OnlineCBT
Chemistry · THERMODYNAMICS
हिन्दीFor the combustion of liquid benzene at 298 K according to \( \text{C}_6\text{H}_6(l) + \frac{15}{2}\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 3\text{H}_2\text{O}(l) \), the difference between the heat evolved at constant pressure (\( q_p \)) and at constant volume (\( q_v \)) is equal to:
Correct answer · Option 1
-1.5 RT
Explanation
\( q_p - q_v = \Delta H - \Delta U = \Delta n_g RT \). Here, \( \Delta n_g = 6 - 7.5 = -1.5 \). Thus, \( q_p - q_v = -1.5 RT \).
