Question

If ${{\text{T}}_{\text{1}}}{\text{,}}{{\text{T}}_{\text{2}}}{\text{,}}{{\text{T}}_3}$ are the temperatures at which the ${{\text{U}}_{{\text{RMS}}}},{{\text{U}}_{{\text{average}}}},{{\text{U}}_{{\text{MP}}}}$ of oxygen gas are all equal to $1500{\text{ m/s}}$ then the correct statement is-
A.${{\text{T}}_{\text{1}}} > {{\text{T}}_{\text{2}}} > {{\text{T}}_{\text{3}}}$
B.${{\text{T}}_{\text{1}}}{\text{ < }}{{\text{T}}_{\text{2}}}{\text{ < }}{{\text{T}}_{\text{3}}}$
C.${{\text{T}}_{\text{1}}}{\text{ = }}{{\text{T}}_{\text{2}}}{\text{ = }}{{\text{T}}_{\text{3}}}$
D. None of these

Answer 1

Hint: Maxwell- Boltzmann distribution tells us about the distribution of speeds for a gas at a certain temperature.
Velocities according to Maxwell-Boltzmann distribution are given as-
Root mean square velocity=$\sqrt {\dfrac{{3KT}}{m}} $ , Average velocity=$\sqrt {\dfrac{{8KT}}{{\pi m}}} $ and most probable velocity=$\sqrt {\dfrac{{2KT}}{m}} $
Where T is the temperature, m is the mass of the gas and K is Boltzmann constant. Here the value of all the velocities is same and since the gas is also same so the value of K and m will remain same in all three velocities. Put the values in the formula and solve to get the value of temperature ${T_1}$ , ${T_2}$ and ${T_3}$ .

Complete step by step answer:
Given that ${{\text{T}}_{\text{1}}}{\text{,}}{{\text{T}}_{\text{2}}}{\text{,}}{{\text{T}}_3}$ are the temperatures at which the velocities${{\text{U}}_{{\text{RMS}}}},{{\text{U}}_{{\text{average}}}},{{\text{U}}_{{\text{MP}}}}$ of oxygen gas are all equal to $1500{\text{ m/s}}$ then we have to find the order of relation between the temperatures.
According to Maxwell-Boltzmann distribution of molecular speeds, the value of molecular speeds is-
Root mean square velocity=$\sqrt {\dfrac{{3KT}}{m}} $
Average velocity=$\sqrt {\dfrac{{8KT}}{{\pi m}}} $
And most probable velocity=$\sqrt {\dfrac{{2KT}}{m}} $
Where T is the temperature, m is the mass of the gas and K is Boltzmann constant.
Here the value of K and m will be same for all velocities as the gas is same and the value of all velocities will also be same at the temperatures ${{\text{T}}_{\text{1}}}{\text{,}}{{\text{T}}_{\text{2}}}{\text{,}}{{\text{T}}_3}$ respectively.
Then we get-
$ \Rightarrow 1500 = 1.73\sqrt {\dfrac{{K{T_1}}}{m}} $
$ \Rightarrow 1500 = 1.60\sqrt {\dfrac{{K{T_2}}}{m}} $
$ \Rightarrow 1500 = 1.41\sqrt {\dfrac{{K{T_3}}}{m}} $
On solving we get,
$ \Rightarrow \sqrt {\dfrac{{K{T_1}}}{m}} = \dfrac{{1500}}{{1.73}}$
$ \Rightarrow \sqrt {\dfrac{{K{T_2}}}{m}} = \dfrac{{1500}}{{1.60}}$
$ \Rightarrow \sqrt {\dfrac{{K{T_3}}}{m}} = \dfrac{{1500}}{{1.41}}$
On simplifying further we get,
$ \Rightarrow \sqrt {\dfrac{{K{T_1}}}{m}} = 867.05$
$ \Rightarrow \sqrt {\dfrac{{K{T_2}}}{m}} = 937.5$
$ \Rightarrow \sqrt {\dfrac{{K{T_3}}}{m}} = 1063.82$
So it is clear that the value ${T_3}$ is more than the value of ${T_2}$and value of ${T_2}$is more than that of ${T_1}$
So we can say, ${{\text{T}}_{\text{1}}}{\text{ < }}{{\text{T}}_{\text{2}}}{\text{ < }}{{\text{T}}_{\text{3}}}$

Hence the correct answer is B.

Note:
According to this, all the gas molecules do not travel at the same velocity. Each possesses different velocity. The gas has three velocities-
1.Most probable velocity-is the velocity that most of the gas molecules of the same mass possess.
2.Average velocity- is the mean of all the velocities of all the gas molecules.
3.Root mean square velocity-is the square-root of the mean of the squares of velocities of all the gas molecules.