Step-by-Step Solution:

1. Understand the Calorimetry Process

The combustion is carried out in a constant volume calorimeter (also known as a bomb calorimeter). In such a process, the heat released by the reaction ($q_{reaction}$) is absorbed by the calorimeter and its contents. The heat change measured at constant volume corresponds to the change in internal energy ($\\Delta U$), not the change in enthalpy ($\\Delta H$).

2. Calculate the Heat Absorbed by the Calorimeter ($q_{cal}$)

The heat absorbed by the calorimeter is calculated using its heat capacity ($C_{cal}$) and the change in temperature ($\\Delta T$).

Given:

• Heat capacity of the calorimeter, $C_{cal} = 2.5 \\text{ kJ K}^{-1}$
• Initial temperature, $T_1 = 298.0 \\text{ K}$
• Final temperature, $T_2 = 298.45 \\text{ K}$

First, calculate the temperature change: $\\Delta T = T_2 - T_1 = 298.45 \\text{ K} - 298.0 \\text{ K} = 0.45 \\text{ K}$

Now, calculate the heat absorbed by the calorimeter: $q_{cal} = C_{cal} \\times \\Delta T = (2.5 \\text{ kJ K}^{-1}) \\times (0.45 \\text{ K}) = 1.125 \\text{ kJ}$

3. Determine the Heat of Combustion at Constant Volume ($q_v$)

The heat released by the combustion reaction is equal in magnitude but opposite in sign to the heat absorbed by the calorimeter. $q_{reaction} = -q_{cal} = -1.125 \\text{ kJ}$

Since the process is at constant volume, this heat released is the change in internal energy for the amount of gas burned: $\\Delta U = q_v = -1.125 \\text{ kJ}$

This value is for the combustion of 3.5 g of the gas.

4. Calculate the Molar Internal Energy of Combustion ($\\Delta_c U$)

To find the molar value, we first need to calculate the number of moles ($n$) of the gas that was combusted.

Given:

• Mass of the gas, $m = 3.5 \\text{ g}$
• Molecular weight of the gas, $M = 28 \\text{ g mol}^{-1}$

Number of moles: $n = \\frac{m}{M} = \\frac{3.5 \\text{ g}}{28 \\text{ g mol}^{-1}} = 0.125 \\text{ mol}$

Now, calculate the molar internal energy of combustion ($\\Delta_c U$): $\\Delta_c U = \\frac{\\Delta U}{n} = \\frac{-1.125 \\text{ kJ}}{0.125 \\text{ mol}} = -9 \\text{ kJ mol}^{-1}$

5. Calculate the Enthalpy of Combustion ($\\Delta_c H$)

The relationship between enthalpy change ($\\Delta H$) and internal energy change ($\\Delta U$) is given by: $\\Delta H = \\Delta U + \\Delta n_g RT$ where $\\Delta n_g$ is the change in the number of moles of gas in the balanced chemical reaction, R is the ideal gas constant, and T is the temperature.

To find $\\Delta n_g$, we need the chemical formula of the gas. Gases with a molecular weight of 28 g/mol include carbon monoxide (CO) and ethene (C$_2$H$_4$).

• If the gas is CO: $CO(g) + \\frac{1}{2}O_2(g) \\rightarrow CO_2(g)$. Here, $\\Delta n_g = 1 - (1 + 0.5) = -0.5$.
• If the gas is C$_2$H$_4$: $C_2H_4(g) + 3O_2(g) \\rightarrow 2CO_2(g) + 2H_2O(l)$. Here, $\\Delta n_g = 2 - (1 + 3) = -2$.

The term $\\Delta n_g RT$ would be: $RT = (8.314 \\times 10^{-3} \\text{ kJ K}^{-1} \\text{ mol}^{-1})(298 \\text{ K}) \\approx 2.48 \\text{ kJ mol}^{-1}$.

• For CO, $\\Delta n_g RT = (-0.5)(2.48) = -1.24 \\text{ kJ mol}^{-1}$.
• For C$_2$H$_4$, $\\Delta n_g RT = (-2)(2.48) = -4.96 \\text{ kJ mol}^{-1}$.

This would make $\\Delta_c H$ different from $\\Delta_c U$. For example, if the gas was CO, $\\Delta_c H = -9 - 1.24 = -10.24 \\text{ kJ mol}^{-1}$.

However, the question asks for a single numerical (integer) value. The value calculated for the molar internal energy change, $\\Delta_c U$, is exactly -9 kJ mol$^{-1}$. Given the ambiguity of the gas's identity and the integer answer format, it's highly probable that the question intends for us to report the magnitude of the molar internal energy change, or to assume $\\Delta H \\approx \\Delta U$. This is a common simplification in introductory problems or indicates the term $\\Delta n_g RT$ is to be ignored.

Therefore, we take the numerical value of the molar internal energy of combustion.

Numerical value = $|\\Delta_c U| = |-9 \\text{ kJ mol}^{-1}| = 9$.

Final Answer The numerical value for the enthalpy of combustion of the gas is taken as the magnitude of the internal energy of combustion, which is 9 kJ mol$^{-1}$.