Higher Secondary Physics: June 2021

In the previous chapter, we completed a discussion on Thermodynamics. In this chapter, we will see Kinetic theory

• Robert Boyle discovered that, if temperature of a gas is kept constant, it’s pressure will be inversely proportional to it’s volume
• This can be demonstrated as follows:
• Consider a gas inside a cylinder fitted with a movable piston
   ♦ If we move the piston downwards, the volume of the gas decreases
         ✰ At the same time, pressure of the gas increases
   ♦ If we move the piston upwards, the volume of the gas increases
         ✰ At the same time, pressure of the gas decreases
◼ This inverse relation is known as Boyle’s Law
• Robert Boyle made this discovery in the seventeenth century. Soon, other laws such as Charles Law, Avogadro's hypothesis etc., were discovered. All those laws were discovered in the seventeenth century. But precise mathematical explanations for those laws could be given only in the nineteenth century
• The mathematical explanations were based on the fact that, matter is made up of atoms and molecules. This fact about matter was not known to the scientists of the seventeenth century
• A sample of a gas is made up of a large number of atoms or molecules
   ♦ Noble gases like neon and argon exist as individual atoms
         ✰ For such gases, atoms and molecules implies the same
   ♦ Ordinary gases like oxygen and nitrogen exist as individual molecules
◼ Except when they are very close to each other, there is practically no interactions between the molecules in a gas sample
• This is because, compared to their sizes, the distances between those molecules are very large
• In this chapter, we will see how precise mathematical explanations can be given for the gas laws

Behavior of gases

• Some basics can be written in 7 steps:
1. Since the interactions between gas molecules are negligible, properties of gases are easier to understand. This is because:
• In the case of solids and liquids, we will have to deal with equations related to mutual forces between molecules
• But for gases, such equations are absent
2. Consider a gas sample at normal pressure and normal temperature

The word ‘normal’ needs to be given importance. The reason can be written in 2 steps:
(i) If the pressure is very high, the volume will be so low that, the molecules come into close contacts with each other
   ♦ Then they will begin to interact with each other
   ♦ In such a situation, our equations will not be valid
(ii) If the temperature is very low, the gases will become liquids
   ♦ The molecules will then be in close contact with each other
   ♦ They will begin to interact with each other
   ♦ Then also, our equations will not be valid

◼ When the pressure and temperature are normal, any given gas sample will satisfy the following relation:
PV = KT
   ♦ Where:
         ✰ P is the pressure
         ✰ V is the volume
         ✰ T is the temperature
3. This relation can be explained in 2 steps:
(i) PV = KT can be rearranged as: $\mathbf\small{\rm{\frac{PV}{T}=K}}$
   ♦ When temperature is T₁, let pressure be P₁ and volume be V₁
   ♦ When temperature is T₂, let pressure be P₂ and volume be V₂
so on . . .
(ii) So the pressure, temperature and volume can vary
• But the combination will always be constant
• That is: $\mathbf\small{\rm{\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}=\frac{P_3V_3}{T_3}=K}}$
4. But this K is constant for the given sample only
• If we take another (larger or smaller in volume) sample of the same gas, it will be having another value of K
• Scientists discovered that, this K is equal to Nk_B
   ♦ Where:
         ✰ N is the number of molecules of in the sample
         ✰ k_B is a constant called Boltzmann constant
         ✰ This k_B is same for all gases
5. It is clear that, the variation of K from sample to sample, is due to the difference in N
• We can write: Eq.13.1: PV = (Nk_B)T
◼ Now we can consider different samples
• Let in the first sample, N be N₁
In that sample,
   ♦ When temperature is T₁, let pressure be P₁ and volume be V₁
• Let in the second sample, N be N₂
In that sample,
   ♦ When temperature is T₂, let pressure be P₂ and volume be V₂
• Let in the third sample, N be N₃
In that sample,
   ♦ When temperature is T₃, let pressure be P₃ and volume be V₃
so on . . .
◼ Then the relation in 3(ii) becomes:
$\mathbf\small{\rm{\frac{P_1V_1}{N_1T_1}=\frac{P_2V_2}{N_2T_2}=\frac{P_3V_3}{N_3T_3}=k_B}}$
6. Based on the above result in (5), we can give a satisfactory explanation for Avogadro's hypothesis. It can be written in steps:
(i) Take two gas samples
• For the first sample, let the P, V and T be P₁, V₁ and T₁ respectively
   ♦ Then we can write: $\mathbf\small{\rm{\frac{P_1V_1}{T_1}=N_1k_B}}$
• For the second sample, let the P, V and T be P₂, V₂ and T₂ respectively
   ♦ Then we can write: $\mathbf\small{\rm{\frac{P_2V_2}{T_2}=N_2k_B}}$
(ii) Suppose that, the two samples are experiencing the same pressure and temperature. Also suppose that, the two samples have the same volume
• Then we can write: P₁ = P₂, V₁ = V₂, T₁ = T₂
(iii) Then N1k_B will become equal to N2k_B
• It follows that, N₁ = N₂
• This is Avogadro's hypothesis
7. Avogadro in the seventeenth century said that:
If two samples have the same pressure, volume and temperature, they will be having the same number of particles
• The above steps from (1) to (6) give the mathematical proof

Let us see the practical applications of Avogadro’s hypothesis. It can be written in 7 steps:
1. Taking two samples:
   ♦ The first sample is of gas A
   ♦ The second sample is of another gas B
2. The two samples must contain the same number of molecules
• Here a problem arises. It can be explained in 2 steps:
(i) Labs at some part of the world may prepare the two samples in such a way that, there are 20000 molecules in each of them
(ii) Labs at some other part of the world may prepare the two samples in such a way that, there are 50000 molecules in each of them
◼ In order to avoid such differences, scientists decided to fix up a number
• They said:
Each of the two samples should have exactly one mole of molecules
   ♦ One mole is 6.023 × 10²³ numbers
   ♦ It is just like: One dozen is 12 numbers
   ♦ So one mole of molecules is 6.023 × 10²³ molecules
• So the problem is solved
3. Next, the two samples must be subjected to the same pressure
• Here also a problem arises. It can be explained in 2 steps
(i) Some labs may subject each of the two samples to a pressure of 3 bar
(ii) Some other labs may subject each of the two samples to a pressure of 500 torr
◼ In order to avoid such differences, scientists decided to fix up a pressure
• They said:
Each of the two samples should be subjected to exactly 1 atm pressure
• So the problem is solved
4. Next, the two samples must be at the same temperature
• Here also a problem arises. It can be explained in 2 steps
(i) Some labs may keep each of the two samples at a temperature of 298 K
(ii) Some other labs may keep each of the two samples at a temperature of 300 K
◼ In order to avoid such differences, scientists decided to fix up a temperature
• They said:
Each of the two samples should exactly be at a temperature of 273 K
• So the problem is solved
5. The S.T.P:
   ♦ The temperature 273 K is accepted world wide as standard temperature
   ♦ The pressure 1 atm is accepted world wide as standard pressure
◼ Together, they are known as: Standard temperature and pressure
   ♦ It is abbreviated as S.T.P
6. So all the samples in all the labs will be:
   ♦ Containing one mole each
   ♦ Subjected to 1 atm pressure
   ♦ At a temperature of 273 K
◼ Now we come to the interesting result about volume:
If the above three conditions are satisfied, each sample will be occupying a volume of 22.4 liters
7. The Avogadro number:
• One mole of gaseous molecules (belonging to any gas oxygen, nitrogen, argon etc.,) at 273 K and 1 atm pressure will be occupying a volume of 22.4 liters
◼ Conversely,
• 22.4 liters of any gas at 273 K and 1 atm pressure, will be containing one mole (6.023 × 10²³) of molecules
◼ Since this fact was discovered by the Italian scientist Amedeo Avogadro, the number 6.023 × 10²³ is called Avogadro number in his honor
   ♦ The symbol for Avogadro number is N_A
• So we can write: N_A = 6.023 × 10²³

Next we will see the details about the Universal gas constant. It can be written in 7 steps:
1. We have seen that: $\mathbf\small{\rm{\frac{PV}{T}=N\;k_B}}$
• Dividing both sides by N_A, we get: $\mathbf\small{\rm{\frac{PV}{T\;N_A}=\left(\frac{N}{N_A} \right)\;k_B}}$
2. But $\mathbf\small{\rm{\left(\frac{N}{N_A} \right)}}$ is the number of moles in the sample
   ♦ We will denote it as 𝝁
• So we get: $\mathbf\small{\rm{\frac{PV}{T\;N_A}=\mu\;K_B}}$
⇒ PV = 𝝁N_Ak_BT
3. Both N_A and k_B are constants
• So multiplying them will give a new constant. Let us denote this new constant as R
• Then we get: Eq.13.2: PV = 𝝁RT
4. R is called the Universal Gas Constant. Because, it is same for all gases which are at STP
• Let us find the value of R. It can be done in 2 steps:
(i) We have: $\mathbf\small{\rm{R=\frac{PV}{\mu T}}}$
• At STP,
   ♦ P = 1 atm = 101325 N m^-2
   ♦ V = 22.4 liters = (22.4 × 10³) cm³ = (22.4 × 10³ × 10^-6) = 22.4 × 10^-3 m³
   ♦ T = 273 K
   ♦ 𝝁 = 1 mole
(ii) Substituting these values in (i), we get:
$\mathbf\small{\rm{R=\frac{101325\;(Nm^{-2}) \times 22.4 \times 10^{-3} (m^3)}{1 \; (mole) \times 273 \; (K)}}}$ = 8.314 N m mole^-1 K^-1 = 8.314 J mole^-1 K^-1
5. We have seen that, 𝝁 of a sample is equal to $\mathbf\small{\rm{\left(\frac{N}{N_A} \right)}}$
• But it is easier to calculate 𝝁 using the relation: $\mathbf\small{\rm{\mu=\left(\frac{M}{M_0} \right)}}$
   ♦ Where
         ✰ M is the mass of the sample
         ✰ M₀ is the molar mass of the gas
• We already know this relation from our chemistry classes
6. The equation PV = 𝝁RT can be written in another form. It can be derived in 2 steps
(i) We have Eq.13.1: PV = (Nk_B)T
• This can be rearranged as: $\mathbf\small{\rm{P=\left( \frac{N}{V} \right)k_BT}}$
(ii) Now, $\mathbf\small{\rm{\left( \frac{N}{V} \right)}}$ is the number density, that is, the number of molecules per unit volume. We can denote it as n
• Thus we get Eq.13.3: P = k_BnT
7. Let us see yet another form. It can be derived in 2 steps:
(i) We have: Eq.13.2: PV = 𝝁RT
• This is same as: $\mathbf\small{\rm{PV=\left( \frac{M}{M_0} \right)RT}}$
This can be rearranged as: $\mathbf\small{\rm{P=\left( \frac{M}{VM_0} \right)RT}}$
(ii) Now, $\mathbf\small{\rm{\left( \frac{M}{V} \right)}}$ is the density. We can denote it as ρ
• Thus we get Eq.13.4: $\mathbf\small{\rm{P=\frac{\rho RT}{M_0}}}$

Next we can discuss about the Kinetic theory of an Ideal gas. Before taking up that discussion, we need to obtain a good understanding about Boyle's Law, Charles Law, Ideal gas equation etc., We have discussed them in some previous chapters in Chemistry. The links are given below:

1. Boyle's Law and Charles law

2. Avogadro's hypothesis and Details about STP

3. Ideal gas equation, Partial pressure and mole fraction

• Now we will see some solved examples. The link is given below:

Solved examples 13.1 to 13.11

In the next section, we will see Kinetic theory of an Ideal gas

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