Question

The velocities of sound in an ideal gas at temperature T${}_1$ and T${}_2$ are found to be $V_1$ and $V_2$ respectively. If the r.m.s. velocities of the molecules of the same gas at the same temperature T${}_1$ and T${}_2$ are $v_1$ and $v_2$, respectively, then:

• A. $v_2=v_1(\frac{V_1}{V_2})$
• B. $v_2=v_1(\frac{V_2}{V_1})$
• C. $v_2=v_1\left(\sqrt{\frac{V_2}{V_1}}\right)$
• D. $v_2=v_1\left(\sqrt{\frac{V_1}{V_2}}\right)$

Answer 1

Answer: (B)

$v_2=v_1(\frac{V_2}{V_1})$

$v_{rms}=\sqrt{\frac{3RT}{M_{mol}}}$

So, $\frac{v_{1rms}}{v_{2rms}}=\sqrt{\frac{3R{T}_1}{M_{mol}}}\times \sqrt{\frac{M_{mol}}{3R{T}_2}}$

Or, $\frac{v_{1rms}}{v_{2rms}}=\sqrt{\frac{T_1}{T_2}}$ -----(1)

Now, $V\propto \sqrt{T}$

So, $\frac{V_1}{V_2}=\sqrt{\frac{T_1}{T_2}}$ -----(2)

Form (1) and (2),
$\frac{v_1}{v_2}=\frac{V_1}{V_2}$

$\therefore$ $v_2=v_1\left(\frac{V_2}{V_1}\right)$