In adiabatic conditions, 2 mole of CO_2 gas at 300 K is expanded...

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}