In the previous section we saw the causes of deviation from ideal gas behaviour. In this section, we will see liquefaction
• We have seen that,
♦ At high pressures, the gases deviate from ideal behaviour
♦ At low pressures, the gases do not deviate much from ideal behaviour
✰ That means, the gases can be expected to behave ideally at low pressures
• We have seen the reason:
♦ At low pressures, the inter molecular attractive forces are low
■ In a similar way, temperature also affects the behaviour of gases
♦ This can be explained in 6 steps
1. Consider a gas at low pressure. We expect it to behave ideally
2. But if the temperature is low, the molecules will be having low kinetic energy
♦ So they will be moving slowly
3. In such a situation, the molecules may get attracted to each other
• That means, at low temperatures, the following assumption in the kinetic molecular theory is not valid:
♦ There is no inter molecular forces in gases
• That means, at low temperatures, deviation from ideal behaviour may occur
4. Thus we can write:
■ For ideal behaviour, low pressure and high temperature are required
5. It is not easy to 'vary the temperature' of a gas sample
• But we can easily 'vary the applied pressure' by adjusting the piston
6. So we fix the temperature at a convenient trial value
• Then we note down the various pressure values at which ideal behaviour is obtained
• If at that temperature, ideal behaviour can be obtained for a large range of pressure values, it is called Boyle temperature or Boyle point
Now we will see liquefaction of gases. It can be written in 15 steps:
1. We have learnt about isotherms when we discussed Boyle’s law in a previous section (Details here)
♦ Fig.5.27(a) below, shows five isotherms of CO2
2. The red (50 oC) and green (31.1 oC) curves have the familiar shapes that we saw before
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| Fig.5.27 |
• We will now study those three curves in detail
3. Consider the bottom most blue curve. It is named as PQRS in fig.b
• Let us travel from P to S along the path PQRS
4. First segment is PQ
• From P to Q, it is a straight line, sloping downwards towards the right
♦ That means, as we move from right to left along PQ, the pressure increases and volume decreases
5. The next section is QR
• From Q to R, it is a horizontal straight line
♦ That means, as we move from right to left along QR, the volume decreases but the pressure remains constant
• How can pressure remain constant even when volume decreases?
• The answer can be written in 6 steps:
(i) Consider the pressure at Q
♦ Let us denote it as pQ
(ii) The gas offers a resistive pressure of pQ
♦ This resistance is the reading in the pressure gauge
(iii) We try to increase the pressure by pushing down the piston
♦ The volume decreases, but the pressure is not increasing. It remains at pQ
(iv) This is because, the gas is not putting up any 'increased resistance'
♦ The resistance put up by the gas remains constant at pQ
(v) The extra pressure that we provide through the piston is absorbed by the gas
♦ This absorbed pressure is used for the condensation of the gas into liquid
♦ The condensation begins at Q
(vi) As we move from Q to R, more and more gas is turned into liquid
♦ When we reach R, all the gas is converted into liquid
• Now we know the reason for the 'horizontal nature of QR
6. The next section is RS
• To the left of R, there is no gas. Only liquid
• Once all the gas is converted, the liquid begins to put up resistance
♦ The pressure gauge will begin to show increased readings
• We see that, RS is a steep line
• Since it is a steep line, we can write:
♦ To the left of R, even for large increase in pressure, the decrease in volume will be small
♦ This is obvious because, liquids are very difficult to compress
7. So we have seen the details about PQRS
• The second curve from bottom (the magenta curve) is named as ABCD
♦ It has a shape similar to that of PQRS
• So all the details will be the same. We need not write them again
• However, some interesting points can be written in 3 steps:
(i) Beginning of condensation:
♦ For PQRS, the condensation begins at Q
♦ For ABCD, the condensation begins at B
(ii) Comparison of pressures:
♦ The pressure pQ is smaller than pressure pB
(iii) Comparison of temperatures
♦ For PQRS, the temperature is 13.1 oC
♦ For ABCD, the temperature is 21.5 oC
■ Based on the above three steps, we can write:
♦ It is easier to liquefy a gas when the temperature is lower
✰ 'Easier' indicates that, a lower pressure is sufficient for obtaining liquefaction
8. So we have seen two curves which have a horizontal segment
• We can draw several such curves
• For curves in the upper regions, the length of the horizontal segment decreases
♦ That means, as temperature increases, the length of the horizontal segment decreases
• Finally, the ‘horizontal segment’ will reduce to a ‘point’
♦ This is shown by the third curve (the white curve) EFG
• We see that, instead of a horizontal segment, the curve EFG has a single point ‘F’
9. This ‘F is an important point. It can be explained in 9 steps:
(i) F lies on the isotherm 30.98
♦ At F, there is a 'slight bend' for the isotherm
♦ This ‘slight bend’ indicates the merger of horizontal segments (like QR and BC) into a single point
(ii) So we can write:
♦ Isotherms below 30.98:
✰ All of them will have a horizontal segment
♦ Isotherms above 30.98:
✰ All of them will be smooth
(iii) That means, the isotherm 30.98 is a transition
• The gas must be at a temperature 'equal to or less than 30.98 oC' if liquefaction is to be achieved
(iv) Every point on the isotherm 30.98 will be at a temperature of 30.98 oC
• But each of those points will be having it’s own pressure value
(v) Consider the following points:
• The points lying on the isotherm 30.98
♦ And to the right of F
✰ These points will be having pressure values less than pF
✰ The gas will not liquefy at those pressure values
(vi) Consider the following points:
• The points lying on the isotherm 30.98
♦ And to the left of F
✰ These points will be having pressure values greater than pF
✰ The gas will liquefy at any of those pressure values
(vii) But we do not need high pressures
♦ pF is sufficient to achieve liquefaction
(ix) So the temperature and pressure at F are important
■ The temperature at F is called critical temperature (Tc)
♦ Since F lies on the isotherm 30.98, we know that:
♦ Tc of carbon dioxide is 30.98 oC
♦ Above 30.98 oC, CO2 will always be in the gaseous state
♦ 30.98 oC is the highest temperature at which liquid carbon dioxide is observed
■ Volume of one mole of the gas at critical temperature is called critical volume (Vc)
■ The pressure at critical temperature is called critical pressure (pc)
■ The critical temperature, pressure and volume are called critical constants
10. We have seen two curves which have a horizontal segment: PQRS and ABCD
• We can draw several curves with such horizontal segment
• As mentioned earlier, for the curves in the upper regions, the length of the horizontal segment decreases
• That means, as temperature increases, the length of the horizontal segment decreases
• Finally, the ‘horizontal segment’ will reduce to a ‘point’
♦ In our present case, the 'point' is F
11. Though not as important as F, the points like B, C, R and Q are also salient points
• This can be explained in 5 steps:
(i) Points like B, C, R and Q are the ends of the horizontal segments
(ii) Condensation begins at the right end points of the horizontal segments
♦ Further to the right of these points, there is only gas
(iii) Condensation ends at the left end points of the horizontal segments
♦ Further to the left of these points, there is only liquid
(iv) So in between the end points of the horizontal segments, gas and liquid coexist
(v) We can join those end points through a smooth curve
♦ We will get a dome shaped region. This is shown in fig.5.28(a) below:
♦ Within this dome shaped region, gas and liquid coexist
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| Fig.5.28 |
• We now know how to carry out the process. We will write it in 4 steps:
(i) First cool the gas so that the temperature falls below Tc of that gas
(ii) Then increase the pressure by compressing it
(iii) By the above two steps, the gas will begin to condense little by little
♦ Beginning of the condensation is marked by the right end of the horizontal line
♦ End of the condensation is marked by the left end of the horizontal line
(iv) When the left end is reached, all gas would be converted to liquid
13. It is obvious that, during ‘some part of the process’, both liquid phase and gaseous phase will coexist
♦ We have seen that, this ‘some part of the process’ will be within the dome shaped region
14. We can think of another method in which only one phase (either liquid phase or gaseous phase) is present
• This can be explained in 3 steps:
(i) Consider the process in which a gas is slowly being converted into a liquid
(ii) Take ‘any instant during the process’
• At that instant, examine the contents inside the cylinder
♦ All the contents inside the cylinder must be in the gaseous state
♦ OR
♦ All the contents inside the cylinder must be in the liquid state
(iii) In other words
♦ The contents should never be in the form of a 'mixture' of gas and liquid
♦ It must be 'pure gas' or 'pure liquid'
15. Such a process can be achieved in steps:
Step A:
• Consider fig.5.28(b) above
♦ Let the temperature of the gas inside the cylinder be 21.5 oC
♦ Let the pressure be pA
♦ Let the volume be VA
• This situation is represented by point A
♦ A is outside and to the right of the dome
♦ So obviously, only gas is present
(i) Increase the temperature of the gas to 31.1 oC
♦ All the while this ‘increase in temperature’ is effected, the volume must remain constant at VA
♦ That is., the piston must remain at the same position
(ii) As the volume is kept same and temperature is increased, the pressure of the gas will increase
♦ The gas will reach a new point X
(iii) It is easy to find the position of X:
♦ Since the new temperature is 31.1 oC, X will lie on the isotherm 31.1
♦ Since the volume is the same, X will lie on the vertical through VA
♦ So the point of intersection of the isotherm and the vertical will be the position of X
(iv) pX will be the new pressure
(v) The path AX is indicated by the hatched thick yellow line
Step B:
(i) Decrease the volume from VA to VY
♦ This can be achieved by pushing the piston down
(ii) All the while this ‘decrease in volume’ is effected, temperature must remain constant at 31.1 oC
(iii) As the temperature is kept same and volume is decreased, the pressure will increase
♦ The pressure will become pY
♦ We can see that, pY is greater than pX
(iv) The path XY is indicated by the hatched thick yellow line
(v) As the temperature remains constant (at 31.1 oC) during this step, the path XY will lie along the isotherm 31.1
Step C:
(i) Decrease the temperature
♦ All the while this ‘decrease in temperature’ is effected, the volume must remain constant at VY
♦ That is., the piston must remain at the same position
(ii) Since the volume remains the same, the next point Z will lie on the vertical line through Y
♦ So we move down through that vertical line
(iii) As we move down, the temperature falls because, all points that we meet, will be below the isotherm 31.1
(iv) As we move further down, we will cross the isotherm 30.98
(v) As soon as we cross that isotherm 30.98, all the gas will be turned to liquid
(vi) The path YZ is indicated by the hatched thick yellow line
(vii) So the total path we travel is: AXYZ
(viii) Thus we avoid the dome shaped region and achieve liquefaction
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