Question

A sample of oxygen at $NTP$ has volume $V$ and a sample of hydrogen at $NTP$ has volume $4V$. Both the gases are mixed and the mixture is maintained at $NTP$. If the speed of sound in hydrogen at $NTP$ is $1270m/s$, calculate the speed of sound (in m/s) in the mixture

Answer 1

Answer: 635

If $V_H$ and $V_m$ are the velocities in hydrogen and mixture respectively, then
$\frac{v_m}{v_H}=\sqrt{\frac{{\rho }_H}{{\rho }_m}}\dots \left(i\right)$
Density of mixture, ${\rho }_m=\frac{{\rho }_o{V}_o+{\rho }_H{V}_H}{V_o+V_H}$
where ${\rho }_0$ and $V_0$ are the density and volume of the oxygen
${\rho }_m=\frac{{\rho }_H{V}_H\left(1+\frac{{\rho }_o}{{\rho }_H}\times \frac{V_o}{V_H}\right)}{V_H\left(1+\frac{V_o}{V_H}\right)}$
or $\frac{{\rho }_m}{{\rho }_H}=\left(\frac{1+\frac{{\rho }_o}{{\rho }_H}\times \frac{V_o}{V_H}}{\left(1+\frac{V_o}{V_H}\right)}\right)$
$=\frac{1+16\times \frac{1}{4}}{1+\frac{1}{4}}=4$
or $\frac{{\rho }_H}{{\rho }_m}=\frac{1}{4}$
From equation $\left(i\right)$,
$v_m=v_H\sqrt{\frac{1}{4}}=\frac{v_H}{2}$
$=\frac{1270}{2}=635m/s$