Reconciling the Temperature and Pressure Relationship in Gasses vs Fluids

Question

I've been trying to wrap my mind around this for the last couple of days to no avail. I suspect that much of my confusion comes from the fact that I have not yet taken a class specifically focused on thermodynamics, so my current perception of this likely holds a number of fundamental misunderstandings.

My question, essentially, is what exactly the relationship between temperature and pressure is in a massive volume of a fluid. To clarify (I hope) the aim of my inquiry, I began thinking about this from the perspective of my rudimentary understanding of the temp/pressure relationship in ideal gasses. From this perspective, temperature and pressure seem to be two sides of the same coin - two (equivalent?) measures of the energy in the system available to do work on its surroundings, because both are determined by the kinetic energy transferred over the sum of particle collisions being observed.

So, in a massive volume of fluid where pressure increases greatly with depth, could that fluid come to state of thermal equilibrium? If it can, how do we explain why the the temperature near the surface at relatively low pressures is the same as the temperature at larger depths and pressures within the same volume? If the pressure is greater at depth, that must mean that there are more particle collisions over a unit time than at low pressures (more energy transferred to an observer at these depths), and would this not also imply a higher temperature???

I hope the aim of my question here is discernible

Answer 1

My question, essentially, is what exactly the relationship between temperature and pressure is in a massive volume of a fluid

Unfortunately, it depends upon which type of pressure you are interested in. There are multiple types of pressure that sum to create the total pressure of a system.

For example, one can examine Bernoulli's equation and see what appear to be three types of pressure: $$ \frac{1}{2} \rho \ v^{2} + \rho \ g \ h + P_{o} = constant \tag{1} $$ where $\rho$ is the mass density of the fluid, $v$ is a bulk flow speed, $g$ is the local acceleration of gravity, $h$ is a relative height from a reference plane (you get to choose this), and $P_{o}$ is the particle pressure at the observation point. Note that the first term in Equation 1 is called the ram (or dynamic) pressure and it only applies to situations where a measurement is made in a reference frame moving relative to the fluid (i.e., in the fluid flow rest frame, there is no ram pressure).

In the case of an ideal gas, we can assume that: $$ P_{o} = \sum_{s} \ n_{s} \ k_{B} \ T_{s} \tag{2} $$ where $n_{s}$ is a number density, $k_{B}$ is the Boltzmann constant, and $T_{s}$ is the average temperature of species $s$.

From this perspective, temperature and pressure seem to be two sides of the same coin - two (equivalent?)

No, temperature and pressure are not equivalent in units or qualitative meaning (e.g., see Equation 2 above).

So, in a massive volume of fluid where pressure increases greatly with depth, could that fluid come to state of thermal equilibrium?

Yes of course, why not? The pressure gradient, in this case, is imposed by gravity and held by gravity. So the system can certainly reach hydrostatic equilibrium.

You can consider the Earth's atmosphere as a fluid as well, in which case the pressure can be modeled as an exponential as I explained in this answer.

If it can, how do we explain why the the temperature near the surface at relatively low pressures is the same as the temperature at larger depths and pressures within the same volume?

But the temperature need not be the same at all depths if the column is long enough. For instance, at great depths in the oceans the temperature of the surrounding liquid can be very low but the total pressure measured would be quite large (due to the $\rho \ g \ h$ term in Equation 1).

Similarly, atmospheres are not composed of a single temperature gas. In the ideal gas case, one can explain the temperature variations with density variations over a limited range of altitudes (solar input introduces complications above ~30 km).

If the pressure is greater at depth, that must mean that there are more particle collisions over a unit time than at low pressures (more energy transferred to an observer at these depths), and would this not also imply a higher temperature???

Again, not necessarily. See previous answers/comments above.