Question

The root mean square velocity of molecules of a gas whose density is $1.4\,Kg{m^{ - 3}}$ at a pressure of $76\,cm$ of mercury.
(Specific gravity of mercury $= 13.6$ and $g = 9.81\,m{s^{ - 2}}$ ) is:
A) $4.66 \times {10^2}\,m{s^{ - 1}}$
B) $2.5 \times {10^2}\,m{s^{ - 1}}$
C) $2.33 \times {10^2}\,m{s^{ - 1}}$
D) $6.66 \times {10^{2\,}}\,m{s^{ - 1}}$

Answer 1

To find the value of the root means square velocity of the gas, use the formula given below and substitute all the required parameters of the gas and the mercury in that and get the value of the answer by simplifying the substituted formula.

Formula used:
The formula of the root means square velocity is given by
${v_{rms}} = \sqrt {\dfrac{{3P{\rho _m}g}}{\rho }}$
Where ${v_{rms}}$ is the root means square velocity of the gas, $P$ is the pressure of the mercury, ${\rho _m}$ is the density of the mercury, $g$ is the acceleration due to gravity and the $\rho$ is the density of the gas.

Complete step by step solution:
It is given that the
Density of the gas, $\rho = 1,4\,Kg{m^{ - 3}}$
Pressure of the mercury, $P = 76\,cm$
Specific gravity of mercury, $s = 13.6$ and
Acceleration due to gravity, $g = 9.81\,m{s^{ - 2}}$
Using the formula of the root means square velocity of the gas which is given below.
${v_{rms}} = \sqrt {\dfrac{{3P{\rho _m}g}}{\rho }}$
Substitute the pressure of the mercury, density of the mercury, acceleration due to gravity and the density of the gas in the above formula.
$\Rightarrow$ ${v_{rms}} = \sqrt {\dfrac{{3 \times 0.76 \times 13600 \times 9.81}}{{1.4}}}$
By doing the basic arithmetic operation in the above step, we get
$\Rightarrow$ ${v_{rms}} = \sqrt {\dfrac{{304188.48}}{{1.4}}}$
By further simplification in the above step, we get
$\Rightarrow$ ${v_{rms}} = 4.66 \times {10^2}\,m{s^{ - 1}}$
Hence the value of the root means square velocity of the gas is ${v_{rms}} = 4.66 \times {10^2}\,m{s^{ - 1}}$.

Thus the option (A) is correct.

Note: The density of the mercury is calculated from the specific gravity of it by using the formula, $s = \dfrac{\rho }{{{\rho _w}}}$. Since the density of the water is $1000$ , multiplying the $13.6$ with the $1000$ , the density of the mercury is obtained as $13600$ .