10 text{ mole} of an ideal gas is undergoing | vedclass

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(C) The process from $P_1$ to $P_2$ occurs at a constant volume $V = 1 	ext{ m}^3$.For an isochoric process, the heat exchanged is given by $\Delta Q

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(C) The process from $P_1$ to $P_2$ occurs at a constant volume $V = 1 	ext{ m}^3$.For an isochoric process, the heat exchanged is given by $\Delta Q = n C_v \Delta T$.Using the ideal gas law $PV = nRT$, we have $nR \Delta T = V \Delta P$.Substituting this into the heat equation: $\Delta Q = \frac{C_v}{R} (nR \Delta T) = \frac{C_v}{R} V (P_2 - P_1)$.Given $n = 10 	ext{ mol}$, $C_v = 21 	ext{ J/K} \cdot 	ext{mol}$, $R = 8.3 	ext{ J/mol} \cdot 	ext{K}$, $V = 1 	ext{ m}^3$, $P_1 = 21.7 	ext{ Pa}$, and $P_2 = 30 	ext{ Pa}$.$\Delta Q = \frac{21}{8.3} 	imes 1 	imes (30 - 21.7) = \frac{21}{8.3} 	imes 8.3 = 21 	ext{ J}$.Thus, $\alpha = 21$.

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