How is pressure related to the rate of momentum change in a gas?

[voko]

Is it correct to say that below this critical temperature gases may be converted into liquid?

It is only below the critical temperature that they can. Check out "supercritical fluid".

 

[rogerk8]

What is the exact relationship between pressure and potential energy of a gas?

 

[dauto]

What potential energy? do you mean the amount of work that can be extracted from the gas if allowed to expand? that depends on the exact expansion process. For instance, an adiabatic expansion will give a different result than an isothermic expansion. The work is not a state function.

 

[rogerk8]

Jesus, I know nothing but please consider this:

$$U=KE+PE$$

and

$$KE=kT≈\frac{mv^2}{2}$$

and

$$\frac{PV}{N}=kT$$

Then

$$U=kT+kT$$

?

 

[voko]

You seem to be looking for the internal energy of gas. For an ideal gas, whose molecules do not interact, it is just the kinetic energy. For non-ideal gas, there are many models of varying complexity. Perhaps the simplest one is the van der Waals model.

 

[Andrew Mason]

Jesus, I know nothing but please consider this:

$$U=KE+PE$$

and

$$KE=kT≈\frac{mv^2}{2}$$

and

$$\frac{PV}{N}=kT$$

Then

$$U=kT+kT$$

?

The average translational kinetic energy is NkT/2 for each translational degree of freedom, so <KE> = 3NkT/2

But that is not necessarily the entire internal kinetic energy. For non-monatomic gases, there may be additional degrees of freedom of movement such as rotational or vibrational modes. These contribute to internal energy BUT NOT to temperature. There is an average KE = NkT/2 associated with each additional degree of freedom.

So, letting f = no. of degrees of freedom, the internal average kinetic energy is <KE> =fNkT/2

But internal energy can include potential energy if there are forces between molecules. It takes energy to increase the average separation of the molecules. So when potential energy increases due to increased average separation of molecules, internal energy increases.

According to the equipartition theorem, which is an important part of the kinetic theory, there is an energy NkT/2 associated with each degree of freedom for potential energy as well.

So U = KE + PE = fNkT/2 where f is the total no. of degrees of freedom for kinetic and potential energy.

AM

 

[rogerk8]

Thank you once again Andrew Mason!

I am thankful for all your help!

Now I know that there is a kinetic energy of NkT/2 per degree of freedom and that this is associated with potential energy as well.

My brand new understanding of internal energy is thus

$$U=KE+PE=fNkT/2$$
But what about PE and pressure?

Roger

 

[Andrew Mason]

Thank you once again Andrew Mason!

I am thankful for all your help!

Now I know that there is a kinetic energy of NkT/2 per degree of freedom and that this is associated with potential energy as well.

My brand new understanding of internal energy is thus

$$U=KE+PE=fNkT/2$$
But what about PE and pressure?

Roger

Do you have a specific question about PE and pressure?

First of all you have to have a non-ideal gas with attractive forces between molecules. For such a gas PE increases with volume (as the space between molecules grows, on average, so does PE). If T remains constant (i.e. KE is constant) what happens to the pressure? (hint: how is pressure related to the rate at which molecules change momentum in striking the container walls? If KE is constant, but the distance between walls increases, what happens to the frequency of molecules colliding with the wall?).

AM

 

[rogerk8]

Do you have a specific question about PE and pressure?

First of all you have to have a non-ideal gas with attractive forces between molecules. For such a gas PE increases with volume (as the space between molecules grows, on average, so does PE). If T remains constant (i.e. KE is constant) what happens to the pressure? (hint: how is pressure related to the rate at which molecules change momentum in striking the container walls? If KE is constant, but the distance between walls increases, what happens to the frequency of molecules colliding with the wall?).

AM

I tried understanding this reading sentence by sentence but I felt stupid until I read your hint.

High pressure has to do with a high rate of momentum change, right?

So if the distance to the walls increases, the frequency must lessen (at the same KE), right?

The answer to your riddle must then be that pressure lessens

Roger
PS
While I obviously do not understand physics in spite of beautiful and simple equations here is a couple of naive thoughts:

$$p=U/V=(KE+PE)/V=fnkT/2...[1]$$
and we are almost back at the Ideal Gas Law. The difference being f=2 (instead of 3 for mono-atomic gases, right?)

But I should probably not stare so much at equations and try to understand instead. Here is by the way another thought

$$KE_{av}=fkT/2...[2]$$
which comes from my Plasma Physics book where f is said to be 3 but there's no N here!