310 J of heat is required to raise the tempe | vedclass

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Question

(B) At constant pressure,the heat supplied is given by $Q_P = n C_P \Delta T$.Given $Q_P = 310 \ J$,$n = 2 \ moles$,and $\Delta T = 35 - 25 = 10 \ K$.

Vedclass · Accepted Answer

$144$

Vedclass · Answer

(B) At constant pressure,the heat supplied is given by $Q_P = n C_P \Delta T$.Given $Q_P = 310 \ J$,$n = 2 \ moles$,and $\Delta T = 35 - 25 = 10 \ K$.$310 = 2 	imes C_P 	imes 10 = 20 C_P$.$C_P = \frac{310}{20} = 15.5 \ J \ mol^{-1} K^{-1}$.Using the relation $C_P - C_V = R$,where $R \approx 8.3 \ J \ mol^{-1} K^{-1}$:$C_V = C_P - R = 15.5 - 8.3 = 7.2 \ J \ mol^{-1} K^{-1}$.At constant volume,the heat required is $Q_V = n C_V \Delta T$.$Q_V = 2 	imes 7.2 	imes 10 = 144 \ J$.

$310 \ J$ of heat is required to raise the temperature of $2 \ moles$ of an ideal gas at constant pressure from $25 \ ^oC$ to $35 \ ^oC$. The amount of heat required to raise the temperature of the gas through the same range at constant volume is $.... \ J$.

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