Why can't the molecules of an ideal gas have the same speed? 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):


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*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. 


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*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) 
 A: "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. 
A: 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.
A: 
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.  
 
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.
A: 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.
