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Q.
What is the efficiency of a carnot engine in which diatomic gas is used. And the gas adiabatically expands in volume from $V$ to $32V$ .
NTA AbhyasNTA Abhyas 2020
Solution:
For an adiabatic process, relation between $T$ and $V$ is
$\left(TV\right)^{\left(\right. \gamma - 1 \left.\right)}=$ constant
$\therefore T_{1}V_{1}^{\left(\right. \gamma - 1 \left.\right)}=T_{2}\left(V_{2}\right)^{\left(\right. \gamma - 1 \left.\right)}$
Also for a diatomis gas , $\gamma =\frac{7}{5}$ .
In an adiabatic expansion, temperature decreases. So, $T_{2}$ is sink temperature and $T_{1}$ is source temperature.
$\Rightarrow T_{1}=T_{2}\left(\frac{V_{2}}{V_{1}}\right)^{\left(\frac{7}{5} - 1\right)}=T_{2}\left(\right.32\left(\left.\right)^{\left(\frac{2}{5}\right)}\therefore given\frac{V_{2}}{V_{1}}=32$
$\Rightarrow \frac{T_{2}}{T_{1}}=\left(\frac{1}{32}\right)^{\frac{2}{5}}=\frac{1}{4}$
Efficiency of Carnot cycle is
$\Rightarrow \eta=1-\frac{T_{2}}{T_{1}}=1-\frac{1}{4}=\frac{3}{4}=0.75$