Question

A gas bulb of ${\text{1L}}$ capacity contains $2.0 \times {10^{21}}$ molecules of nitrogen exerting a pressure of $7.57 \times {10^3}{\text{ N}}{{\text{m}}^{ - 2}}$ . Calculate the root mean square speed and temperature of gas molecules. If the ratio of most probable speed to the root mean square speed is $0.82$, calculate the most probable speed of these molecules at this temperature.

Answer 1

Hint: Most probable speed is represented by the speed of the largest number of particles. The most probable speed determines that a large number of particles are moving with this speed. It is represented by the Maxwell Boltzmann distribution graph.

Complete step by step answer:
The ideal gas equation plays an important role in relating volume, moles, pressure and temperature with each other. If we want to calculate any one missing thing from all of these then it is easy to find that from the ideal gas equation.
Ideal gas equation:
${\text{PV = nRT}}$ --------------------(1)
Where, $P$ is the pressure
$V$ is the volume.
$n$ is the number of moles,
$R$ is a gas constant and
$T$ is temperature.
So the amount of gas present =$\dfrac{{{\text{Number of molecules}}}}{{{\text{Avogadro's number}}}}$
Moles of gas, $n$=$\dfrac{{2.0 \times {{10}^{21}}}}{{6.022 \times {{10}^{23}}}}$
Volume given is ${\text{1L}}$ which is equal to ${\text{1}}{{\text{0}}^{ - 3}}{{\text{m}}^3}$
 Pressure is $7.57 \times {10^3}{\text{ N}}{{\text{m}}^{ - 2}}$
And the value of the gas constant, $R = 8.314j/k - mol$
By putting all these values in equation 1
${\text{T = }}\dfrac{{{\text{1}}{{\text{0}}^{ - 3}} \times 7.57 \times {{10}^3}}}{{\dfrac{{2.0 \times {{10}^{21}}}}{{6.022 \times {{10}^{23}}}} \times 8.314}}$
$ \Rightarrow {\text{T = 274}}{\text{.2K}}$
We have to calculate the most probable speed and we have given that the most probable speed is 0.82 times the root mean square speed.
$ \Rightarrow $ most probable speed, ${{\text{V}}_m}$ =$0.82 \times $ root mean square speed ---------------(2)
And we know that, root mean square speed=$\sqrt {\dfrac{{{\text{3RT}}}}{{\text{M}}}} $ ------------(3)
Where, ${\text{R}}$ is gas constant, ${\text{T}}$ is temperature and ${\text{M}}$is molar mass of gas.
From equation (2) and equation (3)
Most probable speed, ${{\text{V}}_m}$ =$0.82 \times \sqrt {\dfrac{{{\text{3RT}}}}{{\text{M}}}} $
Now substitute all the values to find ${{\text{V}}_m}$.
$ \Rightarrow $ Most probable speed, ${{\text{V}}_m}$ =$0.82 \times \sqrt {\dfrac{{{\text{3}} \times {\text{8}}{\text{.314}} \times 274.2}}{{28 \times {{10}^{ - 3}}}}} $
$ \Rightarrow $ Most probable speed, ${{\text{V}}_m}$ =$0.82 \times 494.22$
$ \Rightarrow $ Most probable speed, ${{\text{V}}_m}$ =$405.26{\text{m}}{{\text{s}}^{ - 1}}$

Note:
The root mean square speed of the particles is the square root of average squared velocity of the molecules in a gas which is dependent on the two factors, molar weight and temperature directly. It has units similar to velocity.