6 g of graphite is burnt in a bomb calorimet | vedclass

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(A) The molar mass of graphite $(C)$ is $12 \ g \ mol^{-1}$.Number of moles of graphite burnt $= \frac{6 \ g}{12 \ g \ mol^{-1}} = 0.5 \ mol$.In a bom

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$20.667$

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(A) The molar mass of graphite $(C)$ is $12 \ g \ mol^{-1}$.Number of moles of graphite burnt $= \frac{6 \ g}{12 \ g \ mol^{-1}} = 0.5 \ mol$.In a bomb calorimeter,the volume is constant,so the heat released is equal to the change in internal energy,$\Delta U$.Given $\Delta H = -248 \ kJ \ mol^{-1}$. For a solid combustion reaction,$\Delta H \approx \Delta U$.Total heat released $(q)$ $= n 	imes \Delta U = 0.5 \ mol 	imes 248 \ kJ \ mol^{-1} = 124 \ kJ$.The temperature change $\Delta T = 31^{\circ} C - 25^{\circ} C = 6 \ K$.Using the relation $q = C_V 	imes \Delta T$,we get $C_V = \frac{q}{\Delta T}$.$C_V = \frac{124 \ kJ}{6 \ K} = 20.667 \ kJ \ K^{-1}$.

$6 \ g$ 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 $31^{\circ} C$. If $\Delta H$ of this reaction is $-248 \ kJ \ mol^{-1}$,find out $C_V$ (in $kJ \ K^{-1}$) of the bomb calorimeter.

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