A quantity of 56 g of nitrogen gas is enclosed in a rigid vessel at a temperature 300 K. The amount — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

$v_{\text{rms}} \propto \sqrt{T}$. To double the rms velocity, the temperature must be quadrupled, so $T_{\text{final}} = 4 \times 300 = 1200\text{ K}$, meaning $\Delta T = 900\text{ K}$. Moles of $\text{N}_2 = 56\text{ g} / 28\text{ g/mol} = 2\text{ mol}$. In a rigid vessel (isochoric), $q = \Delta U = nC_V\Delta T$. For $\text{N}_2$ (diatomic), $C_V = 5/2 R = 2.5 \times 2 = 5\text{ cal/K-mol}$. Thus, $q = 2 \times 5 \times 900 = 9000\text{ cal} = 9\text{ kcal}$.