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(D) The heat energy required at constant volume is given by the formula $Q = n C_v \Delta T$. For a monatomic gas,the molar heat capacity at constant

Vedclass · Accepted Answer

$\frac{5 Q}{3}$

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(D) The heat energy required at constant volume is given by the formula $Q = n C_v \Delta T$. For a monatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2} R$. Given $n_1 = 2 	ext{ moles}$,$\Delta T_1 = 40^{\circ} C - 30^{\circ} C = 10 	ext{ K}$. So,$Q = 2 	imes \frac{3}{2} R 	imes 10 = 30 R$. For a diatomic gas,the molar heat capacity at constant volume is $C_v = \frac{5}{2} R$. Given $n_2 = 4 	ext{ moles}$,$\Delta T_2 = 33^{\circ} C - 28^{\circ} C = 5 	ext{ K}$. Let the required heat be $Q'$. $Q' = n_2 C_v \Delta T_2 = 4 	imes \frac{5}{2} R 	imes 5 = 50 R$. Now,taking the ratio: $\frac{Q'}{Q} = \frac{50 R}{30 R} = \frac{5}{3}$. Therefore,$Q' = \frac{5}{3} Q$.

$2 \text{ moles}$ of a monatomic gas requires heat energy $Q$ to be heated from $30^{\circ} C$ to $40^{\circ} C$ at constant volume. The heat energy required to raise the temperature of $4 \text{ moles}$ of a diatomic gas from $28^{\circ} C$ to $33^{\circ} C$ at constant volume is

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