$v_{rms}$ of an ideal gas is given by $\sqrt{\dfrac{3RT}{M}}$ .

Now I believe that this depends only on temperature and molar mass of the gas. But someone told me to write RT as $\dfrac{PV}{n}$, then the formula would show pressure dependency at constant volume. But I still believe that it shows only direct temperature dependency because pressure of an ideal gas at constant volume can only(?) be increased by increasing the temperature.

Now, I am just wondering if there's a method to increase the pressure of the gas without us raising the temperature. I mean that the pressure should increase and then the temperature should increase as a result of Gay-Lussac's pressure law (not the other way round).

I think then the claim that $v_{rms}$ depends on pressure will be valid.


1 Answer 1


The only free variable would then be n, which you could raise: add more gas particles. To do this without raising T, you'd need the same $v_{rms}$.

Your wish to raise p without raising T, then have T increase seems contradictory, unless you allow for (exothermic) chemical reactions between different particle types, that kick in slowly.


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