Understanding the relationship between Pressure and change of state

Click For Summary
SUMMARY

The discussion centers on the thermodynamic principles governing the relationship between pressure and changes of state, specifically regarding ice melting under pressure and the boiling point of water in a closed system. The key equations referenced include the Clapeyron equation, $$\frac{dp}{dT}=\frac{\Delta H}{T\Delta V}$$, which describes the relationship between pressure and temperature at phase transitions. It is established that increased pressure can lead to temporary melting of ice and that boiling ceases when vapor pressure cannot overcome external pressure, necessitating additional heat to resume boiling.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically phase transitions.
  • Familiarity with the Clapeyron equation and its application in thermodynamics.
  • Knowledge of vapor pressure and its relationship to boiling points.
  • Basic grasp of thermodynamic functions such as entropy and enthalpy.
NEXT STEPS
  • Study the Clapeyron equation in detail to understand its implications for phase changes.
  • Explore the concept of vapor pressure and its role in boiling point determination.
  • Learn about the thermodynamic functions of entropy and enthalpy and their applications in phase transitions.
  • Investigate the principles of regelation and its effects on ice melting under pressure.
USEFUL FOR

Students and professionals in physics and chemistry, particularly those studying thermodynamics, phase transitions, and the behavior of substances under varying pressure and temperature conditions.

JC2000
Messages
186
Reaction score
16
1. My book states that an increase in pressure on ice causes temporary melting. Could you explain the reasoning behind this/ the law that predicts this?

2. My book also states that if water is boiling in a flask and the outlet is blocked then temperature drops and boiling would stop unless more heat is supplied externally. Again what underpins this prediction? I tried reasoning this one out using the Gas Equations. Applying it to the gas forming I felt that an increase in pressure at boiling point would mean a further increase in T (since ##\frac{P_1}{T_1}=\frac{P_2}{T_2})##

Does this have something to do with vapour pressure?
 
Last edited:
Physics news on Phys.org
This kind of questions should be in the homework forum -- I think. Never mind.

re summary:
Not clear on which law is used to predict the effect pressure has on change of state.
the equation of state perhaps :wink:? For ideal gases the ideal gas law :smile:

re 1: check wikipedia regelation

re 2:
JC2000 said:
My book also states that if water is boiling in a flask and the outlet is blocked then temperature drops and boiling would stop
Could you quote verbatim ? I am surprised about the 'temperature drops' but perhaps there is a subtlety involved
 
  • Like
Likes   Reactions: berkeman and JC2000
Re 1: Yes my book mentions regelation. So does this mean the effect of pressure is 'temporary'.

Also regarding the pressure temperature relation. Since pressure in the system increases, this means the temperature of the gas increases. But then I feel there is an inconsistency with the statement below (where the temperature 'drops' or have I confused the change in boiling point with a drop in temperature?

Re 2:"If now the steam outlet is closed for a few seconds to increase the pressure in the flask, you will notice that boiling stops. More heat would be required to raise the temperature (depending on the increase in pressure) before boiling begins again. Thus boiling point increases with increase in pressure. "

Thank you!
 
Much better: 'more heat would be required to increase (maintain) the temperature' I can agree with.

A liquid 'boils' if the vapour pressure of the liquid is higher than the pressure above the liquid. The evaporation takes energy (heat of evaporation), so if no heat is supplied from outside, the temperature drops until the vapour pressure of the liquid is equal to the pressure above the liquid.

Vapour pressure increases with temperature, so increasing the pressure by closing off the outlet stops the boiling once the evaporation has built up enough pressure. This takes some time during which the temperature of the liquid drops (the duration of the adiabatic flash) -- hence my careful wording.
 
  • Like
Likes   Reactions: JC2000
JC2000 said:
Summary: Not clear on which law is used to predict the effect pressure has on change of state.

1. My book states that an increase in pressure on ice causes temporary melting. Could you explain the reasoning behind this/ the law that predicts this?

2. My book also states that if water is boiling in a flask and the outlet is blocked then temperature drops and boiling would stop unless more heat is supplied externally. Again what underpins this prediction? I tried reasoning this one out using the Gas Equations. Applying it to the gas forming I felt that an increase in pressure at boiling point would mean a further increase in T (since ##\frac{P_1}{T_1}=\frac{P_2}{T_2})##

Does this have something to do with vapour pressure?
With regard to item 1, thermodynamics tells us that the effect of pressure on the melting temperature of a substance is described by the equation $$\frac{dp}{dT}=\frac{\Delta H}{T\Delta V}$$where ##\Delta H## is the heat of melting and ##\Delta V## is the specific volume of the liquid minus the specific volume of the solid.

With regard to item 2, thermodynamics tells us that the effect of temperature on the equilibrium vapor pressure of a substance is given by the Clapeyron equation: $$\frac{dp}{dT}=\frac{\Delta H}{T\Delta V}$$where ##\Delta H## is the heat of vaporization and ##\Delta V## is the specific volume of the vapor minus the specific volume of the liquid.
 
  • Like
Likes   Reactions: JC2000 and BvU
BvU said:
Much better: 'more heat would be required to increase (maintain) the temperature' I can agree with.

A liquid 'boils' if the vapour pressure of the liquid is higher than the pressure above the liquid. The evaporation takes energy (heat of evaporation), so if no heat is supplied from outside, the temperature drops until the vapour pressure of the liquid is equal to the pressure above the liquid.

Vapour pressure increases with temperature, so increasing the pressure by closing off the outlet stops the boiling once the evaporation has built up enough pressure. This takes some time during which the temperature of the liquid drops (the duration of the adiabatic flash) -- hence my careful wording.

(a)I see, so when the flask is covered, the vapour pressure builds until it is greater than the atmospheric pressure would have been above the liquid. At this point more energy would be needed to overcome this and continue evaporation? But why would the atmospheric pressure and vapour pressure be antagonistic to each other?

(b)Is there a way to apply the gas laws to this situation and see what they predict? My attempt : Applying the gas laws to the vapour indicates that its temperature must increase as vapour pressure builds. This does not explain the rise in boiling point since that is due to a different phenomenon. (Is my reasoning correct?)

(c)To conclude, is it fair to say that the effect of pressure on boiling point is explained by vapour pressure build up and not the gas laws?

(d)My book defines boiling point as the temperature at which the liquid and gas phase of a substance coexist (or are in thermal equilibrium), from what you said I gather that boiling point can also be defined as the temperature at which the vapour pressure over the liquid is the same as the atmospheric pressure/ pressure of whatever gas is over the liquid. Is it correct to say that there are two ways to define it (or are they synonymous but I have not understood something?)?

(e)Lastly, in the second paragraph of your answer you mention that temperature in the flask drops, why does this happen? Also, regarding the time in which the system is adiabatic, what part of the system is insulated (or is the entire flask insulated?), and by what?

Thanks!
 
Last edited:
Chestermiller said:
With regard to item 1, thermodynamics tells us that the effect of pressure on the melting temperature of a substance is described by the equation $$\frac{dp}{dT}=\frac{\Delta H}{T\Delta V}$$where ##\Delta H## is the heat of melting and ##\Delta V## is the specific volume of the liquid minus the specific volume of the solid.

With regard to item 2, thermodynamics tells us that the effect of temperature on the equilibrium vapor pressure of a substance is given by the Clapeyron equation: $$\frac{dp}{dT}=\frac{\Delta H}{T\Delta V}$$where ##\Delta H## is the heat of vaporization and ##\Delta V## is the specific volume of the vapor minus the specific volume of the liquid.

Thank you for your answer!
Regarding the formulas, I have not encountered them in my course yet (not in Physics or Chemistry). On looking up wikipedia, I understand that the equation represents the slope of the tangent of a P-T curve. But I also have a few questions :
(A) Can this also be derived from the Gas Laws?
(B) If not, then in the context of pedagogy, at what 'level' are these introduced (what is the progression of topics around the topic of these equations)? Apologies if this question is sort of pointless, but I tend to derive some comfort from knowing how topics relate to each other, and thus a set of equations, seemingly out of the blue don't sit very well with me.
(C) Is the equation an alternative means to predict what is happening (as compared to responder @BvU's answer) or are they the same? Thank you for your time!
 
JC2000 said:
Thank you for your answer!
Regarding the formulas, I have not encountered them in my course yet (not in Physics or Chemistry). On looking up wikipedia, I understand that the equation represents the slope of the tangent of a P-T curve. But I also have a few questions :
(A) Can this also be derived from the Gas Laws?
No.
(B) If not, then in the context of pedagogy, at what 'level' are these introduced (what is the progression of topics around the topic of these equations)? Apologies if this question is sort of pointless, but I tend to derive some comfort from knowing how topics relate to each other, and thus a set of equations, seemingly out of the blue don't sit very well with me.
These are usually introduced in first-semester courses in thermodynamics. Are you familiar with thermodynamic functions like entropy, enthalpy, and Gibbs free energy? This all starts from the condition that, at a change of phase, the change in Gibbs free energy is zero, or, equivalently, that the change in entropy is equal to the change in enthalpy divided by the absolute temperature.
(C) Is the equation an alternative means to predict what is happening (as compared to responder @BvU's answer) or are they the same?
The equations I presented quantify the concepts described by @BvU in his response. They provide exact thermodynamic relationships for calculating the relationship between pressure- and temperature variations at phase equilibrium.
 
  • Like
Likes   Reactions: JC2000
Chestermiller said:
These are usually introduced in first-semester courses in thermodynamics. Are you familiar with thermodynamic functions like entropy, enthalpy, and Gibbs free energy? This all starts from the condition that, at a change of phase, the change in Gibbs free energy is zero, or, equivalently, that the change in entropy is equal to the change in enthalpy divided by the absolute temperature.

Yes, I am familiar with the basics of these terms, but I admit I am not very comfortable applying these ideas. But I see where this belongs now! I think I understand this now.
Thank you for your time!
 
  • #10
JC2000 said:
Yes, I am familiar with the basics of these terms, but I admit I am not very comfortable applying these ideas. But I see where this belongs now! I think I understand this now.
Thank you for your time!
The equation for the differential change in free energy of a pure single phase substance is $$dG=-SdT+VdP$$. So for a pure liquid, $$dG_L=-S_LdT+V_LdP$$and for a pure vapor $$dG_V=-S_VdT+V_VdP$$So, along a phase boundary between vapor and liquid, ##d(G_V-G_L)=d\Delta G=0## or$$-S_VdT+V_VdP=-S_LdT+V_LdP$$or, equivalently,$$-\Delta SdT+\Delta VdP=0$$
 
  • Like
Likes   Reactions: JC2000
  • #11
Chestermiller said:
The equation for the differential change in free energy of a pure single phase substance is $$dG=-SdT+VdP$$. So for a pure liquid, $$dG_L=-S_LdT+V_LdP$$and for a pure vapor $$dG_V=-S_VdT+V_VdP$$So, along a phase boundary between vapor and liquid, ##d(G_V-G_L)=d\Delta G=0## or$$-S_VdT+V_VdP=-S_LdT+V_LdP$$or, equivalently,$$-\Delta SdT+\Delta VdP=0$$

From, $$-\Delta SdT+\Delta VdP=0$$
using ##\Delta S = \frac{\Delta H}{T}## the equation you mention is arrived at (?)

Thank you very much! The derivation made things much clearer!
 
  • Like
Likes   Reactions: Chestermiller

Similar threads

  • · Replies 2 ·
Replies
2
Views
7K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
53K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
9
Views
3K
Replies
14
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K