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The net rate of effusion of a gas through orifice is given by the following equation:

$$ r = \frac {\Delta P A}{\sqrt {2 π R T M}}$$

Rest being constant, the rate of effusion is inversely proportional to the square root of the absolute temperature.

But why is it so?

I mean as the absolute temperature increases so does the average kinetic energy of the molecules. So shouldn't the rate of effusion increase?

Can you please clarify this? If possible provide molecular picture to grasp this fact.


Here, $R$ is the universal gas constant, $T$ is the absolute temperature of the system as well as surrounding, $\Delta P$ is the partial pressure difference between surrounding and the system, $A$ is the area of orifice and $M$ is the molar mass of the gas being effused.

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Rest being constant, the rate of effusion is inversely proportional to the square root of the absolute temperature.

But why is it so?

I mean as the absolute temperature increases so does the average kinetic energy of the molecules. So shouldn't the rate of effusion increase?

The reason is, if pressure difference stayed the same, density difference had to decrease. So you have faster molecules, but less of them moving out.

You were assuming that after the temperature increase, $\Delta P$ would stay the same. That is a possible scenario to consider, but it requires that both pressures, inside and outside, change in such a way that $\Delta P$ stays the same. That is a very special behavior that would require changing gas densities as well in a special way.

In practice, by increasing temperature of both parts of the system, pressure inside will increase, and pressure outside can increase too (by a lesser amount), or stay the same (if open to Earth's atmosphere or other large system of unchangeable pressure). In both cases, the pressure differential will increase.

Let's consider the simplest case to analyze: both inside (region 1) and outside (region 2) are closed systems of fixed volume. Pressure differential $\Delta p = p_1 - p_2$ is positive, and $n_1 > n_2$.

When both regions are heated up from $T$ to $T+\Delta T$, densities $n_1,n_2$ won't change (same number of molecules per same volume), but their pressures will: using the ideal gas law, we have

$$ \Delta p_1 = n_1k_B\Delta T, ~~~ \Delta p_2 = n_2 k_B\Delta T. $$

And it is easy to see that pressure inside (region 1) will increase more. The new pressure differential $\Delta p'$ will be $$ \Delta p' = (n_1-n_2)k_B(T + \Delta T). $$

So, going back to the original formula, if temperature increases, the denominator increases as $\sqrt{T}$, but the numerator increases as $T$. So the rate of effusion increases as $\sqrt{T}$. This is easy to interpret: density of molecules stayed the same, but their speed increased as $\sqrt{T}$.

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  • $\begingroup$ Thanks for the answer. Even though I haven't specifically stated it, but by saying "rest being constant" I tried to mean that we take two different systems and surroundings where such is possible, and in that case only I want the reason for the inverse square root relationship btwn temperature and rate of effusion. $\endgroup$
    – user249968
    Jul 13 '20 at 13:43
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    $\begingroup$ The reason is, if pressure difference stayed the same,density difference had to decrease. So you have faster molecules, but less of them moving out. $\endgroup$ Jul 13 '20 at 13:44
  • $\begingroup$ Can you please add this comment to your answer, so that I can accept it? $\endgroup$
    – user249968
    Jul 13 '20 at 14:37
  • $\begingroup$ what do you mean by density difference $\endgroup$ May 24 at 4:14
  • $\begingroup$ @lalittolani density difference is $n_1-n_2$, where $n_1$ is number of particles per unit volume in region 1. $\endgroup$ May 24 at 12:20

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