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In adiabatic conditions, 2 mole of $\mathrm{CO}_2$ gas at 300 K is expanded such that its volume becomes 27 times. Then, the work done is $\left(\mathrm{C}_{\mathrm{V}}=6 \mathrm{cal} \mathrm{mol}^{-1}\right.$ and $\left.\gamma=1.33\right)$
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$1200 cal$
Given, $\mathrm{T}_1=300 \mathrm{~K}, \mathrm{~V}_2=27 \mathrm{~V}_1, \mathrm{n}=1 \mathrm{~mol}$
$C_V=6 \mathrm{cal} \mathrm{mol}^{-1}, \gamma=1.33$.
In adiabatic conditions
$\frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1}$
or, $\frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{1}{27}\right)^{1.33-1}$
$=\left(\frac{1}{27}\right)^{0.33}=\left(\frac{1}{27}\right)^{1 / 3}=\frac{1}{3}$
or, $\mathrm{T}_2=300 \times \frac{1}{3}=100 \mathrm{~K}$
Thus, $\mathrm{T}_2 \lt \mathrm{T}_1$ hence, cooling takes place due to expansion under adiabatic condition.
$\Delta \mathrm{E}=\mathrm{q}+\mathrm{W}$
$\Delta \mathrm{E} \neq \mathrm{W} \quad(\mathrm{q}=0$ for adiabatic change $)$
$\Delta \mathrm{E}=-$ ve because gas expands.
Then, $\quad \mathrm{W}=-\Delta \mathrm{E}=-\mathrm{C}_{\mathrm{V}}\left(\mathrm{T}_2-\mathrm{T}_1\right)$
$=-6(100-300)=1200 \mathrm{cal}$
$C_V=6 \mathrm{cal} \mathrm{mol}^{-1}, \gamma=1.33$.
In adiabatic conditions
$\frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1}$
or, $\frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{1}{27}\right)^{1.33-1}$
$=\left(\frac{1}{27}\right)^{0.33}=\left(\frac{1}{27}\right)^{1 / 3}=\frac{1}{3}$
or, $\mathrm{T}_2=300 \times \frac{1}{3}=100 \mathrm{~K}$
Thus, $\mathrm{T}_2 \lt \mathrm{T}_1$ hence, cooling takes place due to expansion under adiabatic condition.
$\Delta \mathrm{E}=\mathrm{q}+\mathrm{W}$
$\Delta \mathrm{E} \neq \mathrm{W} \quad(\mathrm{q}=0$ for adiabatic change $)$
$\Delta \mathrm{E}=-$ ve because gas expands.
Then, $\quad \mathrm{W}=-\Delta \mathrm{E}=-\mathrm{C}_{\mathrm{V}}\left(\mathrm{T}_2-\mathrm{T}_1\right)$
$=-6(100-300)=1200 \mathrm{cal}$
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