40 , text{calories} of heat is needed to rais | vedclass

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(B) For an ideal monoatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$ and at constant pressure is $C_p = \frac{5}{2}R$.Gi

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(B) For an ideal monoatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$ and at constant pressure is $C_p = \frac{5}{2}R$.Given $R = 2 \, 	ext{cal} \, 	ext{mol}^{-1} 	ext{K}^{-1}$,we have $C_v = \frac{3}{2} 	imes 2 = 3 \, 	ext{cal} \, 	ext{mol}^{-1} 	ext{K}^{-1}$ and $C_p = \frac{5}{2} 	imes 2 = 5 \, 	ext{cal} \, 	ext{mol}^{-1} 	ext{K}^{-1}$.For a constant pressure process,the heat supplied is $Q_p = n C_p \Delta T$.Given $Q_p = 40 \, 	ext{cal}$,$n = 1 \, 	ext{mole}$,and $\Delta T = 30^{\circ}C - 20^{\circ}C = 10 \, 	ext{K}$.Checking the consistency: $Q_p = 1 	imes 5 	imes 10 = 50 \, 	ext{cal}$.However,the problem states $40 \, 	ext{cal}$ is required. This implies the gas might not be perfectly monoatomic or the provided $R$ is for calculation purposes. Using the ratio $C_v/C_p = 1/\gamma$ where $\gamma = 5/3$ for monoatomic gas:$C_v = C_p / \gamma = (C_p 	imes 3) / 5$.From $Q_p = n C_p \Delta T$,we have $40 = 1 	imes C_p 	imes 10$,so $C_p = 4 \, 	ext{cal} \, 	ext{mol}^{-1} 	ext{K}^{-1}$.Then $C_v = 4 	imes (3/5) = 2.4 \, 	ext{cal} \, 	ext{mol}^{-1} 	ext{K}^{-1}$.For constant volume,$Q_v = n C_v \Delta T = 1 	imes 2.4 	imes 10 = 24 \, 	ext{cal}$.

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