March 28, 2024, 09:21:00 PM
Forum Rules: Read This Before Posting


Topic: Why does pressure remain constant during the liquefaction of a gas?  (Read 3142 times)

0 Members and 1 Guest are viewing this topic.

Offline confusedstud

  • Full Member
  • ****
  • Posts: 198
  • Mole Snacks: +3/-0
In the PV isotherm of a real gas, the pressure remains the same during liquefaction. Why is this so?

I thought that during the liquefaction process, the pressure of the gas is P=nRT/V so as we decrease the volume, both V and n decreases such that P remains constant. However since we are measuring the pressure of the entire system, having more liquid wouldn't the liquid pressure cause an increase in the system's pressure as well?

P(Total)=P(gas)+P(liquid) and as shown above P(gas) is constant but as more gas converts to the liquid state wouldn't P(liquid) increase causing the P(Total) to increase?

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 2048
  • Mole Snacks: +296/-12
Re: Why does pressure remain constant during the liquefaction of a gas?
« Reply #1 on: August 11, 2015, 10:20:43 AM »
No. P(total) = P(gas) = P(liquid). The pressures of liquid and gas are not additive, because they are separate phases, and do not occupy the same volume (it is not like the partial pressures in a gas mixture). The gas exerts a pressure P on the liquid, and the liquid exerts a pressure P on the gas (Newton's 3rd law). Both exert a pressure P on the walls of the vessel (separately, in the regions they occupy, not additively) and the vessel exerts a pressure P on the liquid and gas. See diagram.

Offline confusedstud

  • Full Member
  • ****
  • Posts: 198
  • Mole Snacks: +3/-0
Re: Why does pressure remain constant during the liquefaction of a gas?
« Reply #2 on: August 11, 2015, 11:20:41 AM »
No. P(total) = P(gas) = P(liquid). The pressures of liquid and gas are not additive, because they are separate phases, and do not occupy the same volume (it is not like the partial pressures in a gas mixture). The gas exerts a pressure P on the liquid, and the liquid exerts a pressure P on the gas (Newton's 3rd law). Both exert a pressure P on the walls of the vessel (separately, in the regions they occupy, not additively) and the vessel exerts a pressure P on the liquid and gas. See diagram.

So the pressure acting on the wall by the gas is equal to the pressure exerted on the liquid by the gas. So would P(total) consider the gas's pressure only?

What about the pressure exerted by the liquid on the walls?

Lastly, after all the gas had been liquefied what pressure do we consider now?

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 2048
  • Mole Snacks: +296/-12
Re: Why does pressure remain constant during the liquefaction of a gas?
« Reply #3 on: August 12, 2015, 04:53:26 AM »
Quote
So would P(total) consider the gas's pressure only?
There is no other pressure. The pressure is P. P everywhere. P in the gas, P in the liquid, P on the wall. (That's not an instruction, by the way!)
Pressure is an intensive variable. It is the same at all points in a system at equilibrium. Just like temperature. If the temperature of the gas was T, and the temperature of the liquid was T, you wouldn't say that the "total" temperature was 2T, would you? [Note, this assumes your system is small enough for gravitational pressure to be neglected. It wouldn't be true at the bottom of the sea.]
Quote
What about the pressure exerted by the liquid on the walls?
What about it? It is, I repeat, not additional to the pressure of the gas. Where the gas is in contact with the wall, the pressure is exerted by the gas; where the liquid is i contact, by the liquid, but the pressure is always P.
Quote
Lastly, after all the gas had been liquefied what pressure do we consider now?
As before, the pressure of the system. Only now there is no gas, we may refer to it as the pressure of the liquid.

Offline confusedstud

  • Full Member
  • ****
  • Posts: 198
  • Mole Snacks: +3/-0
Re: Why does pressure remain constant during the liquefaction of a gas?
« Reply #4 on: August 12, 2015, 06:50:03 AM »
Quote
So would P(total) consider the gas's pressure only?
There is no other pressure. The pressure is P. P everywhere. P in the gas, P in the liquid, P on the wall. (That's not an instruction, by the way!)
Pressure is an intensive variable. It is the same at all points in a system at equilibrium. Just like temperature. If the temperature of the gas was T, and the temperature of the liquid was T, you wouldn't say that the "total" temperature was 2T, would you? [Note, this assumes your system is small enough for gravitational pressure to be neglected. It wouldn't be true at the bottom of the sea.]
Quote
What about the pressure exerted by the liquid on the walls?
What about it? It is, I repeat, not additional to the pressure of the gas. Where the gas is in contact with the wall, the pressure is exerted by the gas; where the liquid is i contact, by the liquid, but the pressure is always P.
Quote
Lastly, after all the gas had been liquefied what pressure do we consider now?
As before, the pressure of the system. Only now there is no gas, we may refer to it as the pressure of the liquid.

Hmm but this seems weird wouldn't the liquid's pressure be greater than the gas pressure? In the picture you sent it showed that the P is the same even when there was only a liquid-wall interface which I don't quite understand..

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 2048
  • Mole Snacks: +296/-12
Re: Why does pressure remain constant during the liquefaction of a gas?
« Reply #5 on: August 12, 2015, 08:21:43 AM »
Quote
wouldn't the liquid's pressure be greater than the gas pressure?
Why?

Sponsored Links