Question

A gaseous mixture consists of $16g$ of helium and $16g$ 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

Answer: (B)

$C_v=\frac{n_1{C}_{v_1}+n_2{C}_{v_2}}{n_1+n_2}$
For helium, $n_1=\frac{16}{4}=4$ and ${\gamma }_1=\frac{5}{3}$
$n_2=\frac{16}{32}=\frac{1}{2}$ and ${\gamma }_2=\frac{7}{5}$
$C_{v_1}=\frac{R}{{\gamma }_1-1}$
$=\frac{R}{\frac{5}{3}-1}=\frac{3}{2}R$
$C_{v_2}=\frac{R}{{\gamma }_2-1}=\frac{R}{\frac{7}{5}-1}=\frac{5}{2}R$
$\therefore C_v=\frac{4\times \frac{3}{2}R+\frac{1}{2}.\frac{5}{2}R}{4+\frac{1}{2}}$
$=\frac{6R+\frac{5}{4}R}{\frac{9}{2}}$
$=\frac{29R\times 2}{9\times 4}=\frac{29R}{18}$
Now, $C_v=\frac{R}{\gamma -1}$
$\Rightarrow \gamma -1=\frac{R}{C_v}$
$\gamma =\frac{R}{C_v}+1=\frac{R}{\frac{29}{18}R}+1$
$\Rightarrow \frac{C_p}{C_v}=\frac{18}{29}+1$
$=\frac{18+29}{29}=1.62$