\(6 \mathrm{~g}\) of graphite is burnt in a bomb calorimeter at \(25^{\circ} \mathrm{C}\) and \(1…
\(6 \mathrm{~g}\) of graphite is burnt in a bomb calorimeter at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. The temperature of water increased from \(25^{\circ} \mathrm{C}\) to \(31^{\circ} \mathrm{C}\). If \(\Delta H\) of this reaction is \(-248 \mathrm{~kJ}\) mol, find out \(\mathrm{C}_V\) (in \(\mathrm{kJ} \mathrm{K}^{-1}\) ) of bomb calorimeter.
20.667
41.33
1488
0.145
Solution
Weight of graphite \(=6 \mathrm{~g}\)
In bomb calorimeter volume is constant Hence, \(W=0\), (work done)
\(\Delta H=\Delta U=q \text { (Heat) }\)
Given, \(\Delta H=-248 \mathrm{~kJ} / \mathrm{mol}\)
For, \(12 \mathrm{~g}\) of graphite required energy is \(248 \mathrm{~kJ} / \mathrm{mol}\).
For, \(6 \mathrm{~g}\) of graphite, it will be \(=\frac{248 \times 6}{12}\)
\(\begin{aligned}
\begin{aligned}
\text { From, } q & =C_V \Delta T \\
\frac{248 \times 6}{12} & =C_V \times(31-25) \\
\Rightarrow \quad \quad C_V & =\frac{248}{2 \times 6}=20.6 \mathrm{~kJ} / \mathrm{K}
\end{aligned}
\end{aligned}\)