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Question: If \({{\text{T}}_{\text{1}}}{\text{,}}{{\text{T}}_{\text{2}}}{\text{,}}{{\text{T}}_3}\) are the temp...

If T1,T2,T3{{\text{T}}_{\text{1}}}{\text{,}}{{\text{T}}_{\text{2}}}{\text{,}}{{\text{T}}_3} are the temperatures at which the URMS,Uaverage,UMP{{\text{U}}_{{\text{RMS}}}},{{\text{U}}_{{\text{average}}}},{{\text{U}}_{{\text{MP}}}} of oxygen gas are all equal to 1500 m/s1500{\text{ m/s}} then the correct statement is-
A.T1>T2>T3{{\text{T}}_{\text{1}}} > {{\text{T}}_{\text{2}}} > {{\text{T}}_{\text{3}}}
B.T1 < T2 < T3{{\text{T}}_{\text{1}}}{\text{ < }}{{\text{T}}_{\text{2}}}{\text{ < }}{{\text{T}}_{\text{3}}}
C.T1 = T2 = T3{{\text{T}}_{\text{1}}}{\text{ = }}{{\text{T}}_{\text{2}}}{\text{ = }}{{\text{T}}_{\text{3}}}
D. None of these

Explanation

Solution

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=3KTm\sqrt {\dfrac{{3KT}}{m}} , Average velocity=8KTπm\sqrt {\dfrac{{8KT}}{{\pi m}}} and most probable velocity=2KTm\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 T1{T_1} , T2{T_2} and T3{T_3} .

Complete step by step answer:
Given that T1,T2,T3{{\text{T}}_{\text{1}}}{\text{,}}{{\text{T}}_{\text{2}}}{\text{,}}{{\text{T}}_3} are the temperatures at which the velocitiesURMS,Uaverage,UMP{{\text{U}}_{{\text{RMS}}}},{{\text{U}}_{{\text{average}}}},{{\text{U}}_{{\text{MP}}}} of oxygen gas are all equal to 1500 m/s1500{\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=3KTm\sqrt {\dfrac{{3KT}}{m}}
Average velocity=8KTπm\sqrt {\dfrac{{8KT}}{{\pi m}}}
And most probable velocity=2KTm\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 T1,T2,T3{{\text{T}}_{\text{1}}}{\text{,}}{{\text{T}}_{\text{2}}}{\text{,}}{{\text{T}}_3} respectively.
Then we get-
⇒1500=1.73KT1m\Rightarrow 1500 = 1.73\sqrt {\dfrac{{K{T_1}}}{m}}
⇒1500=1.60KT2m\Rightarrow 1500 = 1.60\sqrt {\dfrac{{K{T_2}}}{m}}
⇒1500=1.41KT3m\Rightarrow 1500 = 1.41\sqrt {\dfrac{{K{T_3}}}{m}}
On solving we get,
⇒KT1m=15001.73\Rightarrow \sqrt {\dfrac{{K{T_1}}}{m}} = \dfrac{{1500}}{{1.73}}
⇒KT2m=15001.60\Rightarrow \sqrt {\dfrac{{K{T_2}}}{m}} = \dfrac{{1500}}{{1.60}}
⇒KT3m=15001.41\Rightarrow \sqrt {\dfrac{{K{T_3}}}{m}} = \dfrac{{1500}}{{1.41}}
On simplifying further we get,
⇒KT1m=867.05\Rightarrow \sqrt {\dfrac{{K{T_1}}}{m}} = 867.05
⇒KT2m=937.5\Rightarrow \sqrt {\dfrac{{K{T_2}}}{m}} = 937.5
⇒KT3m=1063.82\Rightarrow \sqrt {\dfrac{{K{T_3}}}{m}} = 1063.82
So it is clear that the value T3{T_3} is more than the value of T2{T_2}and value of T2{T_2}is more than that of T1{T_1}
So we can say, T1 < T2 < T3{{\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.