A diatomic ideal gas initially at 273 K is given 100 cal heat due to which system did 210 J work. Mo — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Heat $q = 100 \text{ cal} = 420 \text{ J}$. Work $W_{by} = 210 \text{ J}$. $\Delta U = q - W_{by} = 420 - 210 = 210 \text{ J}$. For a diatomic gas, $C_v = 5/2 R$. Since $\Delta U = n C_v \Delta T$, we get $210 = n(5/2 R)\Delta T \implies n\Delta T = \frac{84}{R}$. Molar heat capacity $C = \frac{q}{n\Delta T} = \frac{420}{84/R} = 5R$.