Question

Question: The speed of sound in different gases at a certain temperature T depends on the molar mass of the gases. Show that $\frac{{\mathbf{v}}_{\mathbf{1}}}{{\mathbf{v}}_{\mathbf{2}}}\mathbf{=}\sqrt{\frac{{\mathbf{M}}_{\mathbf{2}}}{{\mathbf{M}}_{\mathbf{1}}}}$whereis the speed of sound in a gas of molar mass M₁and v₂is the speed of sound in a gas of molar mass M₂.(Hint: See Problem 91.)

Short answer

Answer

$\frac{v_1}{v_2}=\sqrt{\frac{M_2}{M_1}}$

where, v₁ and v₂ are the velocities of the sound in a gas of molar mass M₁ and M₂ respectively.

Step-by-step solution

Step 1: Given

1. v₁ is the velocity of the sound in a gas of molar mass M₁ .
2. v₂ is the velocity of the sound in a gas of molar mass M₂ .
3. The temperature T is same for v₁ and v₂.

Determining the concept

By using the formula for the speed of sound from problem 19-91, show that,
.$\frac{{\mathbf{v}}_{\mathbf{1}}}{{\mathbf{v}}_{\mathbf{2}}}\mathbf{=}\sqrt{\frac{{\mathbf{M}}_{\mathbf{2}}}{{\mathbf{M}}_{\mathbf{1}}}}$

From the problems19-91, the speed of sound is,

${\mathbf{v}}_{\mathbf{s}}\mathbf{=}\sqrt{\frac{\mathbf{\gamma RT}}{\mathbf{M}}}$

where,v₁ and v₂.are the velocities of the sound in a gas of molar massM₁ and M₂ respectively.

Determining  v1v2=M2M1

From problem19-91, the speed of sound is,

$v_s=\sqrt{\frac{\gamma RT}{M}}$

Where, $\gamma ={\text{C}}_p/{\text{C}}_V$,R is gas constant, Tis the temperature, and M is the molar mass.

Therefore, the velocity v₁ of the sound in a gas of molar mass M₁ is given by,

$v_1=\sqrt{\frac{\gamma RT}{M_1}}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(1\right)$

Similarly, the velocity v₂ of the sound in a gas of molar mass M₂ is given by,

$v_2=\sqrt{\frac{\gamma RT}{M_2}}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(2\right)$

Dividing Eq. (1) by Eq. (2),

$$\begin{gathered} \frac{v_1}{v_2}=\frac{\sqrt{\frac{\gamma RT}{M_1}}}{\sqrt{\frac{\gamma RT}{M_2}}} \\ \frac{v_1}{v_2}=\sqrt{\frac{M_2}{M_1}} \end{gathered}$$

Hence, proved.

The required relation is proved.