Kelvin Effect Covers Reading Material in Chapter 10.3 Atmospheric Sciences 5200 Physical Meteorology III: Cloud Physics Vapor Pressure (E)

Kelvin Effect Covers Reading Material in Chapter 10.3 Atmospheric Sciences 5200 Physical Meteorology III: Cloud Physics Vapor Pressure (e)

� < �

Vapor � = � Pressure � > � Relative humidity � �� = �(�)

Supersaturation � > � (�) � ≡ �� − 100% � ≈ 0.5 − 2% in convective updrafts The Clausius-Clapeyron Equation

• Used to find the relationship between pressure and temperature along phase boundaries. The Clausius-Clapeyron Equation

• Used to find the relationship between pressure and temperature along phase boundaries. • Need to derive: • Latent Heat • Gibbs Energy Latent Heat Specific Enthalpy at constant pressure and temperature

Specific Enthalpy – intensive units ℎ ≡ � + �� Mechanical Internal energy work

Constant Pressure � = �ℎ ≡ �� + � �� = �� = �� + �� = �

� = (�−�) + �(� − �) Latent Heat Specific Enthalpy at constant pressure and temperature

Specific Enthalpy – intensive units ℎ ≡ � + �� Mechanical Internal energy work Constant Temperature �� = � = �� = �(� − �) �

� = �(� − �) Gibbs Energy � = (�−�) + �(� − �)

� = �(� − �) Combine (�−�) + �(� − �) = �(� − �) Rearrange Remember Gibbs energy is not � + � � − � � = � + � � − �(� ) constant for temperature and pressure

Gibbs Energy = G = � + � � − � � We assumed isothermal, isobaric change in phase to get G1 = G2 here. Clausius – Clapeyron equation

�� − sdT = �� − sdT

� − s = � − s

�� s − s = �� − �

Recall: � = �(� − �) �� s − s � = = �� − � �(� − �) Clausius – Clapeyron equation for the atmosphere

�� s − s � = = Assumption: �>>>� �� − � �(� − �) Specific Volume of liquid water �� = Specific Volume of water vapor �� (�) 1 � Substitute in the ideal gas law for water vapor = � �� = �� = ∗ � � � �� 1 1 = � � = � exp − � � � � � � Saturation Vapor Pressure (Liquid Water) (T) [Pa] s e

T [ oC ] Vapor Pressure (e)

� < �

Vapor � = � Pressure � > � The forces on the hydrogen bonding in the liquid give a net inward attractive force to the molecules on the boundary between the liquid and the vapor.

The net inward force, divided by the distance along the surface, is called surface tension, σ. Its units are N/m or J/m2.

Sketches of the curvature effect. Left is a flat surface of pure water; right is a curved surface of pure water. Credit: W. Brune If the surface is curved, then the amount of bonding that can go on between any one water molecule on the surface and its neighbors is reduced.

As a result, there is a greater probability that any one water molecule can escape from the liquid and enter the vapor phase. Thus, the evaporation rate increases.

The greater the curvature, the greater the chance that the surface water molecules can escape. Thus, it takes less energy to remove a molecule from a curved surface than it does from a flat surface.

As a result, there is a greater probability that any one water molecule can escape from the liquid and enter the vapor phase. Thus, the evaporation rate increases.

The greater the curvature, the greater the chance that the surface water molecules can escape. Thus, it takes less energy to remove a molecule from a curved surface than it does from a flat surface.

https://www.youtube.com/watch?v=jGsOemwlNCk

https://www.chemistryworld.com/news/evaporation-drives-dancing-droplet-breakdown/2500493.article Gibbs Energy � = (�−�) + �(� − �)

� = �(� − �) Combine (�−�) + �(� − �) = �(� − �) Rearrange � + � � − � � = � + � � − �(�)

Gibbs Energy = G = � + � � − � � No longer true, there is a change in the Gibbs free G = G energy accompanying the 1 2 formation of a single drop Gibbs Energy for formation of a drop

G = � + � � − � � No longer true, there is a Gibbs Energy = change in the Gibbs free energy accompanying the formation of a single drop G1 = G2

Drop radius

∆G = � − � = � � − � + 4�� Surface tension �ℎ��, � = �� ; � �� ℎ� �� ℎ� �� ℎ�� Gibbs Energy for formation of a drop

∆G = � − � = � � − � + 4�� ℎ��, � = �� ; � �� ℎ� �� ℎ� �� ℎ�� 4�� = 3�

4�� ∆G = � − � = � − � + 4�� 3� Gibbs Energy for formation of a drop

∆G = � − � = � � − � + 4�� ℎ��, � = �� ; � �� ℎ� �� ℎ� �� ℎ��

Volume occupied by a 4�� molecule in the liquid � = phase! 3�

4�� ∆G = � − � = � − � + 4�� 3� Gibbs Energy for formation of a drop

∆G = � − � = � � − � + 4�� ℎ��, � = �� ; � �� ℎ� �� ℎ� �� ℎ��

Volume occupied by a 4�� molecule in the liquid � = phase! 3�

4�� ∆G = � − � = � − � + 4�� 3� Gibbs Energy for formation of a drop

4�� ∆G = � − � = � − � + 4�� 3�

NEED TO SOLVE FOR � − �

To put equation into something useful – parameters that we can measure Assumption: �>>>� When deriving Gibbs Energy for formation of a drop Clausius Clapeyron Equation

� = (�−�) + �(� − �)

� = �(� − �)

Change Gibbs Energy = Δg = ∆� + ∆� ∆� − ∆� ∆�

We assume a constant temperature and these two Assumption: �>>>� terms become ZERO

� − � = � − � d� ≅ − � d�

Volume occupied by a Volume occupied by a molecule molecule in the liquid phase! in the particle phase! Gibbs Energy for formation of a drop

Assumption: �>>>�

Change Gibbs Energy = � − � = � − � d� ≅ − � ��

� − � ≅ − � d� Gibbs Energy for formation of a drop

Assumption: �>>>�

Change Gibbs Energy = � − � = � − � d� ≅ − � ��

� − � ≅ − � d�

�� Substitute in the ideal gas law for water vapor � = �

�� − � ≅ − d� � Gibbs Energy for formation of a drop

Assumption: �>>>�

Change Gibbs Energy = � − � = � − � d� ≅ − � ��

� − � ≅ − � d�

�� Substitute in the ideal gas law for water vapor � = � Where:

� � � es= vapor pressure over the flat � − � ≅ − d� = −� ��( ) surface es0 =vapor pressure at equilibrium � � Gibbs Energy for formation of a drop

Assumption: �>>>�

Change Gibbs Energy = � − � = � − � d� ≅ − � ��

� − � ≅ − � d�

�� Substitute in the ideal gas law for water vapor � = � Saturation Ratio

�� − � ≅ − d� = −��(�) � Gibbs Energy for formation of a drop

4�� ∆G = � − � = � − � + 4�� 3�

NEED TO SOLVE FOR � − �

To put equation into something useful – parameters that we can measure Gibbs Energy for formation of a drop

4�� ∆G = � − � = � − � + 4�� 3�

SOLVED FOR � − � = −��(�) Gibbs Energy for formation of a drop

4�� ∆G = � − � = � − � + 4�� 3�

SOLVED FOR � − � = −��(�)

4�� ∆G = � − � = −��(�) + 4�� 3�

SURFACE TENSION TERM DOMINATES

1st TERM DOMINATES Kelvin Equation

�∆� = 0 Find point of maximum value= metastable �∆�

∗ ∗ ∆� �� = �

∗ 2�� = ��(�)

2�� = � exp �� Kelvin Equation

2�� = � exp ��

Vapor pressure over a curved interface always exceeds that of the same substance over a flat surface.

Sketches of the curvature effect. Left is a flat surface of pure water; right is a curved surface of pure water. Credit: W. Brune Must be considered in Kelvin Equation calculations for particles with diameters smaller 2�� than about 200 nm � = � exp ��

Vapor pressure over a curved interface always exceeds that of the same substance over a flat surface.

Sketches of the curvature effect. Left is a flat surface of pure water; right is a curved surface of pure water. Credit: W. Brune