In this video we are going to talk about
two physical properties of a species. That will be critical temperature, and critical
pressure, and these two physical properties when compared to the actual system temperature
and pressure will give us an idea of how well we can use the ideal gas equation of state to model that system or whether or not we should look into other equations of state. Now recall
that the ideal gas equation of state relates the pressure and volume to the amount of moles, some gas constant R and the temperature of that system. Now the nice thing about the equation
of state it is relative easy to use no matter what the unknown is given the other conditions
you should be able to calculate it. Another attractive feature, that the gas equation of state is that we can use this regardless of what the species of the gas is. So that basically assumes
that all gas species regardless of their molecular structure will behave the same. Lastly it
does not matter if it is a single species or a mixture we can use the ideal gas equation of state to model that. Now our problem happens when we start getting to temperatures that
start decreasing and get pretty low or pressures that start getting pretty high, at this point
sometimes again depending on the gas species the ideal gas equation of state maybe pretty far off from modeling the PVT data. So the idea is to use the critical temperature and pressure to give us an idea on how close we may be to an ideal state where we can use the ideal gas equation. So let’s start with a little analogy on how we determine the critical temperature
and pressure. Let’s say we have a piston full of ethanol vapor and we are going to heat
this ethanol and lower the pressure so that all the ethanol is a vapor, and what we are
going to do is condense that vapor to a liquid at some specific pressure, and so we will
just say that at some temperature 1, we have a pressure 1 that results in condensation.
So now what we are going to do is increase the temperature a little higher. Which will
result in a needing a higher pressure to reach condensation. So now we have increased. We
will go increase to temperature 2, which is going to result in a pressure 2, again condensation
occurs, and we have both liquid and vapor. Eventually we are going to get a temperature
that we are going to call the critical temperature, were no matter what the pressure is there
is no condensation, and right before this occurs the density if the liquid is virtually
the same same as the density in the vapor. So the last recordings that we can measure
the density of the liquid and the vapor and eventually they are the same. We are going
to take that pressure as the critical pressure, and as I mentioned that critical temperature
and those are the two physical properties of that species. When the temperature and
the pressure is greater then the critical temperature and critical pressure of the
species we are are said to have a super critical fluid meaning that it does not necessarily
behave like a liquid or a gas, and these values of the critical temperature and critical pressure
could typically be looked up in any physical properties hand book in the material and energy balances textbook. We can see that ethanol critical temperature is around 516.3 Kelvin, which
correlates with 243 degree Celsius approximately and the critical pressure reported in atmospheres
is 63 atm. So go through a little bit more and test what we have just gone through
let’s go ahead and take a look at a a conceptual problem that deals with this critical state.
Let’s take a look at the following question. As the temperature of water decreases from
its critical temperature of 374 degrees Celsius, which of the following is true? We are given
4 answers. At this point I will say pause the video and try to answer this on your own. Alright, so now let’s go through each of these to see whether or not A, B,C, or D is true.
Since we are decreasing from its critical temperature of 374 we do not form a super critical fluid.
If we increase above the critical temperature we possibly could form a super critical fluid if we also increase the pressure of the system above the critical pressure. So C is false.
Now as the temperature decreases for the system the saturation of pressure also decreases,
meaning it is going to take a lower pressure of the system to reach condensation. So thus lower pressures are needed for condensation, and B is false. So that leaves us with A or
D. Now if we think of the density of liquid water or the density of the vapor phase of the water, we should know that as temperature decreases the density of the liquid water is going to
increase not decrease, and the density of the vapor phase also decreases. Since there
is a lower temperature we need a lower volume for that same amount of fluid. So A is false
in this case. That leaves us with 1 answer, which is D none of the above. So hopefully
this gives you a good idea of these two parameters and further videos using different equations
of state will definitely use these parameters to help go through those calculations.

• ### Satoshi Chomsky

Speaking too fast for me. But i can replay….

• ### Tamather Elmahdi

THANK YOU!!! <3

good

• ### Elf Nanno

I didn't understand why the B is false

• ### Nora Hernandez

Great video! I think you meant to say that the density of the vapor will also decrease as the temperature decreases not that it is increase.

• ### Renz Bareo

i can't understand why answer A is invalid.

• ### Dev Gupta

Now I understand great