A gas (Molar mass = 280 g mol^–1 ) was burnt in excess O₂ in a constant volume calorimeter and during combustion the temperature of calorimeter increased from 298.0 K to 298.45 K. If the heat capacity of calorimeter is 2.5 kJ K^–1 and enthalpy of combustion of gas is 9 kJ mol^–1 then amount of gas burnt is _______ g. (Nearest Integer)

Let x g is burnt

moles$=\frac{\mathrm{x}}{280}$

heat released by $\frac{\mathrm{x}}{280}\text{ mole }=2.5\times 0.45 \mathrm{kJ}$

heat released by 1 mole$=\frac{2.5\times 0.45\times 280}{\mathrm{x}}\mathrm{kJ}$

$\begin{array}{l}\mathrm{\Delta H}=\mathrm{\Delta U}+\mathrm{\Delta ngRT}\\ \mathrm{\Delta H}\simeq \mathrm{\Delta U}\\ 9=\frac{2.5\times 280\times 0.45}{\mathrm{X}}\\ \mathrm{x}=35\mathrm{g}\end{array}$

Explanation:

Given Data:

Initial Temperature: 298.0 K

Final Temperature: 298.45 K

Heat Capacity of Calorimeter: 2.5 kJ/K

Enthalpy of Combustion: 9 kJ/mol

Molar Mass of Gas: 280 g/mol

Step 1: Calculate the Heat Absorbed by the Calorimeter

Step 2: Determine the Moles of Gas Combusted

Step 3: Calculate the Mass of Gas Combusted

Final Answer:

The amount of gas burnt is 35 grams.