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

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Solution

The correct option is A
9

Given, $C_{v} = 2.5 k J / K = 2500 J / K$

$Δ T = T_{2} - T_{1} = 298.45 - 298 = 0.45 K$ .

$Δ H$ , due to combustion of $3.5 g$ of gas

= $C_{v} \times Δ T = 2500 \times 0.45 = 1125 J$

$1 m o l e = 28 g$ (Mol wt of gas = 28 g/mol)
Hence, $Δ H$ due to combustion of $1 m o l$ of gas

= $\frac{1125}{3.5} \times 28 = 9000 J = 9 k J / m o l$

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Similar questions

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

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 ________.

Q. In a constant-volume calorimeter,

3.5 g

of 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

298.45 K

due to the combustion process. Given that the heat capacity of the calorimeter is

2.5 k J K^{- 1}

, the numerical value for the enthalpy of combustion of the gas in

k J m o l^{- 1}

In a constant volume calorimeter 5g of a gas with molecular weight 40 was burnt in excess of oxygen at 298K. The temperature of the calorimeter was found to increase from 298K to 298.75K due to combustion process. Given that the heat capacity of the calorimeter is 2.5kJK^-1 , the numerical value for the change in internal energy of combustion of the gas in kJ /mol is?

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The correct option is A 9 Given, Cv = 2.5 kJ/K = 2500 J/K ΔT = T2 − T1 = 298.45 − 298 = 0.45 K. ΔH, due to combustion of 3.5 g of gas = Cv × Δ T = 2500 × 0.45 = 1125 J 1 mole = 28 g (Mol wt of gas = 28 g/mol) Hence, ΔH due to combustion of 1 mol of gas = 11253.5 × 28 = 9000 J = 9 kJ/mol