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This question arises from two statements about the kinetic theory of gases in the Indian NCERT chemistry textbook, chapter 5, p. 148.

These are (emphasis by me):

  • Collisions of gas molecules are perfectly elastic. This means that total energy of molecules before and after the collision remains same. There may be exchange of energy between colliding molecules, their individual energies may change, but the sum of their energies remains constant. If there were loss of kinetic energy. the motion of molecules will stop and gases will settle down. This is contrary to what is actually observed.

  • At any particular time, different particles in the gas have different speeds and hence different kinetic energies. This assumption is reasonable because as the particles collide. we expect their speed to change. Even if initial speed of all the particles was same, the molecular collisions will disrupt this uniformity. Consequently. the particles must have different speeds. which go on changing constantly. It is possible to show that though the individual speeds are changing, the distribution of speeds remains constant at a particular temperature.

Now going along the line of the second statement (i.e.,assuming that initially all particles having the same speed initially) and making additional assumption that ideal gas has same kind of molecules we can show that (via conservation of kinetic energy and momentum) the speed of the particle before and after collision is the same.

  • So why this isn't the case for the statement in the book? Are there some other factors acting in there that change the situation?

  • Is the case (of same speed) possible for real gases? (Assuming that ideal one is possible)

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  • $\begingroup$ When your book wrote all particles having the same speed initially ? $\endgroup$
    – user240696
    Dec 10, 2019 at 12:38
  • $\begingroup$ @Knight I have highlighted it now. $\endgroup$
    – user238497
    Dec 10, 2019 at 12:39
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    $\begingroup$ That’s your understanding, book doesn’t assert that. $\endgroup$
    – user240696
    Dec 10, 2019 at 12:41
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    $\begingroup$ @aditya NCERT Chemistry class 11 part -1 page 149 $\endgroup$
    – user238497
    Dec 10, 2019 at 17:31
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    $\begingroup$ @aditya_stack Read that again. The book isn't asserting the molecules start with the same speed. It's saying that even if they did, this would quickly change into a distribution of speeds. It's saying the exact opposite of what you're claiming it's saying :) $\endgroup$
    – Luaan
    Dec 11, 2019 at 8:39

4 Answers 4

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No you cannot.

we can show that (via conservation of kinetic energy and momentum) the speed of the particle before and after collision is the same.

Only in the center of mass of two elastically colliding particles the momentum remains the same. Each pairwise collision has a different center of mass. In the laboratory frame, which is the frame one is trying to model the ideal gas, all the momentum might be taken by one of the particles, leaving the other motionless in the lab. This happens with billiard ball collisions all the time. See this analysis. So even if one made an experimental setup with all the particles of the ideal gas with the same speed, after the first scatter, speeds will change because they will not all be head on, there will be angles, and then the laboratory versus center of mass argument prevails.

The distribution functions for the ideal gas were given by Maxwell using simple and reasonable assumptions. Boltzman refined this.

It is a model, i.e. a theoretical formula, that has been validated by data over and over again.

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    $\begingroup$ This answer is of course correct in dimensions d>1. For interests sake there are systems in one dimension where (quasi)particle collisions preserve the speeds of the incident particles. Such systems are known as "integrable". These systems do not randomise the speeds of particles in the initial state, or more precisely, they are not ergodic. $\endgroup$
    – ComptonScattering
    Dec 11, 2019 at 3:35
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    $\begingroup$ I wonder if this is correct. The billiard example is simply invalid: Of course the balls do not have initially the same speed; they are all at rest but one. Then you can have the funniest things happen. But can you show to me that two balls of equal mass and size (not sure whether that matters) and equal speed with respect to the observer's inertial system can elastically collide and come away with different speeds? This is what the question boils down to, and I think it is impossible without violating the conservation laws. $\endgroup$
    – Peter - Reinstate Monica
    Dec 11, 2019 at 10:35
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    $\begingroup$ @Peter-ReinstateMonica Yes, they can. If a ball with velocity v moving in the x direction collides elastically with an identical ball with velocity v moving in the y direction, first one will stop while second one will move with velocity √2 v along the 45 degree line. Momentum and kinetic energy is conserved but final speeds are different. $\endgroup$
    – aditya_stack
    Dec 11, 2019 at 11:33
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    $\begingroup$ @Peter-ReinstateMonica Initial kinetic energy of first particle: $1/2 m v^2$, initial kinetic energy of second particle $1/2 m v^2$. Total initial kinetic energy: $m v^2$. Final kinetic energy of first particle: $0$, final kinetic energy of second particle: $1/2 m (\sqrt{2}v)^2$. Total final kinetic energy: $m v^2$. $\endgroup$
    – JiK
    Dec 11, 2019 at 12:56
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    $\begingroup$ @aditya_stack Why will the first one stop and the second continue, rather than the second one stop and the first continue? $\endgroup$
    – Taemyr
    Dec 11, 2019 at 14:15
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Even if initial speed of all the particles was same, the molecular collisions will disrupt this uniformity.

I want to show this happens based on a diagram (on the left) from one of Maxwell's papers.

enter image description here

The centre and right diagrams are for two particles with the same mass travelling at the same speed $u_1 = u_2$ and for ease of presentation it is assumed to be a 2D collision.

The velocities in the laboratory frame of these two particles which are about to collide are $\vec u_1$ and $\vec u_2$ and represented in the vector diagram in the centre as $\vec{OA}$ and $\vec {OB}$.

The velocity of the centre of mass of the two particle system relative to the laboratory frame is represented by $\vec {OG} =\vec V = \frac 12 (\vec u_1 + \vec u_2)$.

The velocities of the two particles in the centre of mass frame are $\vec v_1 = \vec u_1 - \vec V = \frac 12(\vec u_1 -\vec u_2)$ and $\vec v_2 = \vec u_2 - \vec V = \frac 12(\vec u_2 -\vec u_1) = - \vec v_1$.

If the collision is elastic then after collision the magnitude of the velocities of the particles stays the same $(v_1 =v_1'=v_2=v_2')$ and if the collision is not head-on only the direction in which the particles are travelling will change by angle $\gamma$.
The velocities in the centre of mass frame of the particles after the collision, $\vec v'_1$ and $\vec v'_2$, are represented by the vectors $\vec {Ga}$ and $\vec {Gb}$ in the right hand diagram.

The velocities in the laboratory frame of the two particles after the collision are represented by the two vectors $\vec {Oa}=\vec w_1 = v_1' +V$ and $\vec {Ob}=\vec w_2 = v_2' +V$ and they are obviously not equal in magnitude.

So a collision changes the speeds of the particles and you can imagine this happening time and time again and in 3D.

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"we can show that (via conservation of kinetic energy and momentum) the speed of the particle before and after collision is the same."

I think you'll find that for a head-on elastic collision between balls of equal mass, the particles swap velocities. But this is not the case for non-head-on (oblique) collisions. A naïve assumption would be that in this case the velocity swap occurs for components along the line joining the centres at collision, but that each ball retains its own velocity component perpendicular to this line. In that case the speeds of the particles, $\sqrt{v_{perp}^2 + v_{par}^2}$ will in general change, and not merely exchange.

I'm not that impressed by your textbook. In perfectly elastic collision, KINETIC energy is conserved. Energy is conserved however inelastic the collision.

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  • $\begingroup$ It's actually not too hard to prove your "naïve assumption" if you assume that the forces acting between the particles are central along the lines connecting their centers of mass. However, this may not be the case for complicated molecules. $\endgroup$
    – Michael Seifert
    Dec 10, 2019 at 14:20
  • $\begingroup$ Indeed. I could have called the central force assumption naïve. $\endgroup$
    – Philip Wood
    Dec 10, 2019 at 15:06
  • $\begingroup$ Well I think your answer is more perfect that the marked correct answer. I have learned a lot from your short answer. A personal thanks from me. $\endgroup$
    – user240696
    Dec 10, 2019 at 15:52
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    $\begingroup$ Thank you. I used to have exactly the same worries as the original poster, and was pleased when I realised what was going on! $\endgroup$
    – Philip Wood
    Dec 10, 2019 at 16:23
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Well, molecules of the gas can have same speed only if they will be static -- or at absolute zero($0K$ or $-273°C$)[but it is only possible in theory].

It is simple physics of vector movement -- sum of forces applied towards one molecule from surrounding molecules which are also in th state of motion(sum of kinetic energy in limited volume applied towards one molecule).

Although it is not my area of expertise, this is what logic dictates to my understanding.

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  • $\begingroup$ not really an answer to the question. $\endgroup$
    – Bill N
    Dec 11, 2019 at 18:44
  • $\begingroup$ I don't think this explains anything. If it does, it went miles over my head. $\endgroup$
    – Dawood ibn Kareem
    Dec 11, 2019 at 21:28

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