A quantity of 70 calories of heat is required to raise the temperature of 2 mole of an ideal gas at — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

$q_p = n C_p \Delta T \implies 70 = 2 \times C_p \times 10 \implies C_p = 3.5 \text{ cal/mol K}$. From Mayer's relation, $C_v = C_p - R$. Using $R \approx 2 \text{ cal/mol K}$, $C_v = 3.5 - 2 = 1.5 \text{ cal/mol K}$. The required heat at constant volume $q_v = n C_v \Delta T = 2 \times 1.5 \times 10 = 30 \text{ calories}$.