## Related questions with answers

The speed of sound in different gases at a certain temperature T depends on the molar mass of the gases. Show that

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

where

$v_1$

is the speed of sound in a gas of molar mass

$M_1$

and

$v_2$

is the speed of sound in a gas of molar mass

$M_2$

.

Solution

VerifiedFrom the $Problem$ $91$ we have given relation for speed of sound:

$\begin{align*} v_s=\sqrt\frac{\gamma\cdot{R}\cdot{T}}{M} \end{align*}$

So for the speed of sound in both gases we have:

$\begin{align*} v_1&=\sqrt\frac{\gamma\cdot{R}\cdot{T}}{M_1}\\ v_2&=\sqrt\frac{\gamma\cdot{R}\cdot{T}}{M_2}\\ \frac{v_1}{v_2}&=\frac{\sqrt\frac{\gamma\cdot{R}\cdot{T}}{M_1}}{\sqrt\frac{\gamma\cdot{R}\cdot{T}}{M_2}}\\ \left( \frac{v_1}{v_2}\right)^2&=\frac{\frac{\gamma\cdot{R}\cdot{T}}{M_1}}{\frac{\gamma\cdot{R}\cdot{T}}{M_2}}\\ \left( \frac{v_1}{v_2}\right)^2&=\frac{M_2}{M_1}\\ \frac{v_1}{v_2}&=\sqrt\frac{M_2}{M_1} \end{align*}$

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