In the previous section, we saw the basic details about Charles' law. We saw the kelvin scale also. In this section, we will see Gay Lussac's law. We will also learn about Avogadro law
• Works done by Gay Lussac revealed a basic fact. It can be written as:
■ If the volume and mass of a gas remains constant, then:
♦ The pressure of that gas will increase if it’s temperature is increased
♦ The pressure of that gas will decrease if it’s temperature is decreased
■ His findings were published as: Gay Lussac’s Law
■ The law states that:
At constant volume, the pressure of a fixed mass of a gas is directly proportional to it’s absolute temperature
• We can write an explanation in 15 steps
(Recall that, we will have already seen ‘absolute temperature’ in the previous section)
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 volume V1 of the sample
3. Measure the initial pressure p1 of the sample
4. Measure the initial temperature T1 of the sample
♦ For our present case, temperatures must be measured in the Kelvin scale
5. Increase the temperature of the sample to T2
♦ Measure the new pressure p2
6. Increase the temperature of the sample to T3
♦ Measure the new pressure p3
7. A number of Temperature-pressure readings can be taken in this way
• We get a set of readings: (T1, p1), (T2, p2), (T3, p3), . . .
■ For all the readings, the volume V 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
8. Gay Lussac found out that, there is a definite relation between the applied temperature and resulting pressure
• He found out that, the pressure is directly proportional to temperature
('Directly proportional' means, when one quantity increases, the other quantity also increases and vice versa)
• That is: p ∝T
• This can be written as: p = k3 × T
♦ Where k3 is the constant of proportionality
9. Applying the equation to the first reading, we get: p1=k3×T1
$\mathbf\small{\rm{\Rightarrow \frac{p_1}{T_1}= k_3}}$
• Applying the equation to the second reading, we get: p2=k3×T2
$\mathbf\small{\rm{\Rightarrow \frac{p_2}{T_2}= k_3}}$
• Applying the equation to the third reading, we get: p3=k3×T3
$\mathbf\small{\rm{\Rightarrow \frac{p_3}{T_3}= k_3}}$
so on . . .
10. Since all results are k3, we get: $\mathbf\small{\rm{\frac{p_1}{T_1}= \frac{p_2}{T_2}=\frac{p_3}{T_3}}}$ . . . so on . . .
11. Let us plot the relation: p = k3 × T
♦ We can plot T1, T2, T3, . . . along the x-axis
♦ We can plot p1, p2, p3, . . . along the y-axis
• The resulting graph is the blue line shown in fig.5.15(a) below:
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| Fig.5.15 |
■ In this graph, all the readings were taken when the volume of the sample was V1. If there is any change in this volume, 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
12. The shape of the graph in fig.5.15(a) is: straight line. This is expected because:
12. The shape of the graph in fig.5.15(a) is: straight line. This is expected because:
p = k3 × T is of the form: y = mx
12. Next we repeat the experiment on the same sample
♦ That is., we take the readings (T1, p1), (T2, p2), (T3, p3), . . . again
♦ But this time, the volume must be another convenient value V2 all the while
♦ This time, the graph is the green line in fig.5.15 (b) above
13. We repeat the experiment again
♦ That is., we take the readings (T1, p1), (T2, p2), (T3, p3), . . . again
♦ Again this time, the volume must be another convenient value V3 all the while
♦ This time, the graph is the magenta line in fig.5.15 (b) above
14. We repeat the experiment one more time
♦ That is., we take the readings (T1, p1), (T2, p2), (T3, p3), . . . one more time
♦ Again this time, the volume must be another convenient value V4 all the while
♦ This time, the graph is the red line in fig.5.15 (b) above
15. So in fig.5.15(b):
♦ All the readings in the blue line are taken when the volume is V1
♦ All the readings in the green line are taken when the volume is V2
♦ All the readings in the magenta line are taken when the volume is V3
♦ All the readings in the red line are taken when volume is V4
■ Since the volume is constant for any line, each line in the p-T graph in fig.5.15(b) is known as an isochore
13. The constants will be different
♦ The constant k3 of the blue line
✰ will be different from
♦ The constant k3 of the green line
✰ will be different from
♦ The constant k3 of the magenta line
✰ will be different from
♦ The constant k3 of the red line
• We obtain different lines in the fig.5.15(b) because, the constants are all different
♦ If the constants were the same, the lines would overlap
14. Interrelation between equations:
• Previously, we have seen two equations:
♦ $\mathbf\small{\rm{p_1V_1=p_2V_2}}$ from Boyle’s law
✰ This equation connects Pressure and volume
♦ $\mathbf\small{\rm{\frac{V_1}{T_1}=\frac{V_2}{T_2}}}$ from Charles’ law
✰ This equation connects volume and temperature
• Now we see a third equation from Gay Lussac’s law: $\mathbf\small{\rm{\frac{p_1}{T_1}=\frac{p_2}{T_2}}}$
✰ This equation connects pressure and temperature
■ The three equations are interconnected
15. We can derive Gay Lussac’s law from Boyle’s law and Charles’ law
• It can be demonstrated in 4 steps:
(i) Boyle's law is an equality: $\mathbf\small{\rm{p_1V_1=p_2V_2}}$
• If we multiply both sides of an equality with the same factor, the equality will not change
(ii) Let us multiply both sides by $\mathbf\small{\rm{\frac{V_1}{T_1}}}$
• We get: $\mathbf\small{\rm{(p_1V_1)\times \frac{V_1}{T_1}=(p_2V_2)\times \frac{V_1}{T_1}}}$
(iii) But from Charles’ law, $\mathbf\small{\rm{\frac{V_1}{T_1}}}$ is $\mathbf\small{\rm{\frac{V_2}{T_2}}}$
• So (ii) becomes: $\mathbf\small{\rm{(p_1V_1)\times \frac{V_1}{T_1}=(p_2V_2)\times \frac{V_2}{T_2}}}$
$\mathbf\small{\rm{\Rightarrow \frac{p_1 V_1^2}{T_1}=\frac{p_2V_2^2}{T_2}}}$
(iv) Now, according to Gay Lussac’s law, the volume must remain constant. That is., V1 = V2
• So (iii) becomes $\mathbf\small{\rm{\frac{p_1}{T_1}=\frac{p_2}{T_2}}}$
• This is the mathematical form of Gay Lussac’s law
Avogadro law
• So far, we have seen three laws:
Boyle’s law, Charles’ law and Gay Lussac’s law
• There are four variables, involved in each of the above three laws
♦ They are: mass, volume, pressure and temperature
• But we can plot only two variables in a graph
♦ So two variables were plotted and the remaining two were kept as constants
■ We can write:
♦ In Boyle’s law:
✰ Mass and temperature are constants
✰ Pressure and volume are the variables
♦ In Charles’ law:
✰ Mass and pressure are constants
✰ Volume and temperature are the variables
♦ In Gay Lussac’s law:
✰ Mass and volume are constants
✰ Pressure and temperature are the variables
• We see that, mass was always a constant
♦ We have not yet seen the effects when the mass of a sample is varied
• This is exactly what we are going to see in Avogadro law
• So we can prepare the following table:
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| Table 5.2 |
• Works done by the Italian scientist Amedeo Avogadro revealed a basic fact. It can be written as:
■ If the pressure and temperature of a gas remains constant, then:♦ The volume of that gas will increase if it’s no. of moles (n) is increased
♦ The volume of that gas will decrease if it’s no. of moles (n) is decreased
■ Avogadro law states that:
One mole of each gas will occupy a volume of 22.71098 L at standard temperature and pressure
• We can write an explanation in 17 steps. While writing the steps, we will also see what 'standard temperature and pressure' is
1. Take a sample of a gas
• Note down the number of moles (n1) 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 pressure p1 of the sample
3. Note down the temperature T1 of the sample
♦ For our present case, temperatures must be measured in the Kelvin scale
4. Measure the initial volume V1 of the sample
5. Increase the number of moles in the sample to n2
♦ Measure the new volume V2
6. Increase the number of moles in the sample to n3
♦ Measure the new volume V3
7. A number of n-volume readings can be taken in this way
• We get a set of readings: (n1, V1), (n2, V2), (n3, V3), . . .
■ For all the readings, the pressure p must be the same as noted in (2)
■ For all the readings, the temperature T must be the same as noted in (3)
8. Avogadro found out that, there is a definite relation between the 'n' and resulting volume
• He found out that, the volume is directly proportional to n
('Directly proportional' means, when one quantity increases, the other quantity also increases and vice versa)
• That is: V ∝ n
• This can be written as: V = k4 × n
♦ Where k4 is the constant of proportionality
9. Applying the equation to the first reading, we get: V1=k4×n1
$\mathbf\small{\rm{\Rightarrow \frac{V_1}{n_1}= k_4}}$
• Applying the equation to the second reading, we get: V2=k4×n2
$\mathbf\small{\rm{\Rightarrow \frac{V_2}{n_2}= k_4}}$
• Applying the equation to the third reading, we get: V3=k4×n3
$\mathbf\small{\rm{\Rightarrow \frac{V_3}{n_3}= k_4}}$
so on . . .
10. Since all results are k4, we get: $\mathbf\small{\rm{\frac{V_1}{n_1}= \frac{V_2}{n_2}=\frac{V_3}{n_3}}}$ . . . so on . . .
11. Let us plot the relation: V = k4 × n
♦ We can plot n1, n2, n3, . . . along the x-axis
♦ We can plot V1, V2, V3, . . . along the y-axis
• The resulting graph is the blue line shown in fig.5.16(a) below:
 |
| Fig.5.16 |
■ Note that, all points which fall along the blue line has two peculiarities:
♦ All those points were measured when the pressure was constant at p1
♦ All those points were measured when the temperature was constant at T1
11. From this graph, we can obtain a 'special information'. For that, we follow 5 steps:
(i) Draw a vertical white dashed line through n=1
♦ This is shown in fig.5.16(b)
(ii) Mark the point of intersection of this vertical dashed line with the graph
(iii) Draw a horizontal blue dashed line through the point of intersection
(iv) Mark the point of intersection of this horizontal dashed line with the y-axis
(v) This point of intersection will give the 'volume when number of moles is 1'
■ So based on fig.5.16(b), we can write:
When pressure and temperature are p1 and T1, one mole of the gas will occupy a volume V1
12. But there is a problem. It can be written in 4 steps:
(i) We obtained an 'important volume' from fig.5.16(b)
♦ But this volume is true only when pressure and temperature are p1 and T1
(ii) If the pressure or temperature change, we will not obtain the blue line in fig.5.16
♦ The graph will be different
♦ This is clear from fig.5.17(a) below:
 |
| Fig.5.17 |
♦ When pressure and temperature are p1 and T1, we will obtain the blue line
♦ When pressure and temperature are p2 and T2, we will obtain the green line
♦ When pressure and temperature are p3 and T3, we will obtain the red line
♦ so on . . .
(iii) Now consider fig.5.17(b)
• We see that:
♦ When pressure and temperature are p1 and T1, one mole will occupy a volume V1
♦ When pressure and temperature are p2 and T2, one mole will occupy a volume V2
♦ When pressure and temperature are p3 and T3, one mole will occupy a volume V3
(iv) It is clear that:
■ One mole can occupy different volumes
♦ It will depend on the pressure and temperature
13. So different people will be getting different volumes
♦ One might say: The volume occupied by one mole of gas is 25 L
♦ Another might say: The volume occupied by one mole of gas is 18 L
♦ so on . . .
■ In order to overcome this difficulty, scientists decided to strictly specify the pressure and temperature
♦ A pressure of 1 bar
♦ and
♦ A temperature of 273.15 K
♦ Were specified
14. So, if we want to find 'the volume occupied by one mole' we must take the readings while maintaining a constant pressure of 1 bar and a constant temperature of 273.15 K
• 1 bar is called the standard pressure
♦ 1 bar = 105 Nm-2
♦ Another name for Nm-2 is pascal
♦ So 1 bar = 105 pascal
♦ 1 bar is the atmospheric pressure experienced at sea level
• 273.15 K is called the standard temperature
♦ We know that, it is the freezing point of water
♦ In Celsius scale, it is 0 oC
■ Together, they are called: Standard temperature and pressure
■ In short form, it is: STP
15. So labs around the world adopted the new standards
• Experiments were performed while keeping:
♦ The pressure at a constant value of 1 bar
♦ The temperature at a constant value of 273.15 K
■ Following are the results obtained:
• At STP:
♦ 1 mol of Argon will occupy 22.37 L
♦ 1 mol of Carbon dioxide will occupy 22.54 L
♦ 1 mol of Dinitrogen (N2) will occupy 22.69 L
♦ 1 mol of Dioxygen (O2) will occupy 22.69 L
♦ 1 mol of Dihydrogen (H2) will occupy 22.72 L
♦ 1 mol of Ideal gas will occupy 22.71 L
• We see that, most values are close to 22.7 L
16. The volume occupied by 'one mole of a gas' is called the molar volume of that gas
• After conducting a very large number of trials, a value of 22.71098 L is now accepted
■ So we can write: At STP, the molar volume of an ideal gas is 22.71098 L
• Thus we have the final graph shown in fig.5.18 below:
 |
| Fig.5.18 |
17. While doing problems involving 'number of moles', we can bring in 'density' also into the calculations
• This can be explained in 4 steps:
(i) Le 'm' be the mass of the gas sample
♦ Then number of moles in that sample will be given by: $\mathbf\small{\rm{n=\frac{m}{M}}}$
♦ Where 'M' is the molar mass of the gaseous substance
(ii) We have: V = k4 × n
♦ Substituting for 'n', we get: $\mathbf\small{\rm{V=k_4 \times \frac{m}{M}}}$
(iii) Rearranging this, we get: $\mathbf\small{\rm{M=k_4 \times \frac{m}{V}}}$
$\mathbf\small{\rm{\Rightarrow M=k_4 \times d}}$
♦ Where 'd' is the density, which is the obtained by dividing mass by volume
(iv) From this equation it is clear that, density of a gas is proportional to it's molar mass M
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