November 27, 2024, 06:47:13 PM
Forum Rules: Read This Before Posting


Topic: pressure rise due to heating  (Read 2907 times)

0 Members and 1 Guest are viewing this topic.

Offline Princersk31

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-0
pressure rise due to heating
« on: October 21, 2019, 02:32:59 AM »
how can I calculate the pressure change due to heating a liquid in a closed constant volume system? I want to know the pressure change due to vaporization and temperature change.

Offline chenbeier

  • Sr. Member
  • *****
  • Posts: 1337
  • Mole Snacks: +102/-22
  • Gender: Male
Re: pressure rise due to heating
« Reply #1 on: October 21, 2019, 07:37:40 AM »
Use GayLussac law p1/T1 = p2/T2. If the whole liquid is vaporized.

Offline Enthalpy

  • Chemist
  • Sr. Member
  • *
  • Posts: 4036
  • Mole Snacks: +304/-59
Re: pressure rise due to heating
« Reply #2 on: October 21, 2019, 03:42:06 PM »
The pressure depends on the volume available to the vapour. Can you confirm that all the liquid evaporates?

If some liquid remains => Equilibrium with the vapour, plus the initial gas if present.

If no liquid remains:

Much volume more than the liquid => Compute the vapour pressure by ideal gas law.

Little volume more than the liquid => Try to estimate using state equations meant for very high pressure.

No extra volume available => The container is very likely to burst. If not, you can evaluate a pressure if you know the behaviour of this particular gas (or supercritical fluid) at unusual pressure.

Offline Princersk31

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-0
Re: pressure rise due to heating
« Reply #3 on: October 21, 2019, 08:48:15 PM »
not all of the liquid evaporates, some of it remains and at equilibrium, the condensation rate is equal to the evaporation rate.
Now my situation is that I inject a pressure of say 5.454 MPa and then get an equilibrium pressure of 5.720 MPa. How can I calculate the excess pressure due to the increased kinetic activity of the gas in the vapor phase of this closed system?
 I have used the Peng Robinson EOS and got a very high value and of course tried the kinetic temperature way but still got a very high figure. Please tell me how I can fix this.

Offline Princersk31

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-0
Re: pressure rise due to heating
« Reply #4 on: October 21, 2019, 09:08:14 PM »
Also am just assuming the pressure increases due to kinetic activity but any other suggestion is welcome.

Offline Enthalpy

  • Chemist
  • Sr. Member
  • *
  • Posts: 4036
  • Mole Snacks: +304/-59
Re: pressure rise due to heating
« Reply #5 on: October 22, 2019, 12:36:17 PM »
Then it' a matter of liquid-vapour equilibrium.

The equilibrium pressure increases with the temperature. Optimally, you have the experimental curve or table that relates them for your compound. If not, maybe you have Antoine's parameters for your compound, and you can make more or less reasonable predictions. Or you know the heat of vaporization of your compound, and can extrapolate the vapour pressure at one temperature from an other temperature near enough.

This hold if the compound is far from its critical point. Near the critical point, the heat of vaporization drops a lot, so extrapolations are wrong and simple equations like Antoine inaccurate. This is a big interrogation since 5MPa is a near the critical temperature of many compounds.

You also need the partial pressure of any other gas that were in the vessel before the temperature rise.

Sponsored Links