Question

Find rms velocity of nitrogen gas in $cm{s^{ - 1}}$ , if $1.4g$ of nitrogen occupies $2L$ at $1520\,torr$ ?

Answer 1

Maxwell Boltzmann distribution law is used to find velocity of gases at different temperature points including most probable velocity, average velocity and rms velocity. The rms velocity of a gas varies directly with the square root of temperature whereas it is inversely proportional to the square root of the molecular mass of the gas.

Complete answer:
We have been provided following data:
Mass of nitrogen gas $= 1.4g$
Volume of nitrogen gas $= 2L$
Pressure $= 1520\,torr$
To find the rms velocity, first we need to find the temperature at which the nitrogen gas is present using the ideal gas equation.
$PV = nRT$
Where, $P$ is the pressure exerted by a gas in atm, $V$ is the volume occupied by the gas in litres, $n$ is the number of moles of gas, $R$ is the universal gas constant and $T$ is temperature of gas in Kelvin.
So, we need to convert the given value of pressure to atm. We have the following relation between units of torr and atm:
$1{\rm{ torr = }}\dfrac{1}{{760}}{\rm{ atm}}$
Thus, the conversion factor for converting the unit of pressure from torr to atm is $\left( {\dfrac{{1{\rm{ atm}}}}{{760{\rm{ torr}}}}} \right)$
Therefore, pressure of gas in atm = $1520{\rm{ torr }} \times {\rm{ }}\left( {\dfrac{{1{\rm{ atm}}}}{{760{\rm{ torr}}}}} \right)$
$\Rightarrow$ Pressure $= 2{\rm{ atm}}$
We know that the number of moles of gas is the ratio of the mass of the gas present to the molecular mass of the gas. So, number of moles of nitrogen gas present will be as follows:
$n = \dfrac{{{\rm{given mass}}}}{{{\rm{molecular mass}}}}$
Molecular mass of nitrogen gas $= 28{\rm{ g}}$
So, $n = \dfrac{{1.4}}{{28}}$
$\Rightarrow n = 0.05\;{\rm{moles}}$
Substituting the values in ideal gas equation:
$PV = nRT$
$\Rightarrow 2 \times 2 = 0.05 \times 0.0821 \times T$
$\Rightarrow T = \dfrac{4}{{0.0041}}$
$\Rightarrow T = 975.6{\rm{ K}}$
Rms velocity: It is the maximum velocity of a gas at a given temperature. It is given by the formula
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}}$
Substituting the given values in the expression:
${v_{rms}} = \sqrt {\dfrac{{3 \times 8.314{\rm{ kg}}{{\rm{m}}^2}{s^{ - 2}}{K^{ - 1}}mo{l^{ - 1}} \times 975.6{\rm{ K}}}}{{28 \times {{10}^{ - 3}}kgmo{l^{ - 1}}}}}$
$\Rightarrow {v_{rms}} = 0.9327{\rm{ m}}{{\rm{s}}^{ - 1}}$
But we need to find the rms velocity in $cm{s^{ - 1}}$ . Converting the units using proper conversion factor, the rms velocity will be as follows:
$\Rightarrow {{v}_{rms}}=932.7\text{ cm}{{\text{s}}^{-1}}\ \ \ \ \ \ \ [\because 1m=100cm]$
Hence, the rms velocity of nitrogen gas under given conditions is $932.7{\rm{ cm}}{{\rm{s}}^{ - 1}}$ .

Note:
To avoid silly mistakes, try to keep a check on the units of each data given at every step. It is very important to use a proper conversion factor to convert the unit of any value.