The rms speed of an ideal gas is $v_{\text{rms}}$ = $\sqrt{\frac{3RT}{M} }$ and the kinetic energy is $E_\text{k} = \frac32RT$. From this, it is concluded that the speed is mass dependent, while the kinetic energy isn't.

This doesn't make sense to me. I know the speed and kinetic energy for things other than ideal gases are surely mass dependent. The heavier an object is, the slower it is and will have a bigger kinetic energy. Why isn't this true for ideal gases?


As the mass increases, the velocity decreases, yes that is true. But how does the kinetic energy remain the same?

The increment in the mass is balanced by the decrement in the velocity. In other words, the velocity changes as the mass changes in such a manner that the kinetic energy of the molecules of the gas will remain to be same no matter what. It's something like the multiplication of two variables always giving the same constant value. Keeping in mind that this is true in an ideal environment.

  • $\begingroup$ I think that the answer should be pointed out that this assumes ideal gas behavior. $\endgroup$ – MaxW Feb 26 '18 at 19:24
  • $\begingroup$ I added that last bit. Thank you @MaxW. Since his question was "Why isn't this true for ideal gases?" I assumed I was writing for ideal conditions. $\endgroup$ – dr.drizzy Feb 26 '18 at 19:32
  • $\begingroup$ I didn't claim that the kinetic energy doesn't change but that is is independent on the gas mass. as the equation suggests Ek = 1.5RT the only thing that influences the kinetic energy is the temperature. $\endgroup$ – ILoveIL Feb 26 '18 at 20:27
  • $\begingroup$ @ILoveIL Yes, only temperature influences kinetic energy. What is the issue? $\endgroup$ – dr.drizzy Feb 26 '18 at 20:39
  • $\begingroup$ @ILoveIL Two different gases in two separate vessels at the same temperature will have same kinetic energy but not the same Vrms. The difference in masses compensates for different Vrms values. Could you reframe your question again? So that I can understand your problem and help you solve it. $\endgroup$ – dr.drizzy Feb 26 '18 at 20:46

The answer can be $$Speed \propto1/√m$$ And$$ K.E. = 1/2mv^2$$. Finally m will cut, and therefore K.E. does not depend on mass.

  • 2
    $\begingroup$ This directly follows from the formula. I believe the OP was asking for an intuitive approach rather than a formula-based approach. $\endgroup$ – Gaurang Tandon May 15 '18 at 9:37

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