Question

The speed of sound in an ideal gas at $T_{1}K$ and $T_{2}K$ are $v_1$ and $v_2$ respectively. If the root mean square velocity of molecules of same gas at these temperatures are $v_{rms1}$ and $v_{rms2}$ respectively, then.

• A. $c_2=c_1\left(\frac{v_2}{v_1}\right)$
• B. $c_2=c_1\left(\frac{v_1}{v_2}\right)$
• C. $c_2=c_1\left(\sqrt{\frac{v_2}{v_1}}\right)$
• D. $c_2=c_1\left(\sqrt{\frac{v_1}{v_2}}\right)$

Answer 1

The correct option is A $c_2=c_1\left(\frac{v_2}{v_1}\right)$

The speed of sound in an ideal gas at temperature $T$ is given by, $v=\sqrt{\frac{\gamma RT}{M}}$ and root mean velocity is $v_{rms}=\sqrt{\frac{3RT}{M}}$

where $\gamma =$ adiabatic constant, $R=$ gas constant and $M=$ molecular mass of gas.

Thus, $v_{rms}=\sqrt{\frac{3v^2}{\gamma }}$

$\therefore v_{rms}\propto v$

Now, $\frac{v_{rms2}}{v_{rms1}}=\frac{v_2}{v_1}$

or $v_{rms2}=v_{rm{s}_1}\left(\frac{v_2}{v_1}\right)$