Question

The gaseous mixture consists of $16$ of helium and $16$ of oxygen. The ratio $\frac{C_p}{C_v}$ of the mixture is :-

• A. 1.59
• B. 1.62
• C. 1.4
• D. 1.54

Answer 1

The correct option is B 1.62

Given: 16 g of Helium and 16 g of oxygen.

Solution:

We know that for a mixture, $C_v=\frac{n_1{C}_{v1}+n_2{C}_{v2}}{n_1+n_2}$ (1)

Now for helium ,moles is $n_1=\frac{16}{4}=4$

For helium ${\gamma }_1$is $\frac{5}{3}$

So, $C_{v1}=\frac{R}{{\gamma }_1-1}$

$\therefore C_{v1}=\frac{3}{2}R$

For oxygen, number of moles are $n_2=\frac{16}{32}=0.5$

For oxygen ${\gamma }_2$is $\frac{7}{5}$

So, $C_{v2}=\frac{R}{{\gamma }_2-1}$

$\therefore C_{v2}=\frac{5}{2}R$

Putting the value of$n_1,n_2,C_{v1},C_{v2}$ in equation (1), we get

So, $C_v=\frac{29R}{18}$

$C_v=\frac{R}{{\gamma }_{-}1}$

$\Rightarrow \gamma =\frac{R}{C_v}+1$

Putting $C_v=\frac{29R}{18}$ in the above equation, we get

$\gamma =\frac{18}{29}+1$

We know that $\frac{C_p}{C_v}=\gamma$

$\therefore \frac{C_p}{C_v}=\frac{18}{29}+1=1.62$

Hence correct option is B.