Question

Question: At what frequency would the wavelength of sound in air be equal to the mean free path of oxygen molecules at 1 atmpressure and 0.00 ⁰C? The molecular diameter is$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathit{c}\mathit{m}$.

Short answer

Answer

The required value of frequency is$3.5\times {10}^9\text{Hz}$ .

Step-by-step solution

Given data

• Temperature;

$$\begin{gathered} \text{T}=0.00°\text{C} \\ =273.15\text{K} \end{gathered}$$

• Pressure;

$$\begin{gathered} \text{p}=1.0\text{atm} \\ =1.01\times {10}^5\text{Pa} \end{gathered}$$

• Molecular diameter;

$$\begin{gathered} \text{d}=3.0\times {10}^{-8}\text{cm} \\ =3.0\times {10}^{-10}\text{m} \end{gathered}$$

Understanding the concept

The expression for the velocity of sound is given by,

$v=f\lambda$

Here v is the velocity of the sound, F is the frequency,$\lambda$ is the wavelength.

Calculate the frequency at which wavelength of sound in air would be equal to the mean free path of oxygen molecules at 1 atm pressure and 0.00 0C

$L=\lambda$

From the above expression,

,$\text{v}=\text{f}\lambda$

Substitute for into the above equation,

$f=\frac{331}{\lambda }$

Let the mean free path of oxygen molecules be L ,

Given condition is

$L=\lambda$

Hence,

$f=\frac{331}{L}$

But,$L=\frac{1}{\sqrt{2}\pi d^2\left(\frac{N}{V}\right)}$

Also, according to the gas law,

$\frac{N}{V}=\frac{p}{kT}$

Therefore,

$f=\frac{331}{\frac{1}{\sqrt{2}\pi d^2\left(\frac{\text{p}}{\text{kT}}\right)}}$

Substitute $3.0\times {10}^{-10}\text{m}$for d , $1.01\times {10}^5\text{Pa}$ for p , $1.38\times {10}^{-23} J/molecule ·K$for k and$273.15 K$ for T into the above equation.

$$\begin{gathered} f=\frac{331}{\frac{1}{\sqrt{2}\left(3.142\right){(3.0\times {10}^{-10})}^2\left(\frac{1.01\times {10}^5}{1.38\times {10}^{-23}\left(273.15\right)}\right)}} \\ =3.54\times {10}^9 \\ \approx 3.5\times {10}^9\text{Hz} \end{gathered}$$

Therefore, the required value of frequency is$3.5\times {10}^9\text{Hz}$ .