In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is $2.5 k J K^{- 1},$ find the numerical value for the enthalpy of combustion of the gas in $k J m o l^{- 1} .$

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Solution

The correct option is A 9
Number of moles of gas = $\frac{3.5}{28} = 0.125$
Heat evoled = Heat capacity $\times △ T$
= 2.5 $\times$ (298.45 - 298)
= 2.5 $\times$ 0.45 = 1.125 kJ
Heat eveloved per mole = $\frac{1.125}{0.125} = 9 k J$

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Similar questions

Q. In a constant volume calorimeter,

3.5 g

of a gas with molar mass

28

was burnt in excess oxygen at

298 K

. The temperature of the calorimeter was found to increase from

298.0 K

298.45 K

. If the heat capacity of the calorimeter is

2.5 k J / K

, the numerical value of the heat of combustion of the gas in kJ/ mol is ________.

In a constant volume calorimeter, $3.5 g$ of a gas with molecular weight 28 g/mol was burnt in excess of oxygen at $298 K$ . The temperature of the calorimeter was found to Increase from $298 K$ to $298.45 K$ due to the combustion process. Given that the heat capacity of the calorimeter is $2.5 k J / K$ . The Numerical value for the enthalpy of combustion of the gas in kJ/mol is

In a constant volume calorimeter, $3.5 g$ of a gas with molecular weight 28 was burnt in excess of oxygen at $298 k$ . The temperature of the calorimeter was found to Increase from $298 k$ to $298.45 k$ due to the combustion process. Given that the heat capacity of the calorimeter is $2.5 K J / K$ . The Numerical value for the enthalpy of combustion of the Gas in KJ /Mole is