Question

Find the speed of sound in the mixture of 1mole helium and 2 mole oxygen

Answer 1

you forgot to write " at temperature 27°C "

according to Laplace equation of speed of sound in gas is given by
$\bold{\nu=\sqrt{\frac{\gamma RT}{M}}}$

Now, mixture contains 1 mole of Helium and 2 mole of oxygen,
Molecular mass of mixture :
$\bold{M_{mix}=\frac{n_1M_1+n_2M_2}{n_1+n_2}}$
$\bold{M_{mix}}$ = (1 × 4 + 2 × 32)/(1 + 2) = 68/3 g/mol = 68/3 × 10⁻³ Kg/mol

For γ of mixture , we should find Cv and Cp
$\bold{C_v=\frac{n_1C_{v_1}+n_2C_{v_2}}{n_1+n_2}}$
Cv for helium = 3R/2 { because monoatomic }
Cv for oxygen = 5R/2 { because diatomic }
$\bold{C_v}$ = (1 × 3R/2 + 2 × 5R/2)/(1 + 2) = 13R/6
So, $\bold{C_p}$ = Cv + R = R + 13R/6 = 19R/6

Now,
Ymix = Cp/Cv = 19/13
Mmix = 0.068/3 kg/mol
Now, speed of sound in mixture = $\bold{\sqrt{\frac{\frac{19}{13}\times\frac{25}{3}\times300}{\frac{68}{3}\times10^{-3}}}}$
= 400.9 m/s

Answer 2

Without Temperature =27°C we an't proceed with calculations
According to Laplace formula, the speed of sound is given by 
V=√γRT/M----------equation(1)
Molecular weight of mixture of gases:
Mmix=(n1M1+n2M2)/n1+n2
=1x4+2x32/1+2
=68/3 x 10⁻³kg/mol

As Helium is Mono atomic so
 Cv=3R/2
while oxygen is diatomic [Cv=5R/2]
hence(Cv)mix=(n1Cv1+n2Cv2)/n1+n2
=1x(3/2)R+ 2x(5/2)R/1+2
=13R/6

(Cp)mix=(Cv)mix +R
=13/6+R=19/6

γmix=Cp/CV
=19R/6/13R/6
=19/13

Now substituting the values of M and γ in equation 1 with T=200 K and r=8.31J/mol

V=√(19/3)(8.31x300)/68/3 x 10⁻³)
=400.9m/s