Question

The velocity of sound in hydrogen is $1400m{s}^{-1}$. What will be the velocity of sound in a mixture of gases having two parts by volume of hydrogen and one part by volume of oxygen. Maintaing the same pressure?

• A. 371.4 m/s
• B. 471.4 m/s
• C. 571.4 m/s
• D. 671.4 m/s

Answer 1

The correct option is C 571.4 m/s

Let ${\rho }_H$ be the density of hydrogen and $\rho$ the density of oxygen. Then $\rho =16{\rho }_H$.
Let x be the volume of the mixture
Volume of hydrogen in the mixture $=\frac{2x}{3}$
Volume oxygen in the mixture $=\frac{x}{3}$
Mass of hydrogen in the mixture $=\left(\frac{2x}{3}\right)\rho$
Mass of oxygen in the mixture $=\left(\frac{2x}{3}\right){\rho }_0=\left(\frac{x}{3}\right)16\rho$
$Totalmass=\frac{2\rho x}{3}+\frac{16\rho x}{3}=\frac{18\rho x}{3}=6\rho x)$
Density of the mixture =$\frac{mass}{volume}=\frac{16\rho x}{x}=6\rho$
Now $\frac{v_H}{v_m}=\sqrt{\frac{{\rho }_m}{\rho }}$
$v_H$ = velocity of sound in hydrogen = $1400m{s}^{-1}$
$v_m$ = velocity of sound in mixture $=1400m{s}^{-1}$
${\rho }_m=6\rho$
${\rho }_H=\rho$
$\therefore \frac{1400}{v_m}=\sqrt{\frac{6\rho }{\rho }}=2.45$
$v_m=\frac{1400}{2.45}=571.4m{s}^{-1}$
Ans: $571.4m{s}^{-1}$