The temperature of the mixture of one mole of | vedclass

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(B) The molar heat capacity at constant pressure for the mixture is given by: $(C_P)_{mix} = \frac{\mu_1 C_{P1} + \mu_2 C_{P2}}{\mu_1 + \mu_2}$ For He

Vedclass · Accepted Answer

$1200$

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(B) The molar heat capacity at constant pressure for the mixture is given by: $(C_P)_{mix} = \frac{\mu_1 C_{P1} + \mu_2 C_{P2}}{\mu_1 + \mu_2}$ For Helium (monatomic),$C_{P1} = \frac{5}{2}R$. For Hydrogen (diatomic),$C_{P2} = \frac{7}{2}R$. Given $\mu_1 = 1$ mole and $\mu_2 = 1$ mole: $(C_P)_{mix} = \frac{1 	imes \frac{5}{2}R + 1 	imes \frac{7}{2}R}{1 + 1} = \frac{6R}{2} = 3R$. Using $R \approx 2 \, cal/mol \cdot K$,we get $(C_P)_{mix} = 3 	imes 2 = 6 \, cal/mol \cdot ^{\circ}C$. The total heat delivered at constant pressure is: $\Delta Q = (\mu_1 + \mu_2) (C_P)_{mix} \Delta T$ $\Delta Q = (1 + 1) 	imes 6 	imes (100 - 0) = 2 	imes 6 	imes 100 = 1200 \, cal$.

The temperature of the mixture of one mole of helium and one mole of hydrogen is increased from $0^{\circ}C$ to $100^{\circ}C$ at constant pressure. The amount of heat delivered will be ...... $cal$.

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