Question
Two ideal gases have the same value of $\frac{\text{C}_\text{P}}{\text{C}_\text{V}}=\gamma.$ What will be the value of this ratio for a mixture of the two gases in the ratio 1 : 2?
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Answer
Considering two gases, in Gas (1) we have,
$\gamma,$ $Cp_1$(Sp. Heat at const. ‘P’), $Cv_1$(Sp. Heat at const. ‘V’), $n_1$(No. of moles)
$\frac{\text{Cp}_1}{\text{Cv}_1}=\gamma\ \&\ \text{Cp}_1-\text{Cv}_1=\text{R}$
$\Rightarrow\gamma\text{Cv}_1-\text{Cv}_1=\text{R}$
$\Rightarrow\text{Cv}_1(\gamma-1)=\text{R}$
$\Rightarrow\text{Cv}_1=\frac{\text{R}}{\gamma-1}\ \&\ \text{Cp}_1=\frac{\gamma\text{R}}{\gamma-1}$
In Gas(2) we have, $\gamma,$ $Cp_2 $ (Sp. Heat at const. ‘P’), $Cv_2$(Sp. Heat at const. ‘V’), $n_2$ (No. of moles)
$\frac{\text{Cp}_2}{\text{Cv}_2}=\gamma\ \&\ \text{Cp}_2-\text{Cv}_2=\text{R}$
$\Rightarrow\gamma\text{Cv}_2-\text{Cv}_2=\text{R}$
$\Rightarrow\text{Cv}_2(\gamma-1)=\text{R}$
$\Rightarrow\text{Cv}_2=\frac{\text{R}}{\gamma-1}\ \&\ \text{Cp}_2=\frac{\gamma\text{R}}{\gamma-1}$
Given $n_1 : n_2 = 1 : 2$
$dU_1 = nCv_1dT\ \&\ dU_2 = 2nCv_2dT = 3nCvdT$
$\Rightarrow\text{C}_\text{v}=\frac{\text{Cv}_1+2\text{Cv}_2}{3}$
$=\frac{\frac{\text{R}}{\gamma-1}+\frac{\text{2R}}{\gamma-1}}{3}$
$=\frac{3\text{R}}{3(\gamma-1)}=\frac{\text{R}}{\gamma-1}\ \dots(1)$
$\text{C}_\text{p}=\gamma\text{C}_\text{v}=\frac{\gamma\text{R}}{\gamma-1}\ \dots(2)$
So, $\frac{\text{Cp}}{\text{Cv}}=\gamma$ [from (1) & (2)]
$\gamma,$ $Cp_1$(Sp. Heat at const. ‘P’), $Cv_1$(Sp. Heat at const. ‘V’), $n_1$(No. of moles)
$\frac{\text{Cp}_1}{\text{Cv}_1}=\gamma\ \&\ \text{Cp}_1-\text{Cv}_1=\text{R}$
$\Rightarrow\gamma\text{Cv}_1-\text{Cv}_1=\text{R}$
$\Rightarrow\text{Cv}_1(\gamma-1)=\text{R}$
$\Rightarrow\text{Cv}_1=\frac{\text{R}}{\gamma-1}\ \&\ \text{Cp}_1=\frac{\gamma\text{R}}{\gamma-1}$
In Gas(2) we have, $\gamma,$ $Cp_2 $ (Sp. Heat at const. ‘P’), $Cv_2$(Sp. Heat at const. ‘V’), $n_2$ (No. of moles)
$\frac{\text{Cp}_2}{\text{Cv}_2}=\gamma\ \&\ \text{Cp}_2-\text{Cv}_2=\text{R}$
$\Rightarrow\gamma\text{Cv}_2-\text{Cv}_2=\text{R}$
$\Rightarrow\text{Cv}_2(\gamma-1)=\text{R}$
$\Rightarrow\text{Cv}_2=\frac{\text{R}}{\gamma-1}\ \&\ \text{Cp}_2=\frac{\gamma\text{R}}{\gamma-1}$
Given $n_1 : n_2 = 1 : 2$
$dU_1 = nCv_1dT\ \&\ dU_2 = 2nCv_2dT = 3nCvdT$
$\Rightarrow\text{C}_\text{v}=\frac{\text{Cv}_1+2\text{Cv}_2}{3}$
$=\frac{\frac{\text{R}}{\gamma-1}+\frac{\text{2R}}{\gamma-1}}{3}$
$=\frac{3\text{R}}{3(\gamma-1)}=\frac{\text{R}}{\gamma-1}\ \dots(1)$
$\text{C}_\text{p}=\gamma\text{C}_\text{v}=\frac{\gamma\text{R}}{\gamma-1}\ \dots(2)$
So, $\frac{\text{Cp}}{\text{Cv}}=\gamma$ [from (1) & (2)]
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