The expression of RMS velocity clearly has square root of number of gas molecules in the denominator and square root of squares of velocities of the gas molecules in the numerator. Then why RMS velocity does not depend on number of gas molecules? By looking at the expression, it seems as if it does depend on number of gas molecules.

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    $\begingroup$ Define "depend on." Does any average depend on the number of items going into the average? $\endgroup$
    – Zhe
    Apr 14, 2021 at 15:56
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    $\begingroup$ RMS velocity does not make much sense, as gas does not move anywhere. RMS speed converges for n->infty to sqrt(3RT/M). So, it depends on n as the given formula value is the limit value for a large molecule set, similarly as temperature. $\endgroup$
    – Poutnik
    Apr 14, 2021 at 16:30

2 Answers 2


No, it does not depend directly on the number of molecules. The expression is $$c_{rms}=\sqrt{\frac{3RT}{M}}$$ where $T$ is the temperature in K, $M$ is the molar mass in kg/mol and $R$ is of course the gas constant in J mol-1 K-1.

However, it is possible to have a situation where the number of molecule affects the temperature of the gas, which would change the r.m.s. velocity. For example, in Joule-Thompson effect, increasing the pressure causes the gas to heat up. It might be possible to do this by pumping in more gas into an insulated constant volume vessel. (although it sounds a bit contrived to be honest!)


We normally assume that the molecules are in contact with a heat bath at a fixed temperature $T$ and then the Maxwell distibution of speeds in range $v\to v+dv$ is

$$ F(v)dv = 4\pi n\left(\frac{m}{2\pi k_BT}\right)^{3/2}v^2e^{-mv^2/2k_BT}dv$$

It rises due to the $v^2$ term and falls due to the exponential giving a curve with a gentle maximum. The average speed is found from the integral over all possible speeds or

$$\langle v\rangle =\frac{1}{n}\int_0^\infty vF(v)dv=\cdots= \sqrt{\frac{8k_BT}{\pi m}}$$

where $n$ is the number of particles of mass $m$ and $k_B$ is the Boltzmann constant. You can see that $n$ cancels out. So it does not matter if there is one particle measured many, many times or many just measured once, the result is the same.

The rms is found by a similar argument

$$\langle v^2\rangle =\frac{1}{n}\int_0^\infty v^2F(v)dv=\cdots= \sqrt{\frac{3k_BT}{m}}$$

Notes. The average of a normal function $f(x)$ is $\displaystyle \langle x\rangle=\frac{\int xf(x)dx}{\int f(x)dx}$ and in this distribution $\int F(v)dv =n$


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