Two kinds of pressure

[Demystifier]

In statistical physics, the pressure depends on temperature. For example, for ideal gas we have
$$P=NRT/V$$
In fluid mechanics, the pressure does not depend on temperature. For example, for a static fluid in the gravitational field we have
$$P=\rho g h$$
What is the relation between those two kinds of pressure? Is the total pressure the sum of those two? Is there a unified treatment of pressure in which the two kinds of pressure are derived from a single theoretical framework?

EDIT: I have solved the problem in post #20, so you can jump to it.

 

[jbriggs444]

In statistical physics, the pressure depends on temperature. For example, for ideal gas we have
$$P=NRT/V$$

This is an absolute pressure. The pressure of a gas at equilibrium in a sealed container on the walls of that container.

In fluid mechanics, the pressure does not depend on temperature. For example, for a static fluid in the gravitational field we have
$$P=\rho g h$$

This is a pressure difference. The pressure difference between the top and bottom of a container of height $h$ under the assumption of equilibrium for a fluid of uniform density $\rho$ in a uniform gravitational field of strength $g$.

What is the relation between those two kinds of pressure? Is the total pressure the sum of those two?

"Total pressure" includes a contribution from the bulk motion of the fluid. So it includes both thermal energy (hot air) and bulk kinetic energy (a windy day).

[I am no expert. I had to Google up "total pressure"]

 

[Lnewqban]

Pressure is only one.
One equation considers the limitation in volume occupied by the fluid, the second is used for a fluid not limited by a fixed volume.
In the case of the atmosphere, the temperature and density of the air varies with altitude.

 

[Dale]

The formula for the second one is wrong. The law for a fluid in static equilibrium is $$\frac{dP}{dh}=-\rho g$$ For constant $\rho$ and $g$ this has the solution $$P=P_0-\rho g h$$

Many sources omit the $P_0$ term. But it is important for questions like this. With it both equations can be simultaneously satisfied together.

 

[Demystifier]

The formula for the second one is wrong. The law for a fluid in static equilibrium is $$\frac{dP}{dh}=-\rho g$$ For constant $\rho$ and $g$ this has the solution $$P=P_0-\rho g h$$

Many sources omit the $P_0$ term. But it is important for questions like this. With it both equations can be simultaneously satisfied together.

Would you say that the thermal pressure in the first post corresponds to $P_0$?

 

[Dale]

Would you say that the thermal pressure in the first post corresponds to $P_0$?

I think that the thermal pressure in the first corresponds to $P$. So combining them you have $$\frac{NRT}{V}=P_0 - \rho g h$$ Then specifically at $h=0$ you do get $$\frac{NRT}{V}=P_0$$

However, you do have to be careful because the formula applies for constant $g$ and constant $\rho$. If they are not constant then you have to use the differential equation above.

 

[Demystifier]

I think that the thermal pressure in the first corresponds to $P$. So combining them you have $$\frac{NRT}{V}=P_0 - \rho g h$$ Then specifically at $h=0$ you do get $$\frac{NRT}{V}=P_0$$

Does it mean that the concentration $n=N/V$ depends on $h$? Or that $T$ depends on $h$? Neither option seems right to me.

 

[kuruman]

Is the total pressure the sum of those two?

I would say "yes". If you have a glass of water filled to height $h$ from the bottom, the pressure at the bottom is $$P=P_0+\rho gh$$ where $P_0$ is the pressure at the free surface of the water. Under ordinary circumstances, $P_0$ is atmospheric pressure which we usually take for granted. Now place the glass in a pressurized box such that the air occupies volume $V$ at temperature $T$, the pressure at the bottom of the glass is $$P=\frac{NkT}{V}+\rho gh.$$

 

[Dale]

Does it mean that the concentration $n=N/V$ depends on $h$? Or that $T$ depends on $h$? Neither option seems right to me.

If the size of the gas is large enough that $n$ is not constant then $\rho$ is also not constant and you have to solve the differential equation. The above simplified equation essentially only works where $P_0$ is large enough that you can treat $n$ (and $\rho$) as constant.

 

[Demystifier]

If the size of the gas is large enough that $n$ is not constant then $\rho$ is also not constant and you have to solve the differential equation. The above simplified equation essentially only works where $P_0$ is large enough that you can treat $n$ (and $\rho$) as constant.

What are the general fundamental equations from which all this can be systematically derived?

 

[pines-demon]

I think that the thermal pressure in the first corresponds to $P$. So combining them you have $$\frac{NRT}{V}=P_0 - \rho g h$$ Then specifically at $h=0$ you do get $$\frac{NRT}{V}=P_0$$

However, you do have to be careful because the formula applies for constant $g$ and constant $\rho$. If they are not constant then you have to use the differential equation above.

Are these two even compatible? One if for incompressible fluids (liquids) and one is for ideal gases, for example pressure in the atmosphere has an exponential behaviour not a linear one...

 

[Dale]

One if for incompressible fluids (liquids)

$\rho g h$ can also be used in gasses under certain conditions. I think I covered the limitations pretty well already above.

 

[pines-demon]

What are the general fundamental equations from which all this can be systematically derived?

Isn't it (non-trivially) the equations of statistical mechanics?

 

[Demystifier]

Isn't it (non-trivially) the equations of statistical mechanics?

How do you get $\rho g h$ in statistical mechanics? It doesn't even depend on temperature.

 

[jbergman]

I read or someone told me that pressure was just an isotropic force (per area). As such there could be different mechanisms that cause pressure. Is that incorrect?

 

[Baluncore]

I read or someone told me that pressure was just an isotropic force (per area). As such there could be different mechanisms that cause pressure. Is that incorrect?

Not really. Where pressure is applied to an object, it is the shape and orientation of the object that responds to the force. It does not depend on the source or origin of the force or pressure.

The unit of pressure is the pascal, Pa, it has dimensions of force per unit area.
The force acts through the fluid, always at the normal to the bounding surfaces.

 

[256bits]

What is the relation between those two kinds of pressure?

Are not both derivations obtained from Newton's laws of motion?

The basic definition of pressure is just P = F/A, where F is the normal force acting on an area A.
That P is the P used in the ideal gas equation PV = nRT.

 

[A.T.]

I think that the thermal pressure in the first corresponds to $P$. So combining them you have $$\frac{NRT}{V}=P_0 - \rho g h$$ Then specifically at $h=0$ you do get $$\frac{NRT}{V}=P_0$$

What if T = 0 ?

 

[Frabjous]

The first equation is an equation of state, a material property. The second equation is hydrostatic equilibrium. One needs to solve these equations simultaneously. You might check out the literature for standard atmospheres or density of planets/stars.

N/V is essentially density.
If one wants to worry about varying T, you are going to have to throw in thermal transport.

 

[Demystifier]

The first equation is an equation of state, a material property. The second equation is hydrostatic equilibrium. One needs to solve these equations simultaneously. You might check out the literature for standard atmospheres or density of planets/stars.

N/V is essentially density.
If one wants to worry about varying T, you are going to have to throw in thermal transport.

Ah, I get it now, thanks!

Let me evaluate this a bit. The $p=NRT/V$ can be written as
$$p=w\rho$$
where $\rho=mN/V$, $m$ is the mass of a single molecule, and $w=RT/m$. The $p=\rho gh$ should actually be replaced with
$$\frac{dp(z)}{dz}=-g\rho(z)$$
Thus we have two coupled equations for $p(z)$ and $\rho(z)$, their solution is
$$p(z)=w\rho_0 e^{-gz/w}$$
$$\rho(z)=\rho_0 e^{-gz/w}$$
For small $z$ this becomes
$$p(z) \approx w\rho_0 - \rho_0 gz$$
$$\rho(z) \approx \rho_0 - \frac{\rho_0 g}{w}z$$
Defining the concentration $n(z)=\rho(z)/m$, the formula for $p(z)$ above can be written as
$$p(z) \approx RTn_0 - mn_0 gz$$
where the first term is the usual thermal pressure of the ideal gas, while the second term is the usual dependence of pressure on depth.

It's all clear now!

 

[Dale]

How do you get $\rho g h$ in statistical mechanics? It doesn't even depend on temperature.

That comes from Newton's 2nd law in hydrostatic equilibrium in a uniform gravitational field.

 

[Dale]

What if T = 0 ?

Then I don't think that the ideal gas law holds.

 

[pines-demon]

Ah, I get it now, thanks!

Let me evaluate this a bit. The $p=NRT/V$ can be written as
$$p=w\rho$$
where $\rho=mN/V$, $m$ is the mass of a single molecule, and $w=RT/m$. The $p=\rho gh$ should actually be replaced with
$$\frac{dp(z)}{dz}=-g\rho(z)$$
Thus we have two coupled equations for $p(z)$ and $\rho(z)$, their solution is
$$p(z)=w\rho_0 e^{-gz/w}$$
$$\rho(z)=\rho_0 e^{-gz/w}$$
For small $z$ this becomes
$$p(z) \approx w\rho_0 - \rho_0 gz$$
$$\rho(z) \approx \rho_0 - \frac{\rho_0 g}{w}z$$
Defining the concentration $n(z)=\rho(z)/m$, the formula for $p(z)$ above can be written as
$$p(z) \approx RTn_0 - mn_0 gz$$
where the first term is the usual thermal pressure of the ideal gas, while the second term is the usual dependence of pressure on depth.

It's all clear now!

So basically the equation is Bernoulli's principle...

 

[Demystifier]

I made a small error nobody noticed, in all equations the $R$ should be replaced with the Boltzmann constant $k$. The $R$ comes when the number of particles is expressed in moles, which no physicist does, only chemists do that.

Let me also make a few notes for liquids like water. The equation of state for such liquids is
$$\rho(p)=c_s^{-2}p+\rho_0$$
where $c_s$ is the speed of sound and $\rho_0$ is a constant not depending on $p$. If the water is not too deep then it is not compressed much, i.e. $c_s^{-2}p \ll \rho_0$. Thus we can approximate $\rho$ with $\rho_0$, so the equation
$$\frac{dp}{dz}=-g\rho$$
leads to the usual result
$$p=p_0-\rho_0gz$$
The speed of sound $c_s$ in the equation of state depends on temperature, but we dropped the dependence on $c_s$ because the liquid water is nearly incompressible. The $\rho_0$ depends on temperature too, but not that much, and this dependence is very nonlinear with the maximum at 4 Celsius. Thus we see that the pressure of liquid water does not depend much on temperature, which is related to the fact that liquid water is nearly incompressible.

 

[Demystifier]

A deeper reason that I was confused was that it looked to me as if there was two reasons for the existence of pressure. One reason was the thermal motion of molecules, the other reason was gravitational weight. Now I understand that there is only one reason, only one cause of pressure. This is the thermal motion of molecules in the case if ideal gas, or more generally the thing which determines the equation of state $p=f(\rho)$. Gravity is not the cause of pressure. Not only that can be pressure without gravity (e.g. hot gas in space far from Earth), but also there can be gravity without pressure (e.g. cold gas, that is dust, on Moon).

In particular, the equation $dp/dz=-g\rho$ does not describe the cause of pressure. Instead, this is the condition that needs to be satisfied in order for gas to be static in the presence of gravity. Normally, gas wants to fall down due to gravity. So if the gas is static, it means there is some force acting upwards on the gas, which prevents it from falling down. This is the pressure of the other gas sitting immediately beneath the gas that otherwise would fall down. The equation $dp/dz=-g\rho$ is just an expression of equilibrium, which provides that all forces are canceled, so that the fluid can be at rest. Out of equilibrium, e.g. when the fluid falls down, $dp/dz \neq -g\rho$. For example, if a drop of water is in a free fall, then the pressure inside the drop does not change with $z$.