105 , cal of heat is required to raise the te | vedclass

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(B) At constant pressure, the heat absorbed is given by $\Delta Q_P = n C_P \Delta T$. Given $\Delta Q_P = 105 \, cal$, $n = 3 \, moles$, and $\Delta

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(B) At constant pressure, the heat absorbed is given by $\Delta Q_P = n C_P \Delta T$. Given $\Delta Q_P = 105 \, cal$, $n = 3 \, moles$, and $\Delta T = 35^{\circ} C - 30^{\circ} C = 5^{\circ} C$. At constant volume, the heat absorbed is given by $\Delta Q_V = n C_V \Delta T$. We know that $\gamma = \frac{C_P}{C_V}$, which implies $C_P = \gamma C_V$. Therefore, $\frac{\Delta Q_P}{\Delta Q_V} = \frac{n C_P \Delta T}{n C_V \Delta T} = \frac{C_P}{C_V} = \gamma$. Substituting the values: $\frac{105}{\Delta Q_V} = 1.4$. $\Delta Q_V = \frac{105}{1.4} = 75 \, cal$. Since the temperature change $\Delta T$ is the same $(5^{\circ} C)$ in both cases, the heat required at constant volume is $75 \, cal$.

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