# Why is so different the temperatures of a compressed gas compared to a liquid/gas mixture of the same fluid and they are both at the same pressure?

Why does $$\ce{C3H6}$$ gas when compressed (in a centrifugal compressor) comes out with a pressure of $$\pu{16 bar}$$ and $$\pu{90 °C},$$ and when it is a mixture of $$\ce{C3H6}$$ liquid/gas stored inside of a vessel reaches also the $$\pu{16 bar},$$ but at a lower temperature of around $$\pu{38 °C}?$$

I know that when it is a vessel with stored $$\ce{C3H6}$$ liquid, some part of the liquid will vaporize until it reaches the equilibrium with the ambient temperature (vapor pressure). I know that if the vessel is exposed to a ambient temperature of $$\pu{38 °C}$$ and if has enough $$\ce{C3H6}$$ liquid to vaporize the equilibrium will be reach at a pressure of $$\pu{16 bar}.$$

But what I cannot understand is why if it is only gas and if I compress the gas to $$\pu{16 bar},$$ why does it come out with a temperature of $$\pu{90 °C}?$$ Shouldn't it come out with the same temperature of $$\pu{38 °C},$$ like the vessel?

If anyone at least could give me the subject/effect/topic so I can study just to get an answer to this question, I would be very grateful.

Note: This isn't a homework question. I work on a LPG storage facility and the other day I started thinking about this, but I cannot understand why this happens.

• I don't understand. If you compress any gas to 16 bar, the temperature may have any value. You can heat or cool this compressed gas so as to reach any possible temperature. Why choose 90°C or 38°C ? I don't see your point. Dec 17, 2019 at 20:41
• @Maurice my point is that if i can get a pressure of 16 bar with a temperature of 38°C in a liquid/gas mixture inside a storage vessel, why does the gas that goes through the compressor and gets his pressure increase to 16 bar gets a temperature of 90°C after compression?, why is the temperature so different between both, since they get both to 16 bar but at different temperatures? Dec 17, 2019 at 22:58

Generally, compression of any gas transforms mechanical energy to thermal energy and the gas warms up, until heat dissipates.

If you have ever inflated bike tyres, you would notice the air pump gets warm ( MTB ) up to hot ( high pressure road bikes ).

When air drops down by 1500 m ( from 0.85 to 1 bar ) after passing over mountain range, it warms up by adiabatic compression by 15 deg C. Warming up to 90 Deg C to get pressure 16 bar does not seems to be much.

A gas, with or without its liquid phase, has 3 variables: temperature, pressure and volume. You adjust 2 of them and the gas adapt to it by adjusting the 3rd one according laws of gas behaviour.

When you keep gas/liquid mixture at the given temperature and volume, it adjusts to it by reaching its saturated vapour pressure for given temperature by either evaporation either condensation.

When you set a gas to have a particular volume applying a particular pressure ( by pumping it in or out ), it adjusts to it reaching a particular temperature.

For propane gas and adiabatic compression ( without heat exchange with surrounding ):

$$p_1^{1-\gamma} T_1^\gamma=p_2^{1-\gamma} T_2^\gamma$$

If we consider as initial conditions 1 bar and 20 Deg C:

$$293.15^{1.13}=16^{-0.13} \cdot T_2^{1.13 }$$

$$T_2^{1.13 }= 293.15^{1.13} \cdot 16^{0.13}$$

$$t = T_2 - 273.15 = {( 293.15^{1.13} \cdot 16^{0.13} )}^{\frac1{1.13}} - 273.15$$