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(D) For a mixture of gases,the molar specific heats are given by:$C_{P,mix} = \frac{n_1 C_{P1} + n_2 C_{P2}}{n_1 + n_2}$ and $C_{V,mix} = \frac{n_1 C_

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$C_P = 0.65 \ J/g \ K, C_V = 0.421 \ J/g \ K$

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(D) For a mixture of gases,the molar specific heats are given by:$C_{P,mix} = \frac{n_1 C_{P1} + n_2 C_{P2}}{n_1 + n_2}$ and $C_{V,mix} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2}$For Nitrogen ($N_2$,diatomic,$\gamma_1 = 1.4$):$n_1 = \frac{7 \ g}{28 \ g/mol} = 0.25 \ mol$$C_{V1} = \frac{R}{\gamma_1 - 1} = \frac{8.314}{0.4} = 20.785 \ J/mol \ K$$C_{P1} = C_{V1} + R = 20.785 + 8.314 = 29.099 \ J/mol \ K$For Argon ($Ar$,monatomic,$\gamma_2 = 1.67$):$n_2 = \frac{20 \ g}{40 \ g/mol} = 0.5 \ mol$$C_{V2} = \frac{R}{\gamma_2 - 1} = \frac{8.314}{0.67} \approx 12.41 \ J/mol \ K$$C_{P2} = C_{V2} + R = 12.41 + 8.314 = 20.724 \ J/mol \ K$Calculating mixture molar specific heats:$C_{V,mix} = \frac{(0.25 	imes 20.785) + (0.5 	imes 12.41)}{0.25 + 0.5} = \frac{5.196 + 6.205}{0.75} = 15.2 \ J/mol \ K$$C_{P,mix} = \frac{(0.25 	imes 29.099) + (0.5 	imes 20.724)}{0.75} = \frac{7.275 + 10.362}{0.75} = 23.516 \ J/mol \ K$Total mass $M_{total} = 7 + 20 = 27 \ g$. Total moles $n_{total} = 0.75 \ mol$.Average molar mass $M_{avg} = \frac{27}{0.75} = 36 \ g/mol$.Specific heats (per gram):$C_V = \frac{C_{V,mix}}{M_{avg}} = \frac{15.2}{36} \approx 0.421 \ J/g \ K$$C_P = \frac{C_{P,mix}}{M_{avg}} = \frac{23.516}{36} \approx 0.65 \ J/g \ K$

$A$ gaseous mixture contains $7 \ g$ of nitrogen and $20 \ g$ of argon. Assuming the gases are ideal,what are the specific heats $C_P$ and $C_V$ (in $J/g \ K$) for the mixture?

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