How does the squeezing effect reduce the melting point?
I think I'm reading the phase diagram wrong. Looking at the phase diagram here, if we start at the triple point with no air in the head space, and we add air until the head space pressure is 1atm, the vapour pressure should remain...
Taking a step back, this whole time I was under the (wrong) impression that for Case 2 (triple point, head space, no air), if I add air (ideal gas) until the total pressure in the head space is 1atm, the equilibrium state will not change; I did not realize that it actually does decrease by 0.01C...
So regardless of whether head space is present, the melting point when there is 1atm of pressure acting on the liquid (head space or not) is 0C, even if we are dealing with ideal gases in the head space?
My steam tables actually stop at 0.01C and I haven't been able to track down a water...
Sorry, I am still confused about why there is a difference in melting point between Cases 1 and 3.
I used the example here and the vapor pressure is calculated to be 0.00596bar (assuming heat of sublimation and vaporization is close enough).
Thank you
Thank you, this summarizes the cases I was confused about perfectly. My original question is basically how and why freezing point vary between Case 1 and 3, since the system in both cases are under 1atm of pressure albeit the way the pressure is applied is different.
By vapour pressure of...
For the case of a pressure of 1atm applied from a piston, in the phase diagram in my OP, I believe the melting point is 0C and 1atm. In the case of a fixed volume container, the melting/triple point is at 0.01C with the vapour pressure at 0.00611bar. Essentially I see that the water is at two...
Sorry, I was thinking of a different case (1atm pressure applied from by a piston); in that case it would be closer to 0C than 0.01C? Another way of asking my original question would be, why do we need a significantly higher total (air) pressure/force than an applied piston pressure to obtain...
With regards to the first statement, I'm confused on what is meant by the same state. In the phase diagram for 1atm of pressure applied by the piston the freezing point is 0C (exactly) but for the case of cooling in an isolated fixed volume box with 1atm of air, the triple point/freezing point...
1/2. I'm basically wondering if nucleation is a requirement for phase change. For instance, when liquid water is at equilibrium with its vapour, does individual molecules simply enter/leave the interface, or does nuclei have to form to induce the phase change?
3. Thermodynamically, I'm...
I phrased that poorly, it was meant as a question not a statement. I'm was thinking, from the liquid's perspective, if I have air at 1atm, or a piston applying a pressure at 1atm, the liquid "feels" the same force, but the equilibrium state is very different. I get that the liquid has no idea...
I can't get my head around why ideal gas has no effect on equilibrium state or in a sense, hold the water down. The pressure on the liquid (and in the head space) applied by the air at 1atm has a drastically different effect than 1atm applied by a piston even though the pressure felt by the...
In the above examples I assumed that air was an ideal gas and that changing the air pressure did not affect the phase equilibrium (if our volume was constant adding air wouldn't have changed anything). Base on the link you recommended, when real gas effects are considered, increasing the total...
1. Is nucleation a phenomenon that occurs in all phase change (freezing/melting, evaporation/condensation)?
2. I've always read evaporation/condensation described as a liquid-vapour interface phenomenon (water molecules going entering-leaving the interface at equal rates in equilibrium). If...
I think I will increase pressure to 50atm:
the mass of liquid water in the bottom half: 0.499kg
the partial pressure of water vapor in the top half: 70.14kPa
the partial pressure of air in the top half: 4994.86kPa
the mass of water vapor in the top half: 5.2mg
the total mass of water in the...