Question

$6g$ of graphite is burnt in a bomb calorimeter at $25^{\circ}C$ and $1\, atm$ pressure. The temperature of water increased from $25^{\circ}C$ to $30^{\circ}C$. If $\Delta H$ of this reaction is $-248 \,kJ\, mol^{-1}$, find out $C_v$ (in $kJ\, K^{-1}$) of bomb calorimeter.

• A. 20.667
• B. 41.33
• C. 1488
• D. 0.145

Answer 1

Answer: (A)

20.667

Answer 2

Weight of graphite $=6 \,g$

In bomb calorimeter volume is constant

Hence, $W=0$, (work done)

$\Delta H=\Delta U=q$ (Heat)

Given, $\Delta H=-248 \,kJ / mol$

For, $12 \,g$ of graphite required energy is $248 \,kJ / mol$.

For, $6 \,g$ of graphite, it will be $=\frac{248 \times 6}{12}$

From, $q=C_{V} \,\Delta \,T$
$\frac{248 \times 6}{12} =C_{V} \times(31-25)$
$\Rightarrow C_{V} =\frac{248}{2 \times 6}$
$=20.6 \,kJ / K$