Question

Oxygen at 1 atmosphere and ${{\text{0}}^{\text{0}}}\text{C}$ has a density of$\text{1}\text{.4290 g }{{\text{L}}^{\text{-1}}}$. Find the rms speed of the oxygen molecule.

Answer 1

To solve this type of solution using the formula of root mean square speed${{\text{V}}_{\text{rms}}}$. It is related to the density of gas and pressure by the relation ${{\text{V}}_{\text{rms}}}\text{=}\sqrt{\dfrac{\text{3P}}{\text{d}}}$.Use this formula to find out the RMS speed of oxygen at 1 atmosphere and ${{0}^{0}}$ temperature.

Complete step by step answer:
The root-mean-square speed is the measure of the speed of particles in a gas, defined as the square root of the average velocity-squared of the molecules in a gas.
Therefore the ${{\text{V}}_{\text{rms}}}$ can be written as:
${{\text{V}}_{\text{rms}}}\text{=}\sqrt{\dfrac{\text{3P}}{\text{d}}}$
Where d is the density of the gas.
P is the pressure in the Pascal’s
Here we are given the oxygen gas at a 1-atmosphere pressure.
$\text{d=1}\text{.4290 g}{{\text{L}}^{\text{-1}}}$
Let us substitute all values in the ${{\text{V}}_{\text{rms}}}$ equation we get,
${{\text{V}}_{\text{rms}}}\text{=}\sqrt{\dfrac{\text{3}\times \text{(1 atm)}}{(1.4290\text{ g}{{\text{L}}^{\text{-1}}})}}$
Note that we have to convert the pressure which is in atm to the pascal. since we know that,
$1\text{ atm=101}\text{.3}\times \text{1}{{\text{0}}^{3}}\text{ Pa}=1.013\times {{10}^{5}}\text{ Pa}$
Thus we get,
${{\text{V}}_{\text{rms}}}\text{=}\sqrt{\dfrac{\text{3}\times \text{(101}\text{.3}\times \text{1}{{\text{0}}^{\text{3}}}\text{)}}{(1.4290\text{ g}{{\text{L}}^{\text{-1}}})}}$
${{\text{V}}_{\text{rms}}}\text{=}\sqrt{\dfrac{303900\text{ Pa}}{(1.4290\text{ g}{{\text{L}}^{\text{-1}}})}}$
${{\text{V}}_{\text{rms}}}\text{=461}\text{.15 m/s}$
Hence, the root means square speed of oxygen gas is$\text{461}\text{.15 m/s}$.

Additional information:
How fast the molecules move indirectly proportional to their absolute temperature (T) and inversely proportional to the mass of the gas-particle.
${{\text{V}}_{\text{rms}}}\text{=}\sqrt{\dfrac{\text{3kT}}{\text{m}}}$
Since we know that,$\text{k=}\dfrac{\text{R}}{{{\text{N}}_{\text{A}}}}$ where ${{\text{N}}_{\text{A}}}$is Avogadro's number
But ${{\text{N}}_{\text{A}}}\text{=}\dfrac{\text{m}}{\text{M}}$ this, $\dfrac{\text{k}}{\text{m}}\text{=}\dfrac{\text{R}}{\text{M}}$
The root means square speed is modified as the${{\text{V}}_{\text{rms}}}$.
${{\text{V}}_{\text{rms}}}\text{=}\sqrt{\dfrac{\text{3RT}}{\text{M}}}$ (1)
Where, ${{\text{V}}_{\text{rms}}}$ is the root mean square speed
R is the gas constant
T is the absolute temperature in kelvin
M is the molar mass
We know that, for an ideal gas
$\begin{aligned} & \text{PV=RT} \\\ & \text{when,n=1} \\\ \end{aligned}$
On substituting in (1) we get,
${{\text{V}}_{\text{rms}}}\text{=}\sqrt{\dfrac{\text{3PV}}{\text{M}}}$
But, density is given by the ratio of mass to volume.
$\text{d=}\dfrac{\text{M}}{\text{V}}$
Thus the equation becomes,
${{\text{V}}_{\text{rms}}}\text{=}\sqrt{\dfrac{\text{3P}}{\text{d}}}$

Note:
In calculating the rms speed use the pressure in the pascal. Before solving such questions first convert the atmospheric pressure to the pascals by the relation.
$1\text{ atm=101}\text{.3}\times \text{1}{{\text{0}}^{3}}\text{ Pa}$
The condition of ${{0}^{0}}\text{c}$ and of the 1 atm pressure is called the $\text{STP}~$conditions commonly used in scientific calculations.