Question

The speed of sound in hydrogen at NTP is $1270{\mathrm{ms}}^{-1}$. Then the speed in a mixture of hydrogen and oxygen in the ratio of $4:1$ by volume will be:

• A. $371{\mathrm{ms}}^{-1}$
• B. $380{\mathrm{ms}}^{-1}$
• C. $635{\mathrm{ms}}^{-1}$
• D. $950{\mathrm{ms}}^{-1}$

Answer 1

The correct option is C

$635{\mathrm{ms}}^{-1}$

Step 1: Given information

The speed of sound in hydrogen at NTP, ${\mathrm{V}}_{{\mathrm{H}}_2}=1270{\mathrm{ms}}^{-1}$

The ratio of hydrogen and oxygen by volume is $4:1$.

Let,

V is the volume of oxygen.

4V is the volume of hydrogen.

d is the density of hydrogen.

D is the density of the mixture of gases.

The total volume of the mixture of hydrogen and oxygen will be:

$$\begin{gathered} =\mathrm{V}+4\mathrm{V} \\ =5\mathrm{V} \end{gathered}$$

and,

Density of oxygen $$\begin{gathered} =16\times \mathrm{d} \\ =16\mathrm{d} \end{gathered}$$

Step 2: To find

We have to find the speed of the mixture.

Step 3: Calculating the density of the mixture of gases

We know the formula,

$\mathrm{m}=\mathrm{v}\times \mathrm{d}$

Here,

m is the mass.

v is the velocity.

d is the density.

Now,

$\Rightarrow \mathrm{Volume}\mathrm{of}\mathrm{mixture}\times \mathrm{Density}\mathrm{of}\mathrm{mixture}=\mathrm{Volume}\mathrm{of}\mathrm{hydrogen}\times \mathrm{Density}\mathrm{of}\mathrm{hydrogen}+\mathrm{Volume}\mathrm{of}\mathrm{oxygen}\times \mathrm{Density}\mathrm{of}\mathrm{oxygen}$

By substituting the given values, we get

$$\begin{gathered} 5\mathrm{V}\times \mathrm{D}=4\mathrm{V}\times \mathrm{d}+\mathrm{V}\times 16\mathrm{d} \\ \mathrm{V}\left(5\mathrm{D}\right)=\mathrm{V}(4\mathrm{d}+16\mathrm{d}) \\ 5\mathrm{D}=20\mathrm{d} \\ \mathrm{D}=\frac{20\mathrm{d}}{5} \\ \mathrm{D}=4\mathrm{d} \end{gathered}$$

Step 4: Calculation of the speed of the mixture

We know the speed of sound in the gas:

$\mathrm{V}=\sqrt{\frac{\mathrm{\gamma \rho }}{\mathrm{d}}}$

Because sound speed is inversely related to the square root of density, then

$\mathrm{V}\propto \sqrt{\frac{1}{\mathrm{d}}}$

The ratio of the speed of mixture and hydrogen is:

$$\begin{gathered} \frac{{\mathrm{V}}_{\mathrm{mix}}}{{\mathrm{V}}_{{\mathrm{H}}_2}}=\sqrt{\frac{\mathrm{d}}{\mathrm{D}}} \\ \frac{{\mathrm{V}}_{\mathrm{mix}}}{{\mathrm{V}}_{{\mathrm{H}}_2}}=\sqrt{\frac{\mathrm{d}}{4\mathrm{d}}} \\ \frac{{\mathrm{V}}_{\mathrm{mix}}}{{\mathrm{V}}_{{\mathrm{H}}_2}}=\sqrt{\frac{1}{4}} \\ \frac{{\mathrm{V}}_{\mathrm{mix}}}{{\mathrm{V}}_{{\mathrm{H}}_2}}=\frac{1}{2}...\left(1\right) \end{gathered}$$

By substituting the speed of sound in hydrogen in $\left(1\right)$, we get

$$\begin{gathered} \Rightarrow \frac{{\mathrm{V}}_{\mathrm{mix}}}{1270}=\frac{1}{2} \\ {\mathrm{V}}_{\mathrm{mix}}=\frac{1270}{2} \\ {\mathrm{V}}_{\mathrm{mix}}=635\mathrm{m}/{\mathrm{s}}^{-1} \end{gathered}$$

Therefore, option C is the correct choice.