In the previous section, we saw the basic details about inter molecular forces and thermal energy. In this section, we will see gaseous state
• Consider the elements like hydrogen, helium, oxygen, nitrogen etc.,
♦ They exist in the gaseous state even when temperature and pressure are normal
♦ In this section, we will be discussing about such gases
• By altering the normal temperature and pressure, some liquids and solids can be converted into gaseous state. We will not be discussing them here
■ The Anglo-Irish scientist Robert Boyle did extensive research on the properties of gases. He published his findings as the Boyle’s Law
■ The law states that:
At constant temperature, the pressure of a fixed amount of gas varies inversely with it’s volume
• We can write an explanation in 29 steps:
1. Take a sample of a gas
• Note down the number of moles present in it
Some examples:
• If the gas is N2, and if ‘m’ grams of N2 is present in that sample, then:
♦ There will be $\mathbf\small{\rm{\frac{m}{28}}}$ moles of N2 in that sample
• If the gas is CO2, and if ‘m’ grams of CO2 is present in that sample, then:
♦ There will be $\mathbf\small{\rm{\frac{m}{44}}}$ moles of CO2 in that sample
2. Note down the temperature T of the sample
♦ Let it be 200 K
3. Apply a pressure p1 on that sample
♦ Measure the resulting volume V1
• Apply another pressure p2 on that sample at the same temperature
♦ Measure the new volume v2
4. A number of pressure-volume readings can be taken in this way
• We get a set of readings: (p1, V1), (p2, V2), (p3, V3), . . .
■ For all the readings, the temperature T must be the same as noted in (2)
■ The number of moles should also be the same
♦ It is not difficult to keep the ‘number of moles same’
♦ Because, the same sample is used for all readings
5. Boyle found out that, there is a definite relation between the applied pressure and resulting volume
• He found out that, the pressure is inversely proportional to volume
('Inversely proportional' means, when one quantity increases, the other quantity decreases and vice versa)
• That is: $\mathbf\small{\rm{p \propto \frac{1}{V}}}$
• This can be written as: $\mathbf\small{\rm{p=k_1 \times \frac{1}{V}}}$
♦ Where k1 is the constant of proportionality
6. Applying the equation to the first reading, we get: $\mathbf\small{\rm{p_1=k_1 \times \frac{1}{V_1}}}$
$\mathbf\small{\rm{\Rightarrow p_1 V_1=k_1}}$
• Applying the equation to the first reading, we get: $\mathbf\small{\rm{p_2=k_1 \times \frac{1}{V_2}}}$
$\mathbf\small{\rm{\Rightarrow p_2 V_2=k_1}}$
• Applying the equation to the third reading, we get: $\mathbf\small{\rm{p_3=k_1 \times \frac{1}{V_3}}}$
$\mathbf\small{\rm{\Rightarrow p_3 V_3=k_1}}$
so on . . .
7. Since all results are k1, we get: $\mathbf\small{\rm{p_1 V_1=p_2 V_2=p_3 V_3\;.\;.\;.\;so\;on\;.\;.\;.}}$
7. Since all results are k1, we get: $\mathbf\small{\rm{p_1 V_1=p_2 V_2=p_3 V_3\;.\;.\;.\;so\;on\;.\;.\;.}}$
8. Let us plot the relation: $\mathbf\small{\rm{p=k_1 \times \frac{1}{V}}}$
♦ We can plot V1, V2, V3, . . . along the x-axis
♦ We can plot p1, p2, p3, . . . along the y-axis
• The resulting graph is the blue curve shown in fig.5.4(a) below
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| Fig.5.4 |
■ In this graph, all the readings were taken when the temperature of the sample was 200 K. If there is any change in this temperature, while any of the readings are taken, we will not get this shape
■ Also, if there is any change in the 'number of moles in the sample', we will not get this shape
9. The shape of the graph in fig.5.4(a) seems to be peculiar
9. The shape of the graph in fig.5.4(a) seems to be peculiar
• But it is a common shape, seen very often in science and engineering
♦ It is the graph of: $\mathbf\small{\rm{y= A\;constant \times \frac{1}{x}}}$
♦ Note the similarity between [$\mathbf\small{\rm{y= A\;constant \times \frac{1}{x}}}$] and [$\mathbf\small{\rm{p=k_1 \times \frac{1}{V}}}$]
♦ Note the similarity between [$\mathbf\small{\rm{y= A\;constant \times \frac{1}{x}}}$] and [$\mathbf\small{\rm{p=k_1 \times \frac{1}{V}}}$]
• The graph of [$\mathbf\small{\rm{y= 1 \times \frac{1}{x}}}$] is plotted in fig.5.4(b). Here, the value of the 'constant' is 1. Note the similarity between the two graphs
• Whenever two quantities have a 'inverse proportional relationship', the graph will have this shape
10. Next we repeat the experiment on the same sample
♦ That is., we take the readings (p1, V1), (p2, V2), (p3, V3), . . . again
♦ But this time, the temperature must be another convenient value (say 400 K) all the while
♦ This time, the graph is the green curve in fig.5.5 (a) below:
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| Fig.5.5 |
11. We repeat the experiment one more time
♦ That is., we take the readings (p1, V1), (p2, V2), (p3, V3), . . . one more time
♦ Again this time, the temperature must be another convenient value (say 600 K) all the while
♦ This time, the graph is the red curve in fig.5.5 (a) above
12. So in fig.5.5(a):
♦ All the readings in the blue curve are taken when the temperature is 200 K
♦ All the readings in the green curve are taken when the temperature is 400 K
♦ All the readings in the red curve are taken when the temperature is 500 K
■ Since the temperature is constant for any curve, each curve in the p-V graph in fig.5.5(a) is known as an isotherm
13. The constants will be different
♦ The constant k1 of the blue curve
✰ will be different from
♦ The constant k1 of the green curve
✰ will be different from
♦ The constant k1 of the red curve
• We obtain different curves in the fig.5.5(a) because, the constants are all different
♦ If the constants were the same, the curves would overlap
14. It is possible to obtain different curves with [$\mathbf\small{\rm{y= A\;constant \times \frac{1}{x}}}$] also. This is shown in fig.5.5(b)
• In fig.5.5(b), we see that:
♦ For the blue curve, the constant is 1
♦ For the green curve, the constant is 2
♦ For the green curve, the constant is 3
15. Now we know the details about the 'graphical representation of Boyle's Law'
• Let us see another method for making the graphical representation
• The next seven steps from (15) to (21) explains this method:
16. Consider the readings that we obtained in (4):
(p1, V1), (p2, V2), (p3, V3), . . .
• Before plotting them, we convert them into the form:
(p1, V'1), (p2, V'2), (p3, V'3), . . .
♦ Where: $\mathbf\small{\rm{V'_1 = \frac{1}{V_1},\;\;\;\;V'_2 = \frac{1}{V_2},\;\;\;\;V'_3 = \frac{1}{V_3},\;.\;.\;.\;so\;on\;.\;.\;.}}$
• That means, we have to pick out each volume and write it's reciprocal
17. Now, when we apply the Boyle's law, we get:
$\mathbf\small{\rm{p\;=\;k_1 \times \frac{1}{V}\;=\;k_1 \times V'}}$
$\mathbf\small{\rm{\Rightarrow\;p\;\;=\;\;k_1 \times V'}}$
18. So we plot a p vs V' graph
♦ We can plot V'1, V'2, V'3, . . . along the x-axis
♦ We can plot p1, p2, p3, . . . along the y-axis
• The resulting graph is the blue line shown in fig.5.6(a) below:
 |
| Fig.5.6 |
■ In this graph, all the readings were taken when the temperature of the sample was 200 K. If there is any change in this temperature, while any of the readings are taken, we will not get this shape
■ Also, if there is any change in the 'number of moles in the sample', we will not get this shape
19. The shape of the graph in fig.5.6(a) is: straight line
19. The shape of the graph in fig.5.6(a) is: straight line
• This is expected because:
♦ [$\mathbf\small{\rm{p\;\;=\;\;k_1 \times V'}}$] is the equation of a straight line
♦ It is of the form: [y = mx] that we commonly see in analytical geometry classes
20. Next we take the readings obtained in (9)
♦ We convert those readings into the form: (p1, V'1), (p2, V'2), (p3, V'3), . . .
♦ This time, the graph is the green line fig.5.6 (b) above
21. Next we take the readings obtained in (10)
♦ We convert the readings into the form: (p1, V'1), (p2, V'2), (p3, V'3), . . .
♦ This time, the graph is the red line fig.5.6 (b) above
22. So we have seen two methods of plotting the relation between pressure and volume
■ The 'straight line method' has an advantage. It can be explained in 3 steps:
(i) If we obtain straight lines as in fig.5.6 above, we can say that the sample obeys Boyle’s law
(ii) In practice, it is seen that, at high pressures, the ‘straight line character’ is not obtained
(iii) That means, at high pressure, gases deviate from Boyle’s law
23. Let us see some actual readings obtained during an experiment
• The readings are given in the table 5.1 below
♦ The experiment was done on 0.09 mol of CO2
♦ All the readings were taken when the sample was at a temperature of 300 K
 |
| Table 5.1 |
The following 4 details can be written about the table:
(i) The first column gives the pressure values
• The actual values are: 2 × 104, 2.5 × 104 etc.,
• Those are very large values
• For convenience of putting them in the table, each of those values are divided by 104
(ii) The second column gives the volume values
• The actual values are: 112 × 10-3, 89.2 × 10-3 etc.,
• Those are very small values
• For convenience of putting them in the table, each of those values are divided by 10-3
(iii) The third column gives the reciprocals of the values in the second column
• For example: The reciprocal of 112 × 10-3 of is 8.9
(iv) The fourth column gives the product of first and second columns
• The actual values are: 22.4 × 102, 22.3 × 102 etc.,
• Those are large values
• For convenience of putting them in the table, each of those values are divided by 102
24. Based on the table, we can plot two graphs: p-V graph and p-1/V graph
• The p-V graph is shown in fig.5.7 below:
 |
| Fig.5.7 |
• Note that, only the actual values should be used for plotting
♦ The factored values in the table should not be used
• However, suitable scales can be used while plotting
♦ The 'scales' need not be same as the 'factors used in the table'
♦ The scales must be clearly mentioned at the top right corner of the graph
25. The p-1/V graph is shown in fig.5.8 below:
 |
| Fig.5.8 |
• If the line is extended backwards, it will pass through the origin
♦ This is shown by the yellow dashed line
• Indeed, if a line of the form [y = mx] is extended, it will pass through the origin
26. Let us analyse the p-V graph:
• It can be done in 4 steps:
(i) On the y-axis, mark any two convenient points P and Q as shown in fig.5.9 below:
 |
| Fig.5.9 |
(ii) From P, draw a horizontal dashed line to meet the curve at P'
♦ From P’, draw a vertical dashed line to meet the x-axis
♦ This vertical line meets the x-axis at P’’
♦ So the volume at P’’ is the ‘volume corresponding to the pressure at P’
(iii) Repeat the same procedure for the other point Q
♦ Q’’ is the ‘volume corresponding to the pressure at Q’
(iv) We see that:
♦ Q is higher up above P
♦ Q’’ is to the left of P’’
■ So it is clear that, when pressure increases, volume decreases
27. Let us see the effect of ‘decrease in volume’. It can be written in 5 steps
(i) We increase the pressure of a gas by compressing it
• when the gas is compressed, the volume decreases
■ So we can write:
When pressure increases, volume decreases
(ii) Note that, there is no change in the mass of the gas before and after compression
• That means, after compression, the mass occupies a lesser space
(iii) We have: $\mathbf\small{\rm{Density=\frac{Mass}{Volume}}}$
• From this, we get: $\mathbf\small{\rm{d_1=\frac{m}{V_1}\;\;and\;\;d_2=\frac{m}{V_2}}}$
♦ Where:
✰ m is the unchanging mass
✰ d1 and V1 are the initial density and volume
✰ d2 and V2 are the final density and volume
(iv) We see that, the numerator m has no change
• But the denominator V2 is less than V1
• So d2 will be larger than d1
(v) That means, when we compress a gas, it becomes denser
• This is so because, after compression, the molecules have a lesser space to occupy
28. Modification of the Boyle’s law equation:
• We can modify the equation by including 'density' in it
• It can be written in 4 steps:
(i) We have the original equation: $\mathbf\small{\rm{p=k_1 \times \frac{1}{V}}}$
(ii) Also we have: $\mathbf\small{\rm{d=\frac{m}{V}}}$
$\mathbf\small{\rm{\Rightarrow V=\frac{m}{d}}}$
(iii) Substituting for V in (i), we get:
$\mathbf\small{\rm{p=k_1 \times \frac{1}{\frac{m}{d}}}}$
$\mathbf\small{\rm{\Rightarrow p=k_1 \times \frac{d}{m}=\left(\frac{k_1}{m} \right)d}}$
$\mathbf\small{\rm{\Rightarrow d=\left(\frac{m}{k_1} \right)p}}$
(iv) Both m and k1 are constants. So we can put a new constant k’
■ Thus we get: d = k'p
29. Based on the modified equation in (27), we can write the following points:
(i) Take a sample of a gas
♦ Note down the temperature T of the sample
♦ Note down the mass (no. of moles) of the sample
(ii) Keeping T and m constant, apply various pressures: p1, p2, p3, . . .
(iv) Determine the corresponding densities: d1, d2 d3, . . .
(v) Thus we get various pairs: (p1, d1), (p2, d2), (p3, d3), . . .
(vi) We will see that, each pair will satisfy the relation: d = k'p
• That means, the constant k’ is applicable to all pairs
(vii) So it is clear that, density is directly proportional to pressure
In other words: When pressure increases, density also increases
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