
At $300K$, the number of molecules possessing most probable velocity are $100$. At $400K$ the number of such molecules will be:
A. $65$
B. $100$
C. $110$
D. $120$
Answer
574.8k+ views
Hint: We know that the Maxwell-Boltzmann distribution equation is used to find the relation between the number of molecules and their speed at different temperatures. The Maxwell-Boltzmann distribution equation can be written as $f(c) = 4\pi {c^2}{(\dfrac{m}{{2\pi {k_B}T}})^{\dfrac{3}{2}}}{e^{\dfrac{{m{c^2}}}{{^{2{k_B}T}}}}}$.Here, $f(c) = $distribution of the gas molecules moving at different speeds. $m = $mass of the molecule, ${k_B} = $ Boltzmann constant, $T = $absolute temperature,$c = $speed.
Complete step by step answer:
Here in the question we are given that at $300K$, the number of molecules possessing the most probable velocity are $100$ and we have to find the number of such molecules at $400K$. By using the Boltzmann distribution equation $f(c) = 4\pi {c^2}{(\dfrac{m}{{2\pi {k_B}T}})^{\dfrac{3}{2}}}{e^{\dfrac{{m{c^2}}}{{^{2{k_B}T}}}}}$, we can write that $\dfrac{{{n_1}}}{{{n_2}}} = {(\dfrac{{{T_2}}}{{{T_1}}})^{\dfrac{3}{2}}}$. Here ${n_1},{n_2}$ are the number of molecules at ${T_1},{T_2}$ respectively. We are given the value of -
$
{n_1} = 100 \\
{T_1} = 300K \\
{T_2} = 400K \\
$
And we have to find the value of ${n_2}$. So we will now put the given values in the equation $\dfrac{{{n_1}}}{{{n_2}}} = {(\dfrac{{{T_2}}}{{{T_1}}})^{\dfrac{3}{2}}}$,to find the unknown ${n_2}$.
$
\dfrac{{{n_1}}}{{{n_2}}} = {(\dfrac{{{T_2}}}{{{T_1}}})^{\dfrac{3}{2}}} \\
\Rightarrow {n_2} = {n_1}{(\dfrac{{{T_1}}}{{{T_2}}})^{\dfrac{3}{2}}} \\
\Rightarrow {n_2} = 100{(\dfrac{{300}}{{400}})^{\dfrac{3}{2}}} \\
\therefore {n_2} = 65 \\
$
So, from the above explanation and calculation it is clear to us that the number of such molecules at $400K$ is $65$.
So, the correct answer of the given question is option: A. $65$
Additional information:
The concept of Maxwell-Boltzmann distribution was founded by James Clerk Maxwell and Ludwig Boltzmann. It was a revolutionary concept in the field of classical physics and molecular chemistry.
Note:
Always remember that the Maxwell-Boltzmann distribution equation can be written as $f(c) = 4\pi {c^2}{(\dfrac{m}{{2\pi {k_B}T}})^{\dfrac{3}{2}}}{e^{\dfrac{{m{c^2}}}{{^{2{k_B}T}}}}}$. It is very useful to analyse and calculate the relationship between the number of molecules having a particular speed at a given temperature. So, remember this formula because it is very useful in the study of kinetic theory of gas. Always solve the numerical carefully and avoid silly mistakes and calculation errors.
Complete step by step answer:
Here in the question we are given that at $300K$, the number of molecules possessing the most probable velocity are $100$ and we have to find the number of such molecules at $400K$. By using the Boltzmann distribution equation $f(c) = 4\pi {c^2}{(\dfrac{m}{{2\pi {k_B}T}})^{\dfrac{3}{2}}}{e^{\dfrac{{m{c^2}}}{{^{2{k_B}T}}}}}$, we can write that $\dfrac{{{n_1}}}{{{n_2}}} = {(\dfrac{{{T_2}}}{{{T_1}}})^{\dfrac{3}{2}}}$. Here ${n_1},{n_2}$ are the number of molecules at ${T_1},{T_2}$ respectively. We are given the value of -
$
{n_1} = 100 \\
{T_1} = 300K \\
{T_2} = 400K \\
$
And we have to find the value of ${n_2}$. So we will now put the given values in the equation $\dfrac{{{n_1}}}{{{n_2}}} = {(\dfrac{{{T_2}}}{{{T_1}}})^{\dfrac{3}{2}}}$,to find the unknown ${n_2}$.
$
\dfrac{{{n_1}}}{{{n_2}}} = {(\dfrac{{{T_2}}}{{{T_1}}})^{\dfrac{3}{2}}} \\
\Rightarrow {n_2} = {n_1}{(\dfrac{{{T_1}}}{{{T_2}}})^{\dfrac{3}{2}}} \\
\Rightarrow {n_2} = 100{(\dfrac{{300}}{{400}})^{\dfrac{3}{2}}} \\
\therefore {n_2} = 65 \\
$
So, from the above explanation and calculation it is clear to us that the number of such molecules at $400K$ is $65$.
So, the correct answer of the given question is option: A. $65$
Additional information:
The concept of Maxwell-Boltzmann distribution was founded by James Clerk Maxwell and Ludwig Boltzmann. It was a revolutionary concept in the field of classical physics and molecular chemistry.
Note:
Always remember that the Maxwell-Boltzmann distribution equation can be written as $f(c) = 4\pi {c^2}{(\dfrac{m}{{2\pi {k_B}T}})^{\dfrac{3}{2}}}{e^{\dfrac{{m{c^2}}}{{^{2{k_B}T}}}}}$. It is very useful to analyse and calculate the relationship between the number of molecules having a particular speed at a given temperature. So, remember this formula because it is very useful in the study of kinetic theory of gas. Always solve the numerical carefully and avoid silly mistakes and calculation errors.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

10 examples of friction in our daily life

