Question

What is the efficiency of a Carnot engine operating between higher temperature $T_H$ and lower temperature $T_L$?

• A. T_H - T_L
• B.
• C.
• D.

Answer 1

The correct option is C

The work done by the gas over the entire cycle divided by the heat consumed from the hot reservoir is the efficiency of the Carnot engine.

Let us consider each isothermal process and find the work done in each of them.

Process a - b (isothermal)

Work done in an isothermal process is given by the formula

$W=nRT{\ell }_n\left(\frac{v_2}{v_1}\right)$

In this case

$W_1=nRT{\ell }_n\left(\frac{v_n}{v_b}\right)$. . . . (1)

Similarly the work done in process c - d is

$W_3=nRT{\ell }_n\left(\frac{v_c}{v_d}\right)$. . . . (2)

Since both of these process are isothermal

$Q_L=W_3...\left(4\right)and{Q}_H=W_1...$(5)

The Carnot engine is represented below

Applying law of conservation of energy we get

$Q_H-Q_L=W$

Dividing throughout by $Q_H$ we get

$1-\frac{Q_L}{Q_H}=\frac{W}{Q_H}=\eta$. . . . .(6)

Substituting for $Q_H$ and $Q_L$ in(6)

We get

$1-\frac{nR{T}_L{l}_n\left(\frac{v_c}{v_d}\right)}{nR{T}_H{l}_n\left(\frac{v_b}{v_a}\right)}=\eta$

$\Rightarrow \eta =\frac{1-\frac{T_L}{T_H}{l}_n\left(\frac{v_c}{v_d}\right)}{\left(\frac{v_b}{v_a}\right)}$

In process b - c since it is adiabatic the state variables T and V are connected using the relation

$T_b^{\left(\frac{1}{\gamma -1}\right)}\times V_b=T_c^{\left(\frac{1}{\gamma -1}\right)}\times V_c$ ......(8)

and similarly

$T_d^{\left(\frac{1}{\gamma -1}\right)}\times V_d=T_a^{\left(\frac{1}{\gamma -1}\right)}\times V_a$ .......(9)

From (8) and we get

$\frac{v_c}{v_d}=\frac{v_b}{v_a}$

$\Rightarrow l_n\left(\frac{v_c}{v_d}\right)l_n\left(\frac{v_b}{v_a}\right)$ .......(10)

Substituting (10) in (9) we get

$eta=1-T_L$

$\eta =1-\frac{T_L}{T_H}$