\left(8\right)\end{array}$

On comparing equation (7) and (8)

$\begin{array}{r}\frac{V_1}{V_4}=\frac{V_2}{V_3}\\ \Rightarrow \frac{V_3}{V_4}=\frac{V_2}{V_1}---\left(9\right)\end{array}$

Now (9) in (6)

$\eta =1-\frac{T_c}{T_h}$

Hence proved.