> Q. A gaseous mixture consists of $16, g$ of helium and $16, g$ of oxygen. The ratio $frac{C_{p}}{C_{v}}$ of the mixture is : – LIVE ANSWER TODAY

Q. A gaseous mixture consists of $16, g$ of helium and $16, g$ of oxygen. The ratio $frac{C_{p}}{C_{v}}$ of the mixture is :

In This post you will get answer of Q.
A gaseous mixture consists of $16, g$ of helium and $16, g$ of oxygen. The ratio $frac{C_{p}}{C_{v}}$ of the mixture is :

Solution:

$C_{v}=frac{n_{1},C_{v_1}+n_{2},C_{v_2}}{n_{1}+n_{2}}$

For helium, $n_{1}=frac{16}{4}=4$ and $gamma_{1}=frac{5}{3}$

$n_{2}=frac{16}{32}=frac{1}{2} $ and $gamma _{2}=frac{7}{5}$

$C_{v_1}=frac{R}{gamma_{1}-1}$

$=frac{R}{frac{5}{3}-1}=frac{3}{2}R$

$C_{v_2}=frac{R}{gamma_{2}-1}=frac{R}{frac{7}{5}-1}=frac{5}{2}R$

$therefore C_{v}=frac{4timesfrac{3}{2}R+frac{1}{2}. frac{5}{2}R}{4+frac{1}{2}}$

$=frac{6,R+frac{5}{4}R}{frac{9}{2}}$

$=frac{29,Rtimes2}{9times4}=frac{29,R}{18}$

Now, $C_{v}=frac{R}{gamma-1}$

$Rightarrow gamma-1=frac{R}{C_{v}}$

$gamma =frac{R}{C_{v}}+1=frac{R}{frac{29}{18}R}+1$

$Rightarrow frac{C_{p}}{C_{v}}=frac{18}{29}+1$

$=frac{18+29}{29}=1.62$

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