A monoatomic gas expands isobarically. The percentage of heat supplied that increases the thermal en — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

For an isobaric expansion of a monoatomic ideal gas, q = n * $C_p$ * dT = 2.5 * nR * dT. Increase in thermal (internal) energy Δ U = n * $C_v$ * dT = 1.5 * nR * dT. Work done w_by_gas = nR * dT. Percentage of heat increasing internal energy = (1.5 / 2.5) * 100% = 60%. Percentage of heat involved in work = (1 / 2.5) * 100% = 40%. Thus, the ratio is 60:40.