# Why do the temperature drop if we take down the pressure?

## Main Question or Discussion Point

why do the temperature drop if we take down the pressure????

If you have a column with mehane, ethane and butane in a gas liquid phase(at 100 bars),why do the temperature drop if we lower the pressure???

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lets understand this with the help of practical example....
suppose you have a piston and cylinder system contains gas in it and assume that system is isolated from surrounding. Mean no heat and mass transfer in surroundings.
System isolation means the total enthalpy of system remains constant.
so if you compress the gas then all internal energy say heat (enthalpy + other energy) collected in a less volume. So amount of heat per meter cube increased. But in case of expansion the amount of heat per meter cube is decreases.......
Their is no change in total energy occurs.
thanks

Forget the piston and cylinder. I am talking about a coulumn( a fraction coulmn in the gas industry).

If we have a column with liquid and gas(hydrocarbons) at 100bar.. If we now open all valves to depresurise(depresurize to atmosphere) the column down to 20 bar, we will see that the temperature drops. Why??
I have looked at a phase envelope for methane, ethane etc and have seen that when you take down the pressure, you will get more vapour. And i have been told that vapour needs energy and the energy it takes from the surroundings, which in this case is the liquid in the column. And that is why the temperature goes down(i have been told). But do this have anything to do with the molecules(more space between molecules less collision )???????

Forget the piston and cylinder. I am talking about a coulumn( a fraction coulmn in the gas industry).

If we have a column with liquid and gas(hydrocarbons) at 100bar.. If we now open all valves to depresurise(depresurize to atmosphere) the column down to 20 bar, we will see that the temperature drops. Why??
I have looked at a phase envelope for methane, ethane etc and have seen that when you take down the pressure, you will get more vapour. And i have been told that vapour needs energy and the energy it takes from the surroundings, which in this case is the liquid in the column. And that is why the temperature goes down(i have been told). But do this have anything to do with the molecules(more space between molecules less collision )???????

Its because of the evaporation of the liquid. If you look at the vapor pressure of the gas vs temperature you will see that, for any given temperature a liquid will fill the empty volume of its container with vapor at a certain pressure. The higher the temp, the higher the pressure. It takes some energy for the molecules to change from a liquid state to a gaseous state, which lowers the temperature. If you open a valve and release some of the pressure, the liquid will try to return the pressure to what it was before (and then the temperature decreases and the vapor pressure goes down a little). The column you speak of can exist in equilibrium between the gas and liquid at 20 bar only at a lower temperature than 100 bar. This affect is what makes Liquid N2 useful for cooling things.

Helium evaporation is used to get way down to about 3.4 degrees kelvin, below the superconducting temperature of most elemental superconductors.

In response to the piston example, be careful. When you compress a gas with a piston you increase the momentum of the gas particles with each collision between the particles and the moving piston. This increases the temperature according to the kinetic theory of gasses. On the other hand, consider a volume with a divider in the middle, with a volume of gas with temperature T on one side and a vacuum on the other. At time t the divider vanishes. What is the resulting temperature of the gas? The answer is the final temperature equals T. This was a question on my statistical mechanics final which I got wrong...

if you compress the gas then all internal energy say heat (enthalpy + other energy) collected in a less volume. So amount of heat per meter cube increased. But in case of expansion the amount of heat per meter cube is decreases.......
Their is no change in total energy occurs.
thanks
This is not true. In order to compress the gas you must do mechanical work, this adds energy to the system. Heat is not a quantity that you can have per cubic meter, its an amount of energy transferred from one body to another through an interface, and carries a connotation of entropy increase.
More particles with the same energy in a constant volume -> higher pressure, same temp.
Same particles with the same energy in a smaller volume -> higher pressure, same temp.
You did work on the system -> higher pressure, higher temp
You rounded up all of the particles of the system and somehow convinced them to occupy less space without removing entropy from the system -> you violated 2nd law of thermodynamics

Last edited:
HallsofIvy
Homework Helper

"Ideal gas law": PV= nRT where P is the pressure on a gas, V its volume, n the number of molecules, R Avogadro's constant and T is temperature in degrees Kelvin. If your gas is in a rigid container, so that V is constant, and no gas can escape so n is constant, reducing P reduces PV so nRT must reduce which means that T must reduce.

Of course, if your gas is in a rigid container, how do you reduce P without reducing either n? Usually, it is done the other way. You can reduce the temperature from outside so reducing the pressure.

This is not true. In order to compress the gas you must do mechanical work, this adds energy to the system. Heat is not a quantity that you can have per cubic meter, its an amount of energy transferred from one body to another through an interface, and carries a connotation of entropy increase.
More particles with the same energy in a constant volume -> higher pressure, same temp.
Same particles with the same energy in a smaller volume -> higher pressure, same temp.
You did work on the system -> higher pressure, higher temp
You rounded up all of the particles of the system and somehow convinced them to occupy less space without removing entropy from the system -> you violated 2nd law of thermodynamics
thanks its very helpful for me.

thanks....
u made it easy to understand :D

Be careful. The ideal gas law is always a rough approximation. The Redlich-Kwong seemed to be a better approximation of the PV relationships in my experience.

BUT, not all gases cool upon expansion.