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JEE Main 201912 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Evening Shift
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main 2019 · 12 April · Evening Shift · Physics · Kinetic Theory of Gases

The number density of molecules of a gas depends on their distance...

The number density of molecules of a gas depends on their distance r from the origin as, n(r)=n0e−αr4.n\left(r\right) = {n}_{0}{e}^{- \alpha{r}^{4}} . Then the numer of molecules is proportional to:

Options

  1. An0α−3{n}_{0}{\alpha}^{- 3}
  2. Bn0α12\sqrt{{n}_{0}}{\alpha}^{\dfrac{1}{2}}
  3. Cn0α−34{n}_{0}{\alpha}^{\dfrac{- 3}{4}}
  4. Dn0α14{n}_{0}{\alpha}^{\dfrac{1}{4}}

Correct answer

C. n0α−34{n}_{0}{\alpha}^{\dfrac{- 3}{4}}

Step-by-step solution

Density of molecules of gas as a of r is, n(r)=n0e−αr4n ( r ) = {n}_{0}{e}^{- \alpha{r}^{4}}
Total no. of molecules =∫0∞n(r)dv= \mathop{∫}_{0}^{\infty}n ( r ) dv
∫0∞n0e−αr44πr2dr\mathop{∫}_{0}^{\infty}{n}_{0}{e}^{- \alpha{r}^{4}}4\pi{r}^{2}dr
To integrate this, substitute u=α3/4r3u = {\alpha}^{3 / 4}{r}^{3} this will give us dr=13α3/4r2dudr = \dfrac{1}{3{\alpha}^{3 / 4}{r}^{2}}du
Now, the integration will become, 4πn03α3/4∫0∞e−u4/3du\dfrac{4\pi{n}_{0}}{3{\alpha}^{3 / 4}}\mathop{∫}_{0}^{\infty}{e}^{- {u}^{4 / 3}}du
This shows that the numer of molecules is proportional to n0α−3/4{n}_{0}{\alpha}^{- 3 / 4}
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