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 $1270 \,m/s$, calculate the speed of sound (in m/s) in the mixture

Answer 1

If $V_{H}$ and $V_{m}$ are the velocities in hydrogen and mixture respectively, then
$\frac{v_{m}}{v_{H}} = \sqrt{\frac{\rho_{H}}{\rho_{m}}} \ldots\left(i\right)$
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 $\left(i\right)$,
$v_{m}=v_{H} \sqrt{\frac{1}{4}}=\frac{v_{H}}{2}$
$=\frac{1270}{2} = 635\, m/ s$