Question

A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of the gas increases from $V$ to $32\,V$, the efficiency of the engine is
A) $0.25$
B) $0.5$
C) $0.75$
D) $0.99$

Answer 1

In the question, during the adiabatic expansion in part of the cycle volume of the gas is increasing so the heat transfer is zero. So, we take the adiabatic expansion value as the constant. By using the equation of the efficiency of the engine, the engine efficiency has been calculated.
Formula used
The expression for finding the efficiency of the engine cycle is
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$
Where,
${T_1}$be the initial temperature and ${T_2}$ be the final temperature.

Complete step by step solution
Given that the adiabatic expansion,
So adiabatic expansion of the cycle $T{V^\gamma }^{ - 1} = {\text{constant}}$
Now the diatomic ideal gas is used in the engine as the working substance,
Volume $\left( {\dfrac{{{V_2}}}{{{V_1}}}} \right) = 32$
So diatomic gas, $\gamma = \dfrac{7}{5}$
Convert the above equation in terms of ${T_1}$, we get
${T_1} = {T_2}{\left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)^{\gamma - 1}}$
Substitute all the known values in the above equation, we get
${T_1} = {T_2} \times {\left( {30} \right)^{\dfrac{7}{5} - 1}}$
Simplify the above equation, we get
${T_1} = {T_2} \times 4$
So,
${T_1} = 4{T_2}$
Convert the above equation in terms of fractional form, we get
$\dfrac{{{T_2}}}{{{T_1}}} = \dfrac{1}{4}$
$\eta = \left( {1 - \dfrac{{{T_2}}}{{{T_1}}}} \right).............\left( 1 \right)$
Substitute the known values in the equation $\left( 1 \right)$,we get
$\eta = \left( {1 - \dfrac{1}{4}} \right)$
Simplify the above equation, we get
$\eta = \left( {\dfrac{3}{4}} \right)$
Convert the fraction value into the decimal value, we get
$\eta = 0.75$
Therefore, the efficiency of the engine is $0.75$.

Hence, from the above options, option C is correct.

Note: In the question, volume of the gas is increased due to the adiabatic expansion for the diatomic ideal gas so we calculate the temperature of the gas and then substitute those values in the equation of the efficiency.