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My textbook states Boyle's law without a proof. I saw Feynman's proof of it but found it to be too handwavy and at the same time it uses Boltzmann's equipartition theorem from statistical mechanics which is too difficult for me now. So to state roughly what Boyle's law is, it states that at a constant temperature and mass of gas, $$PV=k$$ Where $P$ is pressure and $V$ is the volume and $k$ is constant in this case.

Is there a proof for this that isn't based on any other gas law, perhaps based on Newtonian mechanics?

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    $\begingroup$ Does experimental verification count as "out of nothing" here? $\endgroup$
    – Kyle Kanos
    Feb 7 at 18:59
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    $\begingroup$ "Boyle's law, also referred to as the Boyle–Mariotte law, or Mariotte's law (especially in France), is an experimental gas law that describes how the pressure of a gas tends to decrease as the volume of the container increases." en.wikipedia.org/wiki/Boyle%27s_law $\endgroup$
    – anna v
    Feb 7 at 20:20
  • $\begingroup$ this answer/rant of mine might help physics.stackexchange.com/questions/266077/… $\endgroup$
    – anna v
    Feb 7 at 20:24
  • $\begingroup$ The answer to your title is simply "No!". No physical fact can be derived out of nothing. "Nothing In $\to$ Nothing Out". $\endgroup$
    – ACat
    Feb 9 at 1:42
  • $\begingroup$ @Dvij D.C, I am not suggesting to take the literal meaning but to take it in a sense that how to derive boyle's law without basing on other gas laws. $\endgroup$
    – Rahul
    Feb 9 at 1:44

5 Answers 5

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The law can be derived from the kinetic theory of gases. Several assumptions are made about the molecules, and Newton's laws are then applied. For $N$ molecules, each of mass $m$, moving in a container of volume $V$ with a root mean square speed of $c_{rms}$, the pressure, $p$, exerted on the walls by gas molecules colliding with them is given by $$pV=\tfrac 13 Nmc_{rms}^2.$$ Sir James Jeans (in The Kinetic Theory of Gases) has a simple argument involving molecules exchanging energy with a wall (modelled as spheres on springs!) to show that for gases at the same temperature, $mc_{rms}^2$ is the same. In other words, gas temperature is determined by $mc_{rms}^2$. So for a gas at constant temperature, $c_{rms}$ is constant, and if we keep $N$ constant, too, we deduce that $pV$ is constant.

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The pressure exerted by a gas on the walls of its container is proportional to the frequency with which the molecules strike the walls. The larger the container, the more time on average a molecule spends between collisions with the container walls and thus the lower the pressure.

Consider a cylindrical container. If the walls are smooth, then the magnitude of the molecule's velocity along the cylinder's symmetry axis is fixed and changes sign every time the molecule hits the top or bottom of the cylinder. The time it takes the molecule to travel down the cylinder and back up again is proportional to the height of the cylinder, and so is also, for a fixed cross sectional area, proportional to the volume. This proportionality gives you the constant $PV$, since time is inversely proportional to frequency.

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  • $\begingroup$ There is, I think, an important point that is somewhat obscured in this answer. Extending the cylinder thought experiment a bit: what happens if we double the dimensions of this cylinder in every direction, not just allong the symmetry axis? Then the time for molecules to travel between wall bounces doubles, suggesting that the pressure halves. But the volume of the cylinder increased eight-fold! So this suggest something like $P^3V = const$... Of course, the mistake in my reasoning here is that we fail to take into account the increase of the surface area by a factor of 4. $\endgroup$
    – Marc Paul
    Feb 8 at 13:41
  • $\begingroup$ @MarcPaul exactly, the factor of 4 increase in each wall's surface area, coupled with a factor 2 increase in the travel time between collisions, gives a factor 8 reduction in the pressure, which balances the factor 8 increase in volume. $\endgroup$ Feb 8 at 13:49
  • $\begingroup$ I agree, but I think this could be made clearer in the answer. The first line says "The pressure exerted by a gas on the walls of its container is proportional to the frequency with which the molecules strike the walls. " It is the frequency of striking a fixed area of wall that matters, not the frequency of striking the total wall area, and reformulating that line a bit could resolve this ambiguity. Also, in your cylinder experiment, why do we not have to account for the change in surface area of the cylinder? I think I know why, but it would be helpful if you explain it in the answer. $\endgroup$
    – Marc Paul
    Feb 8 at 14:15
  • $\begingroup$ @MarcPaul I agree with all of your points. I will try to revise the answer if I get time. Thanks for the input! $\endgroup$ Feb 8 at 16:10
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Yes, it's almost all very intuitive.

Let's figure out the force $F$ that the gas applies to a (flat) section of its container wall, with area $A$.

You know that $F = ma = \frac{dmv}{dt} = \frac{dp}{dt}$, that is the rate at which momentum is transferred from the wall to molecules of the gas.

This is of course proportional to the rate at which molecules collide with the wall, which is in turn proportional to the area of the wall, $A$, because that's how much is exposed to the gas, the density of gas molecules, $n/V$, and the average speed of those molecules, $|v|$, because if they go faster then they collide more often.

It's also proportional to the average momentum imparted per collision. When a particle collides, the component of its momentum normal to the wall is reversed, so the total momentum imparted is proportional to its average momentum, or $m|v|$.

Putting that all together, we have, for some constant $R$:

$$F = \frac{nRAm|v|^2}{V} \Leftrightarrow P=\frac{nRm|v|^2}{V}\Leftrightarrow PV=nRm|v|^2$$

We're almost there. Temperature, $T$, is average kinetic energy per molecule, and $m|v|^2$ certainly has the right units. But remember that $|v|$ is the average speed of each molecule, and (average $|v|$)2 is not the same as average $(|v|^2)$. However, if we assume that the shape of the distribution of velocities doesn't depend on the average speed, or the molecular mass, then these two quantities will be proportional, and we can fold the factor into our constant $R$, finally leaving $PV = nRT$.

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The law can be derived from the theory of an ideal gas, which is based on classical mechanics. Boyle originally mentioned two atomistic explanations for the elasticity of air:

  • that air particles were repulsive

  • that elasticity was due to impacts of air moelcules on each other

Newton actually showed that if the first was true, then the repulsive force would be inversely proportional to the force. However, it is the latter hypothesis that was picked up and formed the basic explanadum for the kinetic theory of gases, and in particular for the ideal theory of gases.

From this theory, we derive the equation of state:

$pV = nRT$

And this incorporates Boyles law: at a fixed temperature, the volume of gas is inversely proportional to pressure.

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    $\begingroup$ Do you mean inversely proportional to the distance? $\endgroup$ Feb 8 at 13:05
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Boyle's Law is based on the equation of state for an ideal gas which is

$$PV=nRT$$

Where $R$ is the universal gas constant.

Boyle's law applies to a constant mass (closed) system where $n$ is constant and temperature is constant, and therefore $PV=constant$.

For two equilibrium states 1 and 2, then

$$P_{1}V_{1}=P_{2}V_{2}$$

Hope this helps.

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    $\begingroup$ That's how I wave my hands (being a Chemical Engineer and all), but I can understand why someone down-voted you. The OP was looking for a theoretical proof. $\endgroup$
    – Flydog57
    Feb 8 at 4:06
  • $\begingroup$ @Flysog57 Why don’t you post an answer of your own if you feel you can do better $\endgroup$
    – Bob D
    Feb 8 at 7:17
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    $\begingroup$ I think you misunderstood my comment. I'm a chemical engineer. I looked at the question and said "why that's just the result of the ideal gas equation". Then I re-read the question and said "what, physicists can prove this somehow?" and read the other answers. I don't have a better answer than yours - I was really just pointing out to you the likely reason why someone downvoted your answer. $\endgroup$
    – Flydog57
    Feb 8 at 14:47

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