Question

The speed of sound in hydrogen is 1,270 ms^-1. What will be the speed of a mixture of oxygen and hydrogen mixed in a volume ratio 1: 4?

Answer 1

Suppose V₁ and V₂ be the volumes of O₂ and H₂ respectively in the mixture. ρ₁ and ρ₂ be their respective densities.

If m₁ and m₂ be the mass of oxygen and hydrogen respectively, then

m₁ = v₁ρ₁

and m₂ = v₂ρ₂

If ρ be the density of mixture, then

ρ = \(\frac{m}{V}\) 

= \(\frac{Total\, mass \,of\, mixture}{Total \,volume}\) 

= \(\frac{v_1ρ_1+v_2ρ_1}{v_1+v_2}\)

= \(\frac{v_2ρ_2(\frac{v_1ρ_1}{v_2ρ_2}+1)}{v_2(\frac{v_1}{v_2}+1)}\)

= \(\frac{ρ_2(\frac{v_1ρ_1}{v_2ρ_2}+1)}{(\frac{v_1}{v_2}+1)}\)

Now \(\frac{V_1}{V_2}=\frac{1}{4}\)(given)

And \(\frac{ρ_1}{ρ_2}=\frac{molcular \,wight\, of\, oxygen}{molecular \,weight\, of\, hydrogen}\) 

 Or \(\frac{ρ_1}{ρ_2}=\frac{32}{2}\) = 16 

So, ρ = \(\frac{(\frac{1}{4}x16+1)}{\frac{1}{4}+1}=\frac{5ρ_2}{\frac{5}{4}}\) = 4ρ₂ 

∴\(\frac{ρ}{ρ_2}=4\) 

Also at v = velocity of sound in the mixture

And v₂ = velocity of sound in the hydrogen

= 1,270 ms^-1

Then, \(v=\sqrt{\frac{\gamma P}{ρ}}\)

And v₂ = \(\sqrt{\frac{\gamma P}{ρ_2}}\) 

\(\frac{v}{v_2}=\sqrt{\frac{ρ_2}{ρ}}\)

= \(\sqrt{\frac{1}{4}}=\frac{1}{2}\) 

v = \(\frac{v_2}{4}=\frac{1}{2}\) × 1270 ms^-1

= 635 ms^-1 

Or v = 635 ms^-1