2
$\begingroup$

Let us take Boyles law to start.

Assumptions:

  • Gas is perfect.
  • In a massless piston that can be expanded with no friction
  • Adiabatic

If we were to decrease the volume of the piston, the pressure inside would go up because the gas molecules would be hitting the sides more often.

If we were to increase the outside pressure on the piston, the volume of the piston would go down until the internal pressure matched the external pressure.

We see here that Boyle's law is perfectly explained by the kinetic model of gases (T held constant).


Let's look at Charles law (P held constant)

If we were to increase the temperature of the molecules, their kinetic energy would increase and would therefore hit the piston with greater force and increase the volume of the piston until the internal and external pressures are equal.

But if we were to spontaneously increase the volume of the piston, the temperature would NOT increase as a result (to maintain the increased volume against the constant external pressure) because heat doesn't spontaneously arise.

So it seems that Charles law only works one way, but not the other. And that T and V are not intrinsically linked like P and V are in Boyles law. Is this true?

$\endgroup$
  • $\begingroup$ physics.stackexchange.com/q/245808 may or may not be helpful. If you push the piston down (ie, decrease the volume), you are doing work on the molecules in the gas. $\endgroup$ – barrycarter Feb 1 '17 at 3:17
  • 1
    $\begingroup$ When you say temperature would NOT increase, you are probably invoking the fact that internal energy of the ideal gas does not change and so neither does its temperature. This means that neither is the gas doing any work nor is any work being done on it. But then tell me, how would the volume of system increase spontaneously? $\endgroup$ – Deep Feb 1 '17 at 6:33
  • $\begingroup$ Boyle's law is for an isothermal change not an adiabatic one. $\endgroup$ – Farcher Feb 1 '17 at 9:51
  • $\begingroup$ "because heat doesn't spontaneously arise." It is a mistake to think that heat must be there for temperature to rise. Other kinds of energy transfer can also cause temperature change, for example work. $\endgroup$ – Steeven Feb 1 '17 at 9:52
3
$\begingroup$

The answer itself is hidden in the second part of your question. P in any gas law refers to the pressure inside the volume of the container [ which is always equal to the external pressure on the container ]

Now, in the second part of your question, the statements

(P held constant)

&

if we were to spontaneously increase the volume of the piston

cannot be true simultaneously.

As soon as you want to increase the volume of the container (irrespective of the method of achieving such a change -- whether by a reversible piston process or via irreversible free expansion), you have to reduce the external pressure on the piston & thereby the pressure inside the container also reduces & then you can apply kinetic theory accordingly [ Note that unless these two P's are equal, you cannot apply any gas law because they are valid only for equilibrium situations. It's very important to keep this condition in mind when you are thinking about such thought experiments. ]

To conclude, all gas laws work irrespective of whether the change in the variables are positive or negative as long as you are being careful to satisfy all conditions including the condition of equilibrium & reversibility.

$\endgroup$
  • $\begingroup$ Since the original comment has now been deleted, I am rewriting the comment in response to which I made this:- "Isn't Charles law violated when I have two chambers separated by a wall & I remove the wall?" Umm..the process you just described is irreversible free expansion of ideal gas. So, once you have removed the wall and allowed enough time to let the system achieve equilibrium the pressure now will not be equal to the initial pressure. So, even here, $dP=0$ condition does not hold true. $\endgroup$ – Aritra Feb 1 '17 at 14:01
0
$\begingroup$

Compressing a gas that is not at absolute zero will increase its temperature because a finite volume of gas at a finite temperature has a set amount of heat energy. when that set amount of heat energy occupies a smaller space its temperature rises. Conversely when it occupies a larger space its temperature goes down. so both laws are reversible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.