In the previous section we saw the basic details about phase diagram. Our discussions were based on fig.11.13. For convenience, that fig. is shown again below:
 |
| Fig.11.13 |
• Now we will see a solved example which will help us to get more information about phase diagrams. Later, we will see the phase diagram of water
Solved example 11.23
Fig.11.14 below shows the P-T diagram (phase diagram) of CO2
 |
| Fig.11.14 |
Based on the diagram, answer the following questions:
(i) At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist in equilibrium ?
(ii) What is the effect of decrease of pressure on the fusion and boiling point of CO2 ?
(iii) What are the critical temperature and pressure for CO2 ? What is their
significance ?
(iv) Is CO2 solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60 °C under 10 atm,
(c) 15 °C under 56 atm ?
(v) CO2 at 1 atm pressure and temperature – 60 °C is compressed isothermally.
Does it go through a liquid phase ?
(vi) What happens when CO2 at 4 atm pressure is cooled from room temperature
at constant pressure ?
(vii) Describe qualitatively the changes in a given mass of solid CO2 at 10 atm
pressure and temperature –65 °C as it is heated up to room temperature at
constant pressure.
(viii) CO2 is heated to a temperature 70 °C and compressed isothermally. What
changes in its properties do you expect to observe ?
Solution:
Part (i):
• From the fig.11.14, we see that, the coordinates of the triple point are: (-56.6, 5.11)
♦ So the required temperature is: -56.6 °C
♦ Required pressure is: 5.11 atm
Part (ii):
1. First we will consider the fusion point
• This point will lie on the yellow curve
• Consider the ‘⨯’ mark on the yellow line in the previous fig.11.13
• When we decrease the pressure, we move downwards along the vertical double headed arrow
♦ So we fall out of the yellow line
• But for the fusion point, we must be exactly on the yellow curve
• So, at a lower pressure, to get back on the yellow curve, we must move towards the left
♦ That means, we have to reduce the temperature
• Thus, when pressure is reduced, the fusion point (temperature at which fusion takes place) of CO2 is reduced
2 Next, we will consider the boiling point
• This point will lie on the green curve
• Consider the ‘⨯’ mark on the green line in the previous fig.11.13
• When we decrease the pressure, we move downwards along the vertical double headed arrow
♦ So we fall out of the green line
• But for the boiling point, we must be exactly on the green curve
• So, at a lower pressure, to get back on the green curve, we must move towards the left
♦ That means, we have to reduce the temperature
• Thus, when pressure is reduced, the boiling point (temperature at which boiling takes place) of CO2 is reduced
Part (iii):
1.From the fig.11.14, we see that, the coordinates of the critical point are: (31.1,73)
♦ So the required temperature is: 31.1 oC
♦ Required pressure is: 73 atm
2. Imagine that, keeping pressure constant at 73 atm, the temperature of CO2 is increased to a value just above 31.1 oC
♦ Then the critical point will move horizontally towards the right
♦ That means, the critical point will move out of the green curve
♦ That means, the critical point is out of the blue (liquid) region
• So we can write:
If the temperature of CO2 is raised above 31.1 oC, it will not liquefy even if we apply very high pressures
Part (iv):
(a) The given point is: (-70, 1)
• The point (-78.5, -1) is already marked in fig,11.14
♦ The given point (-70, -1) has the same y-coordinate
♦ So the given point will lie on a horizontal line through (-78.5, -1)
• Now, -70 is greater than -78.5. So (-70, 1) will lie on the right side of (78.5, -1)
♦ That means, (-70, -1) will lie in the vapour region
(b) The given point is (-60, 10)
• The points (-78.5, 1) and (56.6, 5.11) are already marked in fig.11.14
♦ -60 is greater than -78.5, but less than -56.6
♦ So the given point will lie on a vertical line between (-78.5, 1) and (56.6, 5.11)
• That means, the given point will either be a solid or vapour
• The y-coordinate of the given point is 10
♦ This is greater than 5.11
• So the point will lie in the solid region
(c) The given point is (15, 56)
• The point (20, 56) is already marked in fig,11.14
♦ The given point (15, 56) has the same y-coordinate
♦ So the given point will lie on a horizontal line through (20, 56)
• Now, 15 is lesser than 20. So (15, 56) will lie on the left side of (20, 56)
♦ Also, 15 is close to 20
• That means, (20, 56) will lie in the liquid region
Part (v):
• The given point is (-60, 1)
• ‘Isothermal’ means same temperature
• So, when CO2 is compressed isothermally, the pressure will increase but the temperature will remain constant
♦ So it will move along a vertical line through (-60,1)
• This vertical line will lie in between (-78.5, 1) and (-56.6, 5.11)
• When the sample is compresses, it’s pressure increases above 1 atm
• From (-78.5,1), the position of 1 atm is available to us. It lies on the red curve
• So it is clear that, when pressure rises above 1 atm, the sample will be in the solid region. So it does not go through a liquid phase
Part (vi):
• The given point is (room temperature, 4)
• It is cooled at constant pressure. So it will pass along a horizontal line through (room temperature, 4)
• The y-coordinate of the triple point is 5.11
♦ So the horizontal line lies below the triple point
♦ So, when the sample is cooled, it passes from vapour region to solid region
• That means, deposition of CO2 will take place
Part (vii):
• The given point is (-65, 10)
• It is heated at constant pressure. So it will pass along a horizontal line through (-65, 10)
• The y-coordinate of the triple point is 5.11
♦ So the horizontal line lies above the triple point
• The x-coordinate of the triple point is -56.6
♦ The given x-coordinate '-65' is less than -56.6
♦ So the initial position (-65,10) is in the solid region
• So, when the sample is heated, it first passes from solid to liquid. Then it passes from liquid to vapour
Part (viii):
• The given point can be written as (70, P)
• It is compressed isothermally. So it will pass along a vertical line through (70, P)
• The x-coordinate of the critical point is 31.1
♦ So the vertical line lies to the right of the critical point
• At such a high temperature, the CO2 gas will not liquefy even if we apply very high pressures
 |
| Fig.11.15 |
1. The triple point of water is (0.01, 0.006)
• The temperature of 0.01 oC is very close to our familiar freezing point of 0 oC
• But the pressure of 0.006 atm is very much less than our familiar standard atmospheric pressure of 1 atm
2. It may be noted that, (0.01, 0.006) is the triple point. Solid ice, liquid water and gaseous water vapour will be present at that point
• At (0, 1), only solid ice and liquid water will be present
3. For most substances, the yellow curve starts from the triple point and slopes upwards towards the right. It is a positive slope
♦ That means, as temperature increases, pressure also increases
• But for water, the yellow curve starts from the triple point and slopes upwards towards the left. It is a negative slope
♦ That means, as temperature increases, pressure decreases
4. Let us see the reason for the negative slope. It can be written in 5 steps:
(i) Consider a sample of water at 0 oC
♦ If we heat it, it’s volume will decrease
(ii) This is opposite to what we expect. Because, substances usually increase in volume when heated
(We saw this anomalous behaviour of water in a previous section. Fig.11.5 of section 11.2)
(iii) Normally, if we heat a substance in a closed container, it’s volume will increase and so pressure will also increase
(iv) But if it is water at 0 oC, volume will decrease and so pressure will also decrease
(v) That means, with increase in temperature, we are having a decrease in pressure
• That is the reason for the negative slope of the yellow curve
5. Now, consider the freezing point of water
• We know that, at the normal atmospheric pressure of 1 atm, the freezing point is 0 oC
♦ So it is marked as (0,1)
♦ Note that, the boiling point (100,1) will fall at the same horizontal level
6. Keeping the temperature of ice at 0 oC, increase the pressure slightly
• We will then be moving vertically upwards from (0,1)
• When we move vertically upwards from (0,1), we will fall in the liquid region
• That means, even a slight increase in pressure, will cause the ice to liquefy
7. Ice skating is possible because:
• The pressure due to the weight of the skater, cause the upper layer of ice to change into a thin layer of water
• This thin layer of water acts as a lubricant
8. As soon as the pressure is removed, the ice refreezes again. This phenomenon of refreezing is called regelation
9. Regelation is possible due to the negative slope of the yellow curve
• Let us see what would happen if it was a positive slope
♦ Keeping the temperature of ice at 0 oC, increase the pressure
♦ We will then be moving vertically upwards from (0,1)
♦ When we move vertically upwards from (0,1), we will fall in the solid region
♦ That means: If the slope is positive, 'increase in pressure' will not cause the ice to liquefy
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