In the previous section we saw that, real gases deviate from ideal gas behaviour. In this section, we will see the reasons for the deviation
We can write an analysis in 25 steps:
1. Consider the equation for the ideal gas: $\mathbf\small{\rm{V=\frac{nRT}{p}}}$
2. From this equation, we can calculate the volume occupied by ‘one mole of an ideal gas’
♦ All we need to do is: put n = 1
3. Volume occupied by one mole is the molar volume. We can denote it as: Vm
♦ For an ideal gas, we can denote it as: Vm(ideal)
♦ So we get: $\mathbf\small{\rm{V_{m(ideal)}=\frac{RT}{p}}}$
■ That means:
When pressure is p and temperature is T, one mole of the ideal gas occupies a volume of Vm(ideal)
4. Now consider the real gas which we are investigating
♦ Let the volume occupied by n moles of this real gas be V
♦ The temperature and pressure must be same as those mentioned in (3)
• Then, at that pressure p and temperature T, one mole of the real gas will occupy a volume of $\mathbf\small{\rm{\frac{V}{n}}}$
5. As before, volume occupied by one mole is the molar volume. We can denote it as Vm.
♦ For a real gas, we can denote it as: Vm(real)
♦ So we get: $\mathbf\small{\rm{V_{m(real)}=\frac{V}{n}}}$
6. Taking ratio of the two molar volumes, we get:
$\mathbf\small{\rm{\frac{V_{m(real)}}{V_{m(ideal)}}=\frac{\frac{V}{n}}{\frac{RT}{p}}}}$
$\mathbf\small{\rm{\Rightarrow \frac{V_{m(real)}}{V_{m(ideal)}}=\frac{pV}{nRT}}}$
7. So to find the ratio between Vm(real) and Vm(ideal), all we need to do is: Calculate $\mathbf\small{\rm{\frac{pV}{nRT}}}$
• The advantage of this method can be explained in 3 steps:
(i) We have a gas under investigation
♦ We have n moles of that gas
♦ It is at a temperature T
• When a pressure p is applied on that gas, it occupies a volume V
(ii) We want to compare 'it's molar volume' to the 'molar volume of an ideal gas'
• That is., we want to compare the following two volumes A and B:
A. Volume occupied by one mole of the gas under investigation: Vm(real)
♦ This gas is at pressure p and temperature T
B. Volume occupied by one mole of an ideal gas: Vm(ideal)
♦ This ideal gas is the same pressure p and temperature T
(iii) For making the comparison, we take the ratio: $\mathbf\small{\rm{\frac{V_{m(real)}}{V_{m(ideal)}}}}$
■ If the ratio is 1, we can say:
♦ The volume occupied by 'one mole of the gas under investigation' at p and T
♦ is same as
♦ The volume occupied by 'one mole of the ideal gas' at p and T
■ If the ratio is less than 1, we can say:
♦ The volume occupied by 'one mole of the gas under investigation' at p and T
♦ is less than
♦ The volume occupied by 'one mole of the ideal gas' at p and T
(This is because, a fraction becomes less than 1, when the numerator is less than the denominator)
■ If the ratio is greater than 1, we can say:
♦ The volume occupied by 'one mole of the gas under investigation' at p and T
♦ is greater than
♦ The volume occupied by 'one mole of the ideal gas' at p and T
(This is because, a fraction becomes greater than 1, when the numerator is greater than the denominator)
■ Now, we have an easy method to find this ratio. All we need to do is: Calculate $\mathbf\small{\rm{\frac{pV}{nRT}}}$
■ The ratio $\mathbf\small{\rm{\frac{V_{m(real)}}{V_{m(molar)}}}}$ is called compressibility or compression factor
• It is denoted by the letter Z
■ So we can write: $\mathbf\small{\rm{Z=\frac{V_{m(real)}}{V_{m(molar)}}=\frac{p V}{nRT}}}$
8. Scientists calculated Z at various pressure values
• This can be explained in 4 steps:
(i) Take a sample of gas
♦ Note down it's temperature T
♦ Note down the number of moles n
(ii) Apply a pressure p1
♦ Measure the resulting volume V1
♦ Calculate $\mathbf\small{\rm{Z_1=\frac{p_1 V_1}{nRT}}}$
(ii) Apply a pressure p2
♦ Measure the resulting volume V2
♦ Calculate $\mathbf\small{\rm{Z_2=\frac{p_2 V_2}{nRT}}}$
(iii) Repeat the steps for different pressure values: p1, p2, p3, . . .
♦ Calculate the resulting ratios: Z1, Z2, Z3, . . .
♦ Plot p1, p2, p3, . . . along the x-axis
♦ Plot Z1, Z2, Z3, . . . along the y-axis
(iv) If the gas under investigation is CO2, we will obtain the green curve shown in fig.5.23(a) below:
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| Fig.5.23 |
• Like CO2, we repeat the steps for different gas samples: CH4, O2, H2, and N2.
♦ All those samples must have the same n and T
9. The main features of this graph can be written in 4 steps:
(i) A horizontal dashed line is drawn through Z = 1
• If the 'gas under investigation' is an ideal gas, then, what ever be the pressure, all points (p,Z) belonging to that gas will lie on that horizontal line
• We see that, none of the gas in fig.5.23(a) is ideal
(ii) For N2 and H2:
♦ All Z values are greater than 1
(iii) For O2, CH4 and CO2:
♦ Initially the Z values become lesser and lesser than 1
♦ After reaching a minimum Z, they rise up wards and become closer and closer to 1
♦ After reaching 1, they steadily become greater and greater than 1
• We are trying to find the reasons for 'deviation from ideal behaviour'
♦ The curves in fig.5.23(a) give some valuable clues
♦ The following steps from (10) to () gives a detailed analysis
10. Consider the graph curve of CO2. It is shown separately in fig.5.23(b)
• We see that, at pressures less than 200 bar, the Z value progressively becomes lesser and lesser than 1
♦ It reaches a minimum point B
• Let us analyse the portion AB. It can be written in 3 steps:
(i) All Z values in AB are less than 1
(ii) But $\mathbf\small{\rm{Z=\frac{p V}{nRT}}}$
♦ n and T are constants. So denominator is a constant
(iii) That means the decrease in Z is due to the decrease in the numerator 'pV'
• Decrease in 'pV' means:
♦ Either p or V is falling below the 'expected value'
♦ Also, may be both are falling below the 'expected value'
11. Let us see the reason for the fall in V. It can be written in 6 steps:
(i) In fig.5.24(a) below, a gas is compressed using a cylinder-piston mechanism
♦ The pressure applied is p1
♦ The volume (between the piston and the bottom end of the cylinder) is V1
• The gas molecules are very far apart. They have enough space to move about
 |
| Fig.5.24 |
♦ The volume (between the piston and the bottom end of the cylinder) reduces to V2
(iii) In the case of fig.b, we usually take V2 into the calculations
♦ But the 'self-volume of the molecules' needs to be taken into account
♦ We cannot ignore the self-volume because, the molecules now has lesser space to move about
(iv) The actual volume is less than the 'observed V2'
♦ That means, the 'actual volume' is fallen below the 'observed volume'
(v) So we have seen a reason for the fall in V
♦ Remember that, this reason is valid only when the pressure is very high as in fig.5.24(b)
♦ In fig.5.24(a), ‘error in V’ will not occur
(vi) The kinetic molecular theory will give an ‘erroneous V’ at high pressures
♦ Because, the theory ignores the self-volume of molecules
12. There is one more reason for the fall in V. It can be written in 6 steps:
(i) Consider fig.5.25(b) below
♦ The pressure is very high and so volume is reduced to V2
 |
| Fig.5.25 |
(ii) Since the volume is small, the molecules are close to each other
♦ When the molecules are close to each other, inter molecular attractive forces begin to operate
♦ These attractive forces are strong and hence cannot be ignored
♦ This is indicated by the thick yellow arrows in fig.b
(iii) In fig.a, the arrows are thin, indicating that the forces are negligible
(iv) Due to the strong forces in fig.b, the molecules tend to group together
♦ Due to the grouping, the space occupied by the molecules will be lesser than expected
♦ This will lead to a fall in V
(v) So we have another reason for the fall in V
♦ Remember that, this reason is valid only when the pressure is very high as in fig.5.25(b)
♦ In fig.5.25(a), this ‘error in V’ will not occur because, there is practically no attractive forces
(vi) The kinetic molecular theory will give an ‘erroneous V’ at high pressures
♦ Because, the theory ignores the attractive forces even when the molecules are close to each other
13. So we have seen how the volume becomes less at high pressures
• It is clear that, we have to use a ‘reduced volume’ in the calculations. This can be explained in 2 steps:
(i) For reducing the volume, we use ‘(V-nb)’ instead of V
♦ V is the observed volume
♦ n is the number of moles of the gas
♦ b is a constant which is unique for a gas
✰ For examples, for CO2, the value of b is 0.04267 $\mathbf\small{\rm{\frac{L}{mol}}}$
(ii) The b values for various gases can be obtained from the data book
14. Next, we will see the adjustments that have to be made to p. The basics about the adjustments can be written in 3 steps:
(i) Consider fig.5.25(b) again
♦ The pressure is high at p2 and the volume is low at v2
♦ We saw that, there are strong inter molecular attractions in such a situation
(ii) Consider the molecules near the walls of the cylinder
♦ Because of strong attractions, those molecules are pulled back to the interior portion of the cylinder
♦ So those molecules will not strike the walls with the same force as in fig.a
(iii) So in fig.b, the pressure will be different from the observed value
15. We want to find the actual pressure. This can be explained in 5 steps:
(i) Consider fig.5.26(a) below
• A pressure p1 is applied on the piston
• We would know that, the ‘pressure applied is indeed p1’ because, the reading in the ‘pressure gauge’ would be p1
♦ Some images of 'pressure gauge' can be seen here
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| Fig.5.26 |
(ii) The ‘pressure gauge’ is recording the ‘resistive pressure put up’ by the gas inside the cylinder
• So we can write:
♦ The reading in the pressure gauge [p1]
♦ is equal to
♦ The ‘actual resistive pressure put up’ by the gas [p1(ac)]
(iii) Now consider fig.5.26(b)
• A pressure p2 is applied on the piston
• We would know that, the ‘pressure applied is indeed p2’ because, the reading in the ‘pressure gauge’ would be p2
(iv) In this case, the ‘pressure gauge’ is recording the ‘net resistive pressure put up’ by the gas inside the cylinder
♦ The gas will put up an ‘outward resistive pressure’ of p2(ac)
♦ But, as mentioned in (14), there is a reduction in pressure due to the attractions between molecules
✰ We will denote this 'reduction in pressure due to attraction' as: p2(at)
♦ The pressure gauge will be recording the net pressure
• So we can write:
♦ The reading in the pressure gauge [p2]
♦ is equal to
♦ The net of:
✰ The ‘actual resistive pressure put up’ by the gas [p2(ac)]
✰ and
✰ The reduction in pressure due to attraction [p2(at)]
(v) Thus we get:
p2 = [p2(ac)] - [p2(at)]
⇒ [p2(ac)] = p2 + [p2(at)]
16. So we have seen how the pressure changes at high values
• It is clear that, we have to use a ‘higher pressure’ in the calculations. This can be explained in 2 steps:
(i) For increasing the pressure, we use ‘$\mathbf\small{\rm{\left(p+\frac{an^2}{V^2} \right)}}$’ instead of p
♦ p is the observed pressure
♦ V is the observed volume
♦ n is the number of moles of the gas
♦ a is a constant which is unique for a gas
✰ For examples, for CO2, the value of a is 3.640 $\mathbf\small{\rm{\frac{L^2\; bar}{mol^2}}}$
(ii) The a values for various gases can be obtained from the data book
17. Thus we get the modified gas equation: $\mathbf\small{\rm{\left(p+\frac{an^2}{V^2} \right)(v-nb)=nRT}}$
♦ a and b are called: van der Waals constants
18. So we have two equations:
♦ The original ideal gas equation: pV = nRT
♦ The modified gas equation: $\mathbf\small{\rm{\left(p+\frac{an^2}{V^2} \right)(v-nb)=nRT}}$
• If a gas obey the original equation, all the (p, Z) points will lie on the horizontal dashed line in fig.5.23(a) above
• It is clear that, at high pressures, real gases do not obey the original equation
♦ That is why we get the curves in fig.5.23(a)
19. So we must be able to explain the curves using the modified equation
• We were discussing the green curve (CO2) in fig.5.23(b) above
• The details about portion AB can be written in 5 steps:
(i) In the portion AB, the Z values are less than 1
(ii) In step (10), we wrote that:
• The decrease in Z is due to the decrease in the numerator 'pV'
• Decrease in 'pV' means:
♦ Either p or V is falling below the 'expected value'
♦ Also, may be both p and V are falling below the 'expected value'
(iii) Now, we have just found out that, p cannot decrease. It can only increase
♦ Because, '$\mathbf\small{\rm{\left(\frac{an^2}{V^2} \right)}}$' is added to the observed p
• Also V cannot increase. It can only decrease
♦ Because, 'nb' is subtracted from the observed V
(iv) The 'increase in p' and 'decrease in V' will be occurring simultaneously
• So it is clear that, in the portion AB,
♦ The reduction in V
♦ is 'more than sufficient' to neutralize
♦ The rise in p
(v) Note the difference between:
♦ ‘more than sufficient’
♦ and
♦ ‘just sufficient’
• If it is ‘just sufficient’, the 'two opposing trends' will just cancel each other and the ratio will be unity
20. The next portion is BC. An explanation about this portion can be written in 3 steps:
(i) BC is a rising curve
• Even though it is rising, all Z values are less than unity
(ii) So we can write:
• In the portion BC,
♦ The reduction in V
♦ is 'more than sufficient' to neutralize
♦ The rise in p
(iii) But there is a difference from portion AB
♦ In portion BC, the ‘reduction in V’ is becoming less and less dominant
♦ That is., (V-nb) is becoming less and less dominant
♦ That is why, the curve is rising back to the horizontal dashed line
21. The next portion is CD. An explanation about this portion can be written in 2 steps:
(i) CD is above the horizontal dashed line
• That means, all Z values in the portion CD, are greater than unity
(ii) So we can write:
• In the portion CD,
♦ The rise in p
♦ is 'more than sufficient' to neutralize
♦ The reduction in V
22. Note that, at point C,
♦ The rise in p
♦ is 'just sufficient' to neutralize
♦ The reduction in V
23. So we have seen a satisfactory explanation for the curve of CO2
• The curve of CH4 and O2 has similar shapes. So they can be explained in the same way
24. But for the curves of H2 and N2, all Z values are above unity
• This is because, for all high pressure values,
♦ The rise in p
♦ is 'more than sufficient' to neutralize
♦ The reduction in V
25. So we see that, there are two types of gases:
♦ Those which have some Z values lesser than unity
♦ Those which do not have any Z value lesser than unity
• The difference is due to the ‘difference in attractive forces between molecules’. We will see more details in higher classes
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