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

The correct option is B $1.62$

$\text{Given:-}$ Mass of helium $=16g$

Mass of oxygen $=16g$

$\text{Solution:-}$

For mixture,$C_V=\frac{n_1{C}_{V1}+n_2{C}_{V2}}{n_1+n_2}$

For helium,

Number of moles, $n_1=\frac{16}{4}=4$

and ${\gamma }_1=\frac{5}{3}$

For oxygen,

Number of moles, $n_2=\frac{16}{32}=\frac{1}{2}$

and ${\gamma }_2=\frac{7}{5}$

Then, $C_{V1}=\frac{R}{{\gamma }_1-1}=\frac{R}{\frac{5}{3}-1}=\frac{3}{2}R$

And

$C_{V2}=\frac{R}{{\gamma }_2-1}=\frac{R}{\frac{7}{5}-1}=\frac{5}{2}R$

$\therefore$ For mixture,

$C_V=\frac{4\times \frac{3}{2}R+\frac{1}{2}\times \frac{5}{2}R}{4+\frac{1}{2}}$

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

$=\frac{29R\times 2}{9\times 4}=\frac{29R}{18}$

Now, $C_V=\frac{R}{\gamma -1}\Rightarrow \gamma -1=\frac{R}{C_V}$

or $\gamma =\frac{R}{C_V}+1=\frac{R}{\frac{29}{18}R}+1$

$\frac{C_P}{C_V}=\frac{18}{29}+1=1.62$

$\text{Hence the correct option is B}$