Question

Two ideal gases have the same valuve of $C_p/C_v=\gamma$. What will be the value of this ratio for a mixture of the two gases in the ratio 1 :2 ?

Answer 1

Considering two gases, in gas (1) we have, $C{p}_1$ (Sp. heat at constant P), $C{v}_1$ (Sp. heat at moment V), $n_1$ (No of moles)

$\frac{C{p}_1}{C{p}_1}=\gamma andC{P}_1-C{v}_1=R$

$\Rightarrow \gamma C{v}_1-C{v}_1=R$

$\Rightarrow C{v}_1(\gamma -1)=R$

$\Rightarrow C{v}_1=\frac{R}{(\gamma -1)}andC{p}_1=\gamma \frac{R}{(\gamma -1)}$

In gas (2), we have

$C{P}_2$ (Sp, heat at constant P), $C{v}_2$ (Sp. heat at constant V), $n_2$ (No. of moles)

$\frac{C{p}_2}{C{v}_2}=\gamma andC{p}_2-C{v}_2=R$

$\Rightarrow \gamma C{v}_2-C{v}_2=R$

$\Rightarrow C{v}_2(\gamma -1)=R$

$\Rightarrow C{v}_2=\frac{R}{(\gamma -1)}and$

$C{v}_2=\gamma \frac{R}{\gamma -1}$

Given, $n_1=n_2=1:2$

$d{U}_1=nC{v}_{1}dtandd{U}_2$

= $2nC{v}_{2}dT$

When gases are mixed

$nC{v}_1+2nC{v}_{2}dT=3nCvdT$

Cv = $\frac{C{v}_1+2C{v}_2}{3}$

= $\frac{\frac{R}{(\gamma -1)}+\frac{2R}{\gamma -1}}{3}$

= $\frac{3R}{(\gamma -1)3}=\frac{R}{\gamma -1}$