Question

A diatomic ideal gas is used in a car engine as the working substance. lf during the adiabatic expansion part of the cycle, volume of the gas increases from $V$ to $32V$, the efficiency of the engine is
$\begin{aligned} & A)0.5 \\\ & B)0.75 \\\ & C)0.99 \\\ & D)0.25 \\\ \end{aligned}$

Answer 1

The solution can be obtained by considering the fact that the efficiency of an engine solely depends on the initial and final temperatures of the engine. The process considered is adiabatic process, which refers to the change that occurs within a system due to transfer of energy in the form of work. For a diatomic gas, the ratio of specific heats is equal to $\dfrac{7}{5}$.

Formula used:
$1)\eta =1-\dfrac{{{T}_{2}}}{{{T}_{1}}}$
$2)\gamma =\dfrac{7}{5}$
$3){{T}_{1}}V_{1}^{\gamma -1}={{T}_{2}}V_{2}^{\gamma -1}$

Complete answer:
We know that adiabatic process refers to the change that occurs within a system due to transfer of energy in the form of work. Mathematically, adiabatic process can be expressed as:
$T{{V}^{\gamma -1}}=const$
where
$T$ is the absolute temperature of a thermodynamic system
$V$ is its volume
$\gamma$ is the ratio of specific heats
Let this be equation 1.
If ${{T}_{1}}$ and ${{T}_{2}}$are assumed to be the initial temperature and final temperature respectively for a system undergoing adiabatic process, and if ${{V}_{1}}$ and ${{V}_{2}}$be the initial volume and final volume of the thermodynamic system respectively, then, using equation 1, we can write:
${{T}_{1}}V_{1}^{\gamma -1}={{T}_{2}}V_{2}^{\gamma -1}\Rightarrow {{T}_{1}}={{T}_{2}}{{\left( \dfrac{{{V}_{2}}}{{{V}_{1}}} \right)}^{\gamma -1}}$
Let this be equation 2.
Coming to our question, we are given that
$\left( \dfrac{{{V}_{2}}}{{{V}_{1}}} \right)=\dfrac{32V}{1V}=32$
Let this be equation 3.
Also do we know that for a diatomic gas,
$\gamma =\dfrac{7}{5}$
where
$\gamma$ is the ratio of specific heats
Let this be equation 4.
Substituting equation 3 and equation 4 in equation 2, we have
${{T}_{1}}={{T}_{2}}{{\left( \dfrac{{{V}_{2}}}{{{V}_{1}}} \right)}^{\gamma -1}}\Rightarrow \dfrac{{{T}_{1}}}{{{T}_{2}}}={{(32)}^{\dfrac{7}{5}-1}}\Rightarrow \dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{1}{4}$
Let this be equation 5.
Now, we know that efficiency of engine in which adiabatic process of diatomic gas is carried out is given by
$\eta =1-\dfrac{{{T}_{2}}}{{{T}_{1}}}$
where
$\eta$ is the efficiency of the engine
$\dfrac{{{T}_{2}}}{{{T}_{1}}}$ is the ratio of final absolute temperature to the initial absolute temperature of the gas
Let this be equation 6.
Substituting equation 5 in equation 6, we have
$\eta =1-\dfrac{{{T}_{2}}}{{{T}_{1}}}\Rightarrow \eta =\left( 1-\dfrac{1}{4} \right)=\dfrac{3}{4}=0.75$
Let this be equation 7.
Therefore, from equation 7, we can conclude that the efficiency of the engine is $0.75$.

Hence, the correct answer is option $B$.

Note:
It is to be noted that $\gamma$ is nothing but the ratio of specific heat for constant pressure $({{C}_{P}})$ to the specific heat for constant volume $({{C}_{V}})$ of the gas. Mathematically, $\gamma$ is given by $\gamma =\dfrac{{{C}_{P}}}{{{C}_{V}}}$. For a diatomic gas, it needs to be remembered that the value of $\gamma$ is equal to $\dfrac{7}{5}$.