[અચળ કદે મોલર ઉષ્માક્ષમતા $\bar{c}_{\mathrm{v}}$ છે]
\(\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}=-\mathrm{P}_{\mathrm{ext}}\left(\mathrm{V}_2-\mathrm{V}_1\right)\)
\(\mathrm{V}_2=2 \mathrm{~V}_1\)
\(\frac{\mathrm{nRT}_2}{\mathrm{P}_2}=\frac{2 \mathrm{nRT}_1}{\mathrm{P}_1}\)
\(\mathrm{P}_1=5, \mathrm{~T}_1=298\)
\(\mathrm{P}_2=\frac{5 \mathrm{~T}_2}{2 \times 298}\)
\(\mathrm{n} \frac{5}{2} \mathrm{R}\left(\mathrm{T}_2-\mathrm{T}_1\right)=-1\left(\frac{\mathrm{nRT}_2}{\mathrm{P}_1}-\frac{\mathrm{nRT}_1}{\mathrm{P}_1}\right)\)
\(\mathrm{Put}_1=298\)
\(\text { and } \mathrm{P}_2=\frac{5 \mathrm{~T}_2}{2 \times 298}\)
Solve and we get \(T_2=274.16 \mathrm{~K}\)
\(\mathrm{T}_2 \approx 274 \mathrm{~K}\)
$S{O_2} + \frac{1}{2}{O_2} \to S{O_3}$
$2 \mathrm{Fe}_{(\mathrm{s})}+\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})}, \Delta \mathrm{H}^{\mathrm{o}}=-822 \mathrm{~kJ} / \mathrm{mol}$
$\mathrm{C}_{(\mathrm{s})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{(\mathrm{g})}, \Delta \mathrm{H}^{\mathrm{o}}=-110 \mathrm{~kJ} / \mathrm{mol}$
$3\mathrm{C}_{(\mathrm{s})}+\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})} \rightarrow 2 \mathrm{Fe}_{(\mathrm{s})}+3 \mathrm{CO}_{(\mathrm{g})}$ આપેલા પ્ર્ક્રિયા માટે એન્થાલ્પી ફેરફાર__ _ _$J/mol$ છે.