A volume of 1.642 l sample of a mixture of methane gas and oxygen measured at 298 K and 1.192 atm, w — Thermodynamics and Thermochemistry Chemistry Question
Question
A volume of 1.642 l sample of a mixture of methane gas and oxygen measured at 298 K and 1.192 atm, was allowed to react at constant pressure in a calorimeter which together with its content had a heat capacity of 1260 cal/K. The complete combustion of methane to carbon dioxide and water caused a temperature rise in calorimeter 0.667 K. The volume per cent of methane in original mixture is (Given the heat of combustion of methane is –210 kcal/mole)
💡 Solution & Explanation
Total moles of gas $n = \frac{PV}{RT} = \frac{1.192 \times 1.642}{0.0821 \times 298} = 0.08\text{ moles}$. Total heat released $q = C\Delta T = 1260 \times 0.667 \approx 840.4\text{ cal} = 0.84\text{ kcal}$. Moles of methane burned = $0.84 / 210 = 0.004\text{ moles}$. Volume percentage of methane = $(0.004 / 0.08) \times 100 = 5\%$.