Question

Find the speed of sound in a mixture of 1 mol of helium and 2 mol of oxygen at ${27}^{\circ }C$.
A. $401m{s}^{-1}$
B. $301m{s}^{-1}$
C. $201m{s}^{-1}$
D. $101m{s}^{-1}$

Answer 1

Hint:Calculate the average molar mass of the given mixture. Then calculate the value of ${\gamma }_{mix}$ by calculating the specific heats of the mixture at constant pressure and constant volume. Later use the formula for the speed of sound in a mixture of gases.

Formula used:
$v=\sqrt{{\displaystyle \frac{{\gamma }_{mix}RT}{M_{mix}}}}$, where v is the speed of sound in the given mixture, R is gas constant and T of the temperature of the mixture and $M_{mix}$ is average molar mass.
$M_{mix}={\displaystyle \frac{n_1{M}_1+n_2{M}_2}{n_1+n_2}}$, where $n_1$ and $n_2$ are the numbers of moles of the two gases, $M_1$ and $M_2$ are the molar masses of the two gases.
${\gamma }_{mix}={\displaystyle \frac{C_{P,mix}}{C_{V,mix}}}$
$C_{V,mix}={\displaystyle \frac{n_1{C}_{V,1}+n_2{C}_{V,2}}{n_1+n_2}}$, where $C_{V,1}$ and $C_{V,2}$ are the specific heats of the gases at constant volume.
$C_{P,mix}={\displaystyle \frac{n_1{C}_{P,1}+n_2{C}_{P,2}}{n_1+n_2}}$, where $C_{P,1}$ and $C_{P,2}$ are the specific heats of the gases at constant pressure.
$C_V={\displaystyle \frac{f}{2}}R$ and $C_P={\displaystyle \frac{f}{2}}R$, where f is the number of degrees of freedom of the gas.

Complete step by step answer:
The speed of sound travelling in the given mixture is equal to $v=\sqrt{{\displaystyle \frac{{\gamma }_{mix}RT}{M_{mix}}}}$ ….. (i).
The molar mass of helium gas is equal to $M_1=4gmo{l}^{-1}$ and that of oxygen gas is equal to $M_2=32mo{l}^{-1}$.
As per the given data $n_1=1$ and $n_2=2$.
Therefore, the average molar mass of the mixture is equal to $M_{mix}={\displaystyle \frac{n_1{M}_1+n_2{M}_2}{n_1+n_2}}={\displaystyle \frac{(1)(4)+(2)(32)}{1+2}}={\displaystyle \frac{68}{3}}gmo{l}^{-1}={\displaystyle \frac{68}{3}}\times {10}^{-3}kgmo{l}^{-1}$.
Now, let calculate the value of $C_{P,mix}$ and $C_{V,mix}$.
Helium is a monatomic gas. Therefore, it has three degrees of freedom, i.e. $f_1=3$
This means that $C_{P,1}=({\displaystyle \frac{f_1}{2}}+1)R=({\displaystyle \frac{3}{2}}+1)R={\displaystyle \frac{5}{2}}R$
And $C_{V,1}={\displaystyle \frac{f_1}{2}}R={\displaystyle \frac{3}{2}}R$.

Oxygen is a diatomic gas. Therefore, it has five degrees of freedom, i.e. $f_2=5$
This means that $C_{P,2}=({\displaystyle \frac{f_2}{2}}+1)R=({\displaystyle \frac{5}{2}}+1)R={\displaystyle \frac{7}{2}}R$
And $C_{V,2}={\displaystyle \frac{f_2}{2}}R={\displaystyle \frac{5}{2}}R$.
Therefore, $C_{V,mix}={\displaystyle \frac{n_1{C}_{V,1}+n_2{C}_{V,2}}{n_1+n_2}}$
Substitute the known values.
 $\Rightarrow C_{V,mix}={\displaystyle \frac{(1)\left({\displaystyle \frac{3}{2}}R\right)+(2)\left({\displaystyle \frac{5}{2}}R\right)}{1+2}}={\displaystyle \frac{13}{6}}R$
And
$C_{P,mix}={\displaystyle \frac{n_1{C}_{P,1}+n_2{C}_{P,2}}{n_1+n_2}}$
$\Rightarrow C_{P,mix}={\displaystyle \frac{(1)\left({\displaystyle \frac{5}{2}}R\right)+(2)\left({\displaystyle \frac{7}{2}}R\right)}{1+2}}={\displaystyle \frac{19}{6}}R$.

Now, this means that
${\gamma }_{mix}={\displaystyle \frac{C_{P,mix}}{C_{V,mix}}}={\displaystyle \frac{{\displaystyle \frac{19}{6}}R}{{\displaystyle \frac{13}{6}}R}}={\displaystyle \frac{19}{13}}$
It is given that the temperature of the mixture is ${27}^{\circ }C$.
$\Rightarrow T={27}^{\circ }C=300K$.
The value of gas constant $R=8.31J{K}^{-1}$.
Now, substitute all the known values in equation (i).
 $$\begin{gathered} \Rightarrow v=\sqrt{{\displaystyle \frac{{\displaystyle \frac{19}{13}}\times 8.31\times 300}{{\displaystyle \frac{68}{3}}\times {10}^{-3}}}} \\ \therefore v=400.9\approx 401m{s}^{-1} \end{gathered}$$
This means that the speed of the sound waves in the given mixture is $401m{s}^{-1}$.

Hence, the correct option is A.

Note: Some students can mismatch the formulae for the specific heat capacities of a gas at constant pressure and at constant volume. So they have to be careful with these formulas. Note that in formulae involving temperature, we must always substitute the value of temperature in the unit of Kelvin and not any other given unit.