### SimuLab 2: Freezing and Melting

 You will be able to: a) Study the freezing and melting curve; b)Find the triple point; c)Find the equilibrium Gas pressure near Liquid and solid surfaces.

1. Open Universal Application. Press on Play Movie and select freezing.mov. Press Forward. As soon as the first image appears press Stop.

~ You see a Gas of 1000 particles. There are particles of two types: Red and Blue. In this movie they are identical, so you can think on them as different isotopes of the same element. We will study what will happen with the melting point if we change their chemical properties. Each particle has diameter rAB = rAA = rBB = 4 and interaction range RAB = RBB = RAA = 7.4. Initially the system is in the gaseous phase. Each frame of the movie correspond to 100 computer units. The heat exchange coefficient is 10-4 and the thermostat temperature is 0.01.

2. Watch the temperature, potential energy and total energy graphs. Press Forward.

 Q2.15: Describe what you see.

 Q2.16: Try to define the moments when the liquid state emerges and when crystallization starts and stops.

 Q2.17: Explain the behavior of the temperature graph.

3. Rewind the movie till the frame 550. Open Averaging for Movies from the menu. Press Forward and measure the pressure and temperature until the frame 750. Press Stop.

~ This values must be close to the values of the triple point. However as the crystal is growing (See Sect.2.2.1) those values can be a slightly smaller than the true triple point values.

~ Copy the table below:

 time temperature PotE Total E Pressure 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000

4. Rewind the movie till beginning. Reset the Averaging for Movies window and record the values of temperature, pressure potential and total energies into the table each 50 frames till frame 1000. Press Forward.

 Q2.18: Construct the temperature vs. time graph. What means the plateau? What is the crystallization temperature?

~ This graph corresponds to the crystallization curve.

 Q2.19: Construct the Potential and Total energies vs. time graphs.Explain what you see from the graphs.

 Q2.20: Make the graph of the potential energy as function of the temperature. Explain what you see. What is the value of the latent heat of crystallization Lc? And its value per atom?

• [ ] Now we will study in the next Simulab the reverse process assuming that the process the freezing process was so slow that one can regard it as equilibrium.

5. Close Movie and Open the movie melting.mov.

6. Watch the movie from the beginning to the end. Find the time when the crystallization ends that is the beginning of melting.

~ Copy the table below:

 time temperature Pressure 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000

7. Rewind the movie till beginning. Reset the Averaging for Movies window and record the values of temperature and pressure potential into the table each 50 frames till frame 1000. Press Forward.

 Q2.21: Is the temperature of melting smaller or larger the temperature of freezing?. Explain your answer from the point of view of equilibrium.

8. Find the triple point pressure by averaging the pressure during melting and freezing.

 Q2.22: What are the triple points values for pressure and temperature?

 Q2.23: Construct the graphs of the natural logarithm of pressure vs. 1/T for both process: freezing and melting. Describe what you see.

~ Note that above T = 0.73 both curves coincides. Below T = 0.73 and above the triple point the curve corresponding to heating of the substance has lower pressure than the curve corresponding to the cooling.

 Q2.24: Explain the difference of pressures in that region in terms of the desequilibrium between the rates of condensation and evaporation

~ Note that the region of both curves between T = 0.73 and the triple point is relatively straight, this kind of behavior in that region is called the Arrhenius behavior which is very important in many chemical reaction.

 Q2.25: Find the slopes of this curves.

~ The meaning of this slope is the activation energy of evaporation. It means that an atom leaving the liquid state and entering in the gaseous state must gain a potential energy of 4. For further discussion see Sect. 2.2.1.

~ Below the freezing point we can see the branch of the curve with larger slope. This slope correspond to the equilibrium of gas with crystal. Below T = 4.5 the points are not accurate enough because there are few atoms in the gaseous state in our simulation to produce accurate measurements. Discard those points and

 Q2.26: find the slope of the curve. Compare it with with the potential energy of the atoms in the crystal

~ Note that below the freezing point the equilibrium pressure of Gas near the crystal is lower that for Liquid. Above the freezing point the equilibrium pressure of Gas near liquid is lower that for crystal. Close movie.

### 2.2.1  Post-Lab In-Depth Discussion

#### Latent Heat in phase transitions

As gas start to condense, the latent heat of condensation is released and it has to be taken away, thus only part of the heat that is taken away comes from the kinetic energy of the gas. So the rate of temperature decrease. As almost all the gas condenses, the temperature again start to decrease faster. Once the temperature goes below the freezing point a spontaneous nucleation occurs. At first the nucleus grows fast because the liquid is deeply supercooled (See Sect.2.1.1). The latent heat of crystallization released is so large that the heat bath cannot absorb it at once, so the temperature goes up until the crystallization rate decreases to a value at which the latent heat release rate is equal to the heat exchange rate and the temperature stays constant for a long time until all the liquid crystallizes. Since the heat exchange rate is very small (0.0001) the crystallization is very slow: the liquid and the crystal are almost at equilibrium. As a result a perfect crystal with no defects emerges. Moreover at this temperature the droplet is at equilibrium with the Gas, which as very low density (but still non zero). It means that all three states of matter at this temperature are at equilibrium. Such temperature is called the triple point. Once the liquid crystallizes the release of latent heat stops and the temperature starts to decreases.

#### Vapor Pressure

The equilibrium pressure of the gas is proportional to the evaporation rate. Indeed the low density gas at equilibrium is well described by the ideal gas law. It's density is proportional to the pressure. On the other hand the number of collisions of gas molecules with liquid per second is proportional to the density of gas. Hence the condensation rate is proportional to the gas pressure and as the evaporation rate at equilibrium is equal to the condensation rate, so it is also proportional to the gas pressure. The evaporation rate is proportional to the probability that the molecule in liquid gain the potential energy H that is the difference between the potential energy on the gaseous phase (0) and in the liquid phase. This can occurs if the molecule gains the kinetic energy of 4 in these case. The kinetic energy of molecules are distributed with Maxwell distribution
 p ~ exp(- m v2 2 kB T )

. Hence the probability to leave the liquid phase is proportional to
 exp( -H 2 kB T )

, and thus
 P = P0exp- H kB T ,
where P0 is some proportionality coefficient. Taking logaritms of both sides of this equation:
 ln (P) = ln(P0)- H kB T
(2.1)

So if we plot ln(P) as a function of 1 / T the arrhenius plot is a straight line. In our simulation kB = 1 and thus the slope is -H, where H is the potential energy per atom in the liquid.

#### Simulab 2

Figure 2.3: Temperature as function of time for freezing ° and melting [¯]. The plateau in each case shows the temperature of temperatures of those transition. The dashed-lines are used a guide to show the values of that temperatures in freezing (T = 0.537) and melting (T = 0.546).

A2.15: As the temperature goes down, gas start to condense, it forms a compact droplet surrounded by few molecules. Finally a little crystal appears inside the droplet . At this point the temperature goes up and stays at the plateau as the crystal grows. Once the liquid crystallize the temperature starts to go down again. As the temperature decreases the number of particles in the gaseous phase decreases.

A2.16: The liquid emerges at t = 10000 (100 frames). The crystal at t = 50000 (500 frames) and stops at t = 75000 (750 frames).

A2.17: The temperature goes down because we take away energy from the system, almost at constant rate. At first the rate at which the temperature changes is fast because almost all the heat is taken from the kinetic energy of the Gas See Sect2.2.1.

A2.18: The plateau region corresponds to the crystallization temperature. The crystallization temperature is T = 0.537 (See Fig .2.3)

A2.19: Theses graphs are monoatomic since we take out the heat from the system almost at constant rate (See Fig .2.4).

Figure 2.4: Plot of the total and potential energies as function of time during freezing. Theses graphs decrease monoatomic.

A2.20: In the graph of the potential energy as a function of the temperature we see a region of regular liquid above the critical temperature of crystallization (T = 0.537) (See Fig. 2.5). Below the freezing point the liquid in a supercooled state coexist with the crystal. At the critical point we see a plateau characteristic of a phase transition where liquid and crystal coexist. The height of this plateau is the latent heat of crystallization which value is 1722.

Figure 2.5: Potential energy as function of the temperature during the crystallization °. The dashed line is used as a guide to show the transition temperature Tc. The height of the step is the latent heat of crystallization. Notice the region of supercooled liquid in coexistence with the crystal below Tc.

A2.21: The melting temperature is 0.546 and corresponds to the plateau in Fig .2.3 . During freezing it was 0.537. The difference is 0.01. During melting the liquid was growing. The equilibrium was shifted towards the liquid state. The crystal was superheated. Hence the temperature was higher that during freezing when the equilibrium was shifted towars crystalline state and the liquid was supercooled. One can expect that the true equilibrium temperature is somewhere in between and we can assume that it is 0.542 ±0..05 which is also the triple point temperature because the gas is also at equilibrium.

A2.22: During melting the pressure is 1.81 ·10-4. During freezing it is 1.42 ·10-4. The average pressure is the 1.6 ·10-4 ±0.02. Hence the triple points parameters are T = 0.542 ±0.0005 and P = 1.6 ·10-4 ±0.02.

A2.23: See Fig .2.6

Figure 2.6: Ln plot of pressure vs. 1/T for both process: freezing ° and melting [¯]
. The dashed lines shows the triple point and the solid is used as a guide to show the slope -4. Note that above T = 0.73 both curves coincides.

A2.24: During the fast cooling the condensation rate is higher than the evaporation rate. It means that the density of gas surrounding the liquid is higher than the equilibrium one. During the fast heating the evaporation rate is higher than the condensation rate. It means that the density of gas is lower than the equilibrium one. During cooling, the gas is supercooled or supersatured. During heating the liquid is superheated. The equilibrium pressure must be in between the two curves. Above T = 0.73 the liquid does not exist and both curves coincide.

A2.25: The slope is about -4 See Sect. 2.2.1.

A2.26: The slopes is -6.