There is gas particles inside a piston, if the piston was pushed inward, the volume will decrease and the pressure will increase, so my question is What will happen to the speed of the gas particles?


Kinetic theory of gases connects the macroscopic properties $(eg, P,V,T)$ to the microscopic properties $(eg, v_{rms})$ of a system. The assumptions of KTG apply to my answer.

Let me first state the symbols that I am going to use and the parameter they represent :

$P$ - Pressure of gas , $V$- Volume of gas , $T$ - Temperature of gas , $n$ - number of moles of gas , $M$ - Mass of gas in V volume , $\rho$ - density of gas , and $M'$ - Molecular mass of gas.

According to KTG,

$P=\frac{\rho v_{rms}^{2}}{3}$.

$v_{rms}^{2}=\frac{3PV}{M}$ ,

Using ideal gas equation, $PV=nRT$

$v_{rms}^{2}=\frac{3RT}{M'}$ , beacuse $M=nM'$.

As you can see,

$v_{rms}\propto \sqrt{T}$.

The change is speed of gas particles actually depends on the type of process the system undergoes.

We can have two cases :

1) When $\Delta T =0$:

If $\Delta {T}=0$ (isothermal process, $P \propto \frac{1}{V}$), then there will be no change in the speed the of the gas particles.

In this case when the gas is compressed to half the volume, the pressure is doubled. Therefore for such case, speed of the gas particles remains the same.

2) When $\Delta T \neq 0$ :

If $\Delta T \neq 0 $ (Say, adiabatic process), then the speed of the particles will increase with increase in temperature and vice versa. Adiabatic expansion cools the system and adiabatic compression heats up the system.

Similarly, we can make comparisons , For :

Isobaric processes - $\Delta P=0, V\propto T \propto v_{rms}^{2}$ and

isochoric processes - $\Delta V =0,P\propto T \propto v_{rms}^{2}$.

Conclusion : A general comment cannot be made. The process that the system is undergoing has to be mentioned.

  • $\begingroup$ I don't fully understand. Assume we have a gas - hydrogen particles (single) under some temperature T and Vrms. Now the single hydrogen particles merge to form H2, assume reaction does not yield any energy - the Vrms stays the same while M' doubles. The equation Vrms^2 = 3RT/M' suggests that T will double if the particles merge! Does not seem to make sense, what am I missing? $\endgroup$ Oct 22 '18 at 11:54
  • $\begingroup$ Here, you have three variable $v_{rms}$, T and M' and to make a comment on T you must first specify which variable has to put constant because we can never know how will the change in M' affect $v_{rms}$ and T. T will surely double, as you pointed out, but only if $v_{rms}$ remains the same. $\endgroup$
    – Mitchell
    Oct 22 '18 at 18:05
  • $\begingroup$ And if you double M', you can imagine bigger molecules moving with the same speed (i.e, more momentum or more kinetic energy) as before and having more internal energy. Think of these two systems separately rather than one being derived from the former because the interaction between gases dont matter here. $\endgroup$
    – Mitchell
    Oct 22 '18 at 18:07
  • $\begingroup$ Thanks, what I mean is that we can form larger molecules moving at the same speed at no extra energy cost - the molecules simply bind together without changing speed. Therefore Vrms stays constant (molecules simply bind together) and M' and T of course double. The temperature of the gas suddenly increased without any energy input, is this even possible? $\endgroup$ Oct 23 '18 at 20:45

If the system is thermally isolated then the pressure and the temperature of the gas will increase.
The reason for this is that as the piston moves inward the rebound speed of the molecules will be larger than the speed before rebound.
After the molecules have collided with others this produces an increase in the overall kinetic energy due to the random motion of the molecules i.e. the gas temperature has increased.
The pressure has increased because the rate of collision between the molecules and the walls has increased and the change in momentum of a molecule when it rebounds has also increased.

If the system has walls which are at a constant temperature then on average the kinetic energy of the molecules does not change but the rate at which the molecules hit the walls increases which produces an increase in pressure.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.