13.4 - Specific Heat Capacity of Solids

In the previous section, we saw specific heats of gases. In this section, we will see specific heat capacity of solids.

Details can be written in 4 steps:

1. For a solid atom, there will not be any translation or rotation. There will be vibration only.
• So by applying law of equipartition, We know that:
For each atom at temperature T, this energy will be $\small{K_B\,T}$
• Due to the vibration in all three dimensions, the average energy will be $\small{3K_B\,T}$.

2. In a mole of the solid, there will be $\small{N_A}$ atoms. So the total internal energy (U) of one mole of a solid can be obtained as:
$\small{U\,=\,3 K_B\,T\,N_A\,=\,3 R\,T}$

3. At constant pressure, we know that:
$\small{\Delta Q\,=\,\Delta U\,+\,P \Delta V}$ 
• For a solid, $\small{\Delta V}$ is negligible. So we get:
$\small{\Delta Q\,=\,\Delta U}$   

4. Specific heat is the change in internal energy per unit change in temperature. So we can obtain the specific heat for one mole as:
$\small{C\,=\,\frac{3 R\,T}{T}\,=\,3 R}$

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Using the above equation, we can predict the specific heat of different solids. At ordinary temperature, the predicted values are in agreement with the experimental values. However, carbon is an exception. We will see the reason in higher classes.

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Specific heat capacity of water

Details can be written in 4 steps:
1. We treat water like a solid. We saw that, for each solid atom, the energy is $\small{3K_B\,T}$. For each water molecule there are three atoms. So for a water molecule, the energy is $\small{9K_B\,T}$.

2. In a mole of the water, there will be $\small{N_A}$ molecules. So the total internal energy (U) of one mole of water can be obtained as:
$\small{U\,=\,9 K_B\,T\,N_A\,=\,9 R\,T}$

3. At constant pressure, we know that:
$\small{\Delta Q\,=\,\Delta U\,+\,P \Delta V}$ 
• Since we treat water as a solid, $\small{\Delta V}$ is negligible. So we get:
$\small{\Delta Q\,=\,\Delta U}$   

4. Specific heat is the change in internal energy per unit change in temperature. So we can obtain the specific heat for one mole as:
$\small{C\,=\,\frac{9 R\,T}{T}\,=\,9 R}$

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Using the above equation, we can predict the specific heat of water. The predicted value is in agreement with the experimental value.

• If we use the use the unit “$\small{\text{calorie}\, \text{gram}^{-1} \,\text{K}^{-1} }$”, the specific heat of water is one.
• Let us convert the unit into “$\small{\text{J}\, \text{mol}^{-1} \,\text{K}^{-1} }$”
    ♦ One calorie is 4.179 joules
    ♦ One mole of water is 18 grams.
• So we get:
$\small{1~\text{calorie}\, \text{gram}^{-1} \,\text{K}^{-1}~=~\frac{1(4.179)}{(1/18)}~=~75.222~\text{joule}\, \text{mole}^{-1} \,\text{K}^{-1} }$
• Now, 9R = 9(8.31) = 74.79 $\small{\text{joule}\, \text{mole}^{-1} \,\text{K}^{-1} }$
• So the values are in agreement.

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Next we will see an important aspect about the predicted values. It can be written in 4 steps:

1. We have seen the formulas for calculating the specific heats of various substances. In those formulas, temperature is absent. That means, the predicted values are calculated without taking temperature into account.

2. The experimental values have a marked difference from predicted values, when temperature is low. When experiments are done at temperatures close to zero, the specific heats also approach zero.
• This is because, when a substance freeze, the molecules/atoms in it have nearly zero degrees of freedom.

3. The kinetic theory is based on classical physics. For applying classical physics, the degrees of freedom must remain unchanged at all times.

4. However, quantum mechanics can make more accurate predictions. We will see those details in higher classes.

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Mean free path

• Molecules in a gas are moving with large speeds. Speeds that are comparable to the speed of sound. But consider the case when cooking gas leaks from a faulty cylinder. The person at the other side of the room will become aware of the leakage, only after a short duration of time. That means, the molecules of the gas did not travel at the speed of sound, all the way to the other side of the room.
• Similar is the case of a cloud of smoke. The top of the cloud retains the shape for a considerable duration of time. If the molecules of the smoke travel at the speed of sound, in imaginary straight lines drawn in all directions, we will be seeing instant dispersal of the smoke.
• Scientists have discovered the reason for such behaviour. They found out that, the molecules of the gas/smoke undergo collisions. Due to the collisions, the molecules cannot move in straight lines. The direction changes continuously.

Let us try to make a mathematical model for the ‘possibility for collision’. It can be written in 12 steps:
1. Suppose that, the molecules of a gas are spheres of diameter d. Let us concentrate on one such sphere.
• It is the middle sphere in fig.13.3 (a) below. Let it be traveling with the average speed $\small{\bar{v}}$.

Fig.13.3

2. Consider any near by sphere. If the distance between the centers of the two spheres is less than or equal to ‘d’, then there is a possibility for collision.
• In the fig.13.3 above, the middle sphere can collide with any other sphere,
   ♦ whose edge is on the green dashed line   
   ♦ whose edge is between blue and green dashed lines   
   ♦ whose edge is on the red dashed line   
   ♦ whose edge is between blue and red dashed lines   

3. So we can think about a cylinder whose radius is ‘d’. This cylinder is shown in yellow color in the fig.13.3 above. Note that:
   ♦ Diameter of the spheres is ‘d’
   ♦ Radius of the cylinder is also ‘d’
         ✰ This is shown in the diagram shown in fig.13.3(b).
• Then base area of the cylinder will be: $\small{\pi d^2}$

4. Height of the cylinder is equal to the distance traveled by the middle sphere in time $\small{\Delta t}$, which is equal to $\small{\bar{v}\,\Delta t}$
• So volume of the cylinder will be $\small{\pi d^2\,\bar{v}\,\Delta t}$

5. All spheres whose center is on or within the edge of the cylinder, will have a possibility for collision with the middle sphere.

6. If n is the number of spheres per unit volume, then the number of spheres mentioned in (5) will be, n times the volume of the cylinder, which gives: $\small{n\pi d^2\,\bar{v}\,\Delta t}$
• That means, the number of collisions in the time duration of $\small{\Delta t}$ is $\small{n\pi d^2\,\bar{v}\,\Delta t}$

7. So the number of collisions in one second will be: $\small{\Delta t}$ is $\small{n\pi d^2\,\bar{v}}$

8. Let $\small{\tau}$ be the average time between two successive collisions. Then we can write:
$\small{\tau}$  × Number of collisions in one second = one second
Therefore, $\small{\tau~=~\frac{1}{n\pi d^2\,\bar{v}}}$

9. This average time, multiplied by the average speed will give the average distance between two successive collisions. 
• The average distance between two successive collisions is called mean free path ($\small{l}$).
• So we can write:
$\small{l\,=\,\tau\,\bar{v}\,=\,\frac{\bar{v}}{n\pi d^2\,\bar{v}}}$
$\small{\Rightarrow l\,=\,\frac{1}{n\pi d^2}}$  

10. Let us do a dimensional analysis of the equation derived in (9) above.
   ♦ The dimension of $\small{l}$ is $\small{L}$ 
   ♦ 'n' is number per volume. So it's dimensions are: $\small{L^{-3}}$
   ♦ The dimensions of $\small{d^2}$ is $\small{L^2}$ 
So the dimensions of R.H.S is: $\small{\frac{1}{L^{-3}L^2}~=~L}$

11. We derived the equation in (9) by considering the motion of the middle sphere in fig.13.3 above. We assumed that, all other spheres (molecules) are at rest. But in reality, all molecules are in motion.

• So instead of $\small{\bar{v}}$, we need to use $\small{\bar{v}_r}$, which is the average relative velocity.
• Based on this modification, scientists have derived a more accurate equation: $\small{l\,=\,\frac{1}{\sqrt{2}\,n\pi d^2}}$

12. Let us see an example:
We are given one mole of air molecules at STP. We want to find $\small{\tau}$ and the mean free path.
This can be done in steps:
(i) One mole of air molecules is $\small{6.02 ~\times~10^{23}}$ molecules. At STP, this number of molecules will occupy $\small{22.4~\times~10^{-3}~{\rm{m^3}}}$

(ii) So the number of molecules (n) in one cubic meter, can be calculated as:
$\small{n~=~\frac{6.02 ~\times~10^{23}}{22.4~\times~10^{-3}~{\rm{m^3}}}~=~2.7~\times~10^{25}~{\rm{m^{-3}}}}$

(iii) Let
   ♦ d, the diameter of an air molecule = $\small{2~\times~10^{-10}~{\rm{m}}}$
   ♦ $\small{\bar{v}}$, the average velocity of an air molecule = $\small{485~{\rm{m\,s^{-1}}}}$

(iv) Then from the equation in (8), we get:
$\small{\tau~=~\frac{1}{(2.7~\times~10^{25}~{\rm{m^{-3}}})\pi (2~\times~10^{-10}~{\rm{m}})^2\,(485~{\rm{m\,s^{-1}}} )}~=~6.1~\times~10^{-10}~{\rm{s}}}$

(v) Also, from the equation in (9), we get:
$\small{l\,=\,\tau\,\bar{v}\,=\,6.1~\times~10^{-10}~{\rm{s}}~\times~485~{\rm{m\,s^{-1}}}~=~2.9~\times~10^{-7}~{\rm{m}}}$

(vi) $\small{2.9~\times~10^{-7}~{\rm{m}}}$ is approximately equal to 1500d
• That means, the air molecule is able to travel 1500 times it's diameter between two successive collisions.

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In the next chapter, we will see oscillations.

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13.3 - Specific Heats of Gases

In the previous section, we saw the law of equipartition of energy. In this section, we will apply the law to predict the specific heats of gases.

First we will see monatomic gases. The details can be written in 5 steps:

1. For a monatomic gas molecule, there will not be any rotation or vibration. So we need to consider the translational kinetic energy only.
• Applying law of equipartition, We get:
For each molecule at temperature T, this energy will be $\small{\frac{3}{2}K_B\,T}$

2. In a mole of the gas, there will be $\small{N_A}$ molecules. So the total internal energy (U) of one mole of a monatomic gas can be obtained as:
$\small{U\,=\,\frac{3}{2}K_B\,T\,N_A\,=\,\frac{3}{2}R\,T}$  

3. Specific heat is the change in internal energy per unit change in temperature. So we can obtain the specific heat for one mole at constant volume as:
$\small{C_v\,\text{monatomic gas}\,=\,\frac{3}{2}R}$

4. For an ideal gas, $\small{C_p\,-\,C_v\,=\,R}$
Where $\small{C_p}$ is the specific heat for one mole at constant pressure. We can write:
$\small{C_p\,=\,R\,+\,C_v\,=\,\frac{5}{2}R}$

5. We can write the ratio of specific heats also:
$\small{\gamma\,=\,\frac{C_p}{C_v}\,=\,\frac{5}{3}}$

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Next we will see diatomic gases whose molecules are rigid. The details can be written in 5 steps:

1. For a diatomic gas molecule, assuming that it is rigid, there will not be any vibration. So we need to consider the translational and rotational kinetic energies only.
• Applying law of equipartition, We get:
For each molecule at temperature T, this energy will be $\small{5\times \frac{1}{2}K_B\,T~=~\frac{5}{2}K_B\,T}$

2. In a mole of the gas, there will be $\small{N_A}$ molecules. So the total internal energy (U) of one mole of a diatomic gas can be obtained as:
$\small{U\,=\,\frac{5}{2}K_B\,T\,N_A\,=\,\frac{5}{2}R\,T}$  

3. Specific heat is the change in internal energy per unit change in temperature. So we can obtain the specific heat for one mole at constant volume as:
$\small{C_v\,\text{rigid diatomic gas}\,=\,\frac{5}{2}R}$

4. For an ideal gas, $\small{C_p\,-\,C_v\,=\,R}$
Where $\small{C_p}$ is the specific heat for one mole at constant pressure. We can write:
$\small{C_p\,=\,R\,+\,C_v\,=\,\frac{7}{2}R}$

5. We can write the ratio of specific heats also:
$\small{\gamma\,=\,\frac{C_p}{C_v}\,=\,\frac{7}{5}}$

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Next we will see diatomic gases whose molecules are not rigid. The details can be written in 5 steps:

1. For a diatomic gas molecule, if it is not rigid, there will be vibration.
• Applying law of equipartition, We get:
For each molecule at temperature T, this energy will be $\small{5\times \frac{1}{2}K_B\,T\,+\,K_B\,T~=~\frac{7}{2}K_B\,T}$

2. In a mole of the gas, there will be $\small{N_A}$ molecules. So the total internal energy (U) of one mole of a diatomic gas can be obtained as:
$\small{U\,=\,\frac{7}{2}K_B\,T\,N_A\,=\,\frac{7}{2}R\,T}$  

3. Specific heat is the change in internal energy per unit change in temperature. So we can obtain the specific heat for one mole at constant volume as:
$\small{C_v\,\text{non-rigid diatomic gas}\,=\,\frac{7}{2}R}$

4. For an ideal gas, $\small{C_p\,-\,C_v\,=\,R}$
• We can write:
$\small{C_p\,=\,R\,+\,C_v\,=\,\frac{9}{2}R}$

5. We can write the ratio of specific heats also:
$\small{\gamma\,=\,\frac{C_p}{C_v}\,=\,\frac{9}{5}}$

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Next we will see polyatomic gases. The details can be written in 5 steps:

1. In general, a polyatomic molecule has:
    ♦ 3 translational degrees of freedom
    ♦ 3 rotational degrees of freedom
    ♦ $\small{f}$ number of vibrational modes

• Applying law of equipartition, We get:
For each molecule at temperature T, this energy will be $\small{3\times \frac{1}{2}K_B\,T\,+\,3\times \frac{1}{2}K_B\,T\,+f\,\,K_B\,T~=~(3+f)K_B\,T}$

2. In a mole of the gas, there will be $\small{N_A}$ molecules. So the total internal energy (U) of one mole of a polyatomic gas can be obtained as:
$\small{U\,=\,(3+f)K_B\,T\,N_A\,=\,(3+f)R\,T}$  

3. Specific heat is the change in internal energy per unit change in temperature. So we can obtain the specific heat for one mole at constant volume as:
$\small{C_v\,\text{polyatomic gas}\,=\,(3+f)R}$

4. For an ideal gas, $\small{C_p\,-\,C_v\,=\,R}$
• We can write:
$\small{C_p\,=\,R\,+\,C_v\,=\,(4+f)R}$

5. We can write the ratio of specific heats also:
$\small{\gamma\,=\,\frac{C_p}{C_v}\,=\,\frac{4+f}{3+f}}$

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Using the above equations, we can predict the specific heats of different types of gases. Two examples are given below. In those examples, vibrational modes are ignored:

Example 1:
• Predicted value of $\small{C_v}$ and $\small{C_p}$ for monatomic gas are 12.5 J mol⁻¹ K⁻¹ and 20.8 J mol⁻¹ K⁻¹ respectively.
• Experimental value of $\small{C_v}$ and $\small{C_p}$ for the monatomic gas helium are 12.5 J mol⁻¹ K⁻¹ and 20.8 J mol⁻¹ K⁻¹ respectively.

Example 2:
• Predicted value of $\small{C_v}$ and $\small{C_p}$ for diatomic gas are 20.8 J mol⁻¹ K⁻¹ and 29.1 J mol⁻¹ K⁻¹ respectively.
• Experimental value of $\small{C_v}$ and $\small{C_p}$ for the diatomic gas oxygen are 21.0 J mol⁻¹ K⁻¹ and 29.3 J mol⁻¹ K⁻¹ respectively.

If vibrational modes are included in the calculations, the predicted values become closer to the experimental values.

Now we will see some solved examples.

Solved example 13.14
A cylinder of fixed capacity 44.8 litres contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by 15.0 °C ? (R = 8.31 J mol⁻¹ K⁻¹)
Solution:
1. Helium is a monatomic gas.
• So $\small{C_v\,=\,\frac{3}{2} R}$ and $\small{C_p\,=\,\frac{5}{2} R}$
• Given that, cylinder is of fixed capacity. That means, volume is constant. So we have to use $\small{C_v}$.

2. Heat required
= $\small{C_v~\times~\text{no. of moles}~\times~\text{rise in temperature}}$

3. So our next task is to find the number of moles of helium present in the cylinder.
• Given that, the gas is at STP. We know that, at STP, one mole of an ideal gas will have a volume of 22.4 litres.
• In our present case, the volume is 44.8 litres. So there are two moles.

4. Thus from (2), we get:
Heat required
= $\small{\frac{3}{2} R~\times~2~\times~15}$

= $\small{\frac{3}{2} \left(8.31 \text{J}\, \text{mol}^{-1} \,\text{K}^{-1} \right)~\times~2(\text{mol})~\times~15\left(\text{K} \right)}$

= $\small{373.95 \,\text{J}}$

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In the next section, we will see specific heat capacity of solids.

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13.2 - Law of Equipartition of Energy

In the previous section, we explained the gas laws using kinetic theory. In this section, we will see the law of equipartition of energy.

The details can be written in 17 steps:
1. Consider a gas in thermal equilibrium at temperature T.
If 'm' is the mass of each molecule in that gas, then the average value of energy ($\small{\epsilon_t}$) of each molecule can be obtained as:
$\small{\epsilon_t~=~\frac{1}{2}m\,v_x^2~+~\frac{1}{2}m\,v_y^2~+~\frac{1}{2}m\,v_z^2~=~\frac{3}{2}K_B\,T}$

2. $\small{\frac{3}{2}K_B\,T}$ is three times $\small{\frac{1}{2}K_B\,T}$. We know that, there is no preferred direction. So we can write:
    ♦ $\small{\frac{1}{2}m\,v_x^2~=~\frac{1}{2}K_B\,T}$
    ♦ $\small{\frac{1}{2}m\,v_y^2~=~\frac{1}{2}K_B\,T}$
    ♦ $\small{\frac{1}{2}m\,v_z^2~=~\frac{1}{2}K_B\,T}$

3. Now we will write about the number of coordinates required for a molecule.
• If the molecule is free to move in space, we need three coordinates to locate it.
• If the molecule is constrained to move in a plane, we need two coordinates to locate it.
• If the molecule is constrained to move along a line, we need only one coordinate to locate it.

4. The above information in (3) can be written in another form also:
• If the molecule is free to move in space, it has three degrees of freedom.
• If the molecule is constrained to move in a plane, it has two degrees of freedom.
• If the molecule is constrained to move along a line, it has one degree of freedom.

5. The above information in (3) and (4) can be written in yet another form also. Motion of a body as a whole from one point to another is called translation. So we can write:
• If the molecule is free to move in space, it has three translational degrees of freedom.
• If the molecule is constrained to move in a plane, it has two translational degrees of freedom.
• If the molecule is constrained to move along a line, it has one translational degree of freedom.

6. We just saw that, if a molecule is free to move in space, it has three translational degrees of freedom. Each of those three degrees, will contribute a term towards the total kinetic energy of the molecule.
• The degree of freedom in the x-direction will contribute $\small{\frac{1}{2}m\,v_x^2}$
• The degree of freedom in the y-direction will contribute $\small{\frac{1}{2}m\,v_y^2}$
• The degree of freedom in the z-direction will contribute $\small{\frac{1}{2}m\,v_z^2}$

7. Consider a diatomic gas like oxygen ($\small{O_2}$) or nitrogen ($\small{N_2}$). Each molecule of such a gas will have two atoms.
• In fig.13.2 below, the brown spheres represent atoms.
• Two brown spheres are joined together by a yellow line. • This line represents the bond between the two atoms.
    ♦ The green line is perpendicular to the yellow line.
    ♦ The red line is also perpendicular to the yellow line.
    ♦ The green and red lines are perpendicular to each other.

Fig.13.2

8. In the fig. on the left side, the molecule as a whole, is rotating about the green line. So green line is the axis of rotation. It is marked as (1).
• Due to this rotation, the molecule will have a rotational energy equal to $\small{\frac{1}{2}\,I_1\,\omega_1^2}$
    ♦ $\small{I_1}$ is the moment of inertia of the molecule about axis (1)
    ♦ $\small{\omega_1}$ is the angular speed of the molecule about axis (1)

9. In the fig. on the right side, the molecule as a whole, is rotating about the red line. So red line is the axis of rotation. It is marked as (2).
• Due to this rotation, the molecule will have a rotational energy equal to $\small{\frac{1}{2}\,I_2\,\omega_2^2}$
    ♦ $\small{I_2}$ is the moment of inertia of the molecule about axis (2)
    ♦ $\small{\omega_2}$ is the angular speed of the molecule about axis (2)

10. The molecule can rotate about the yellow line also. But the moment of inertia of the molecule about the yellow line is very small. So this rotation will not contribute much to the total rotational energy. We can safely ignore this rotation.

11. Based on (8) and (9), we can write:
Rotational energy of the molecule is given by:
$\small{\epsilon_r~=~\frac{1}{2}\,I_1\,\omega_1^2~+~\frac{1}{2}\,I_2\,\omega_2^2}$

• It is clear that, a diatomic molecule has two rotational degrees of freedom.
    ♦ One along axis (1)
    ♦ The other along axis (2)

12. So a diatomic molecule has a total energy given by:
$\small{\epsilon_t\,+\,\epsilon_r\,=\,\frac{1}{2}m\,v_x^2\,+\,\frac{1}{2}m\,v_y^2\,+\,\frac{1}{2}m\,v_z^2\,+\,\frac{1}{2}\,I_1\,\omega_1^2\,+\,\frac{1}{2}\,I_2\,\omega_2^2}$

• Note that, for a monatomic molecule, the last two terms will be absent.

13. We have seen the energy contributions from translation and rotation. If a molecule experience vibration, that molecule will posses vibrational energy also.
• Consider the two spheres in fig.13.2 above. If the spheres oscillate along the yellow line, then that molecule as a whole, will posses vibrational energy. It is like a spring with the two spheres at it's ends. We will see more details in later chapters.
• The $\small{O_2}$ molecule is rigid. So this energy does not come into effect at moderate temperatures. But a $\small{CO}$ molecule will posses this energy even at moderate temperatures.

14. For a molecule like $\small{CO}$, the vibrational energy is given by:
$\small{\epsilon_v\,=\,\frac{1}{2} m \left(\frac{dy}{dt} \right)^2\,+\,\frac{1}{2} k y^2}$
    ♦ $\small{k}$ is the force constant of the oscillator
    ♦ $\small{y}$ is the vibrational co-ordinate

15. Now we can write the total energy:
$\small{\epsilon\,=\,\epsilon_t\,+\,\epsilon_r\,+\,\epsilon_v}$

• Where,

$\small{\epsilon_t\,=\,\frac{1}{2}m\,v_x^2\,+\,\frac{1}{2}m\,v_y^2\,+\,\frac{1}{2}m\,v_z^2}$

$\small{\epsilon_r\,=\,\frac{1}{2}\,I_1\,\omega_1^2\,+\,\frac{1}{2}\,I_2\,\omega_2^2}$

$\small{\epsilon_v\,=\,\frac{1}{2} m \left(\frac{dy}{dt} \right)^2\,+\,\frac{1}{2} k y^2}$

16. Now we can analyze the contribution from each type of energy.
(i) Each of the translational degree of freedom, contributes one square term.
(ii) Each of the rotational degree of freedom, contributes one square term.
(iii) Each vibrational frequency, contributes two square terms.

17. Now we can write about the general form of contribution:
• We have seen that, each square term in the translational energy is equal to $\small{\frac{1}{2}K_B\,T}$
• The Scottish physicist James Clerk Maxwell proved that, each square term has an average energy equal to $\small{\frac{1}{2}K_B\,T}$. This is known as the law of equipartition of energy.
• So we can write:
    ♦ Each translational degree of freedom contributes $\small{\frac{1}{2}K_B\,T}$ to the total energy.
    ♦ Each rotational degree of freedom contributes $\small{\frac{1}{2}K_B\,T}$ to the total energy.
    ♦ Each vibrational frequency contributes two times $\small{\frac{1}{2}K_B\,T}$ = $\small{K_B\,T}$ to the total energy.

• We will see the detailed proof of the law in higher classes.

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In the next section, we will apply the law to predict specific heats of gases theoretically.

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Chapter 13.1 - Kinetic Theory of An Ideal Gas

In the previous section, we saw the ideal gas equation PV = 𝝁RT. In this section, we try to explain the gas laws using kinetic theory.

Some basics can be written in 4 steps:
1. A sample of a gas (that we consider for the discussions) will consist of the molecules of that gas only. There will not be any impurities like dust particles, water vapour etc.,
2. Molecules are very far apart. So there will not be any forces of attraction or repulsion between them
3. But molecules do collide with each other and with the walls of the container. All those collisions are elastic. There will not be any energy loss due to the collisions
4. What ever be the type of collisions, the total momentum will be conserved

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• Based on the above points, let us now analyze the collisions of some gas molecules inside a closed container. Through such an analysis, we can derive an expression for pressure. It can be written in 20 steps:
1. The yellow rectangle in fig.13.1(a) below is a closed cubical container
• It contains N molecules of a gas

Fig.13.1

2. One molecule is shown as a green sphere
• It travels towards the left along the red dotted line
   ♦ This red dotted line is parallel to the x-axis

3. The molecule hits the left side wall of the container
• This wall is parallel to the y-axis
   ♦ Remember that, the red dotted line is parallel to the x-axis
   ♦ So the molecule will be hitting the wall in a perpendicular direction

4. Let us see how the molecule bounces back:
• Due to the perpendicular collision, the molecule will bounce back along the same red line
• Also, since the collision is elastic, the molecule will bounce back with the same speed (magnitude of velocity)
◼  So we can write about the velocities:

If $\small{v_{x(1)}}$ is the initial velocity before collision, the velocity after collision will be $\small{-v_{x(1)}}$.

5. Let m be the mass of the molecule. Then the momentum of the molecule can be calculated as:
   ♦ Initial momentum = $\small{mv_{x(1)}}$
   ♦ Final momentum = $\small{-mv_{x(1)}}$

6. Then change in momentum suffered by the molecule
= (final momentum – initial momentum)
= $\small{\left[-mv_{x(1)}~-~mv_{x(1)} \right]~=~-2mv_{x(1)}}$

7. We have seen the principle of conservation of momentum in an earlier chapter (Fig.5.16 of section 5.7)
◼  Based on that principle, the change in momentum suffered by the wall will be $\small{2mv_{x(1)}}$

8. The wall suffers this change in momentum because, the molecule exerts a force on that wall
◼  We know that, force is equal to ‘rate of change of momentum’
   ♦ To find the ‘rate of change of momentum’,
   ♦ we need to divide ‘change of momentum’ by time
• So our next task is to find the time during which this ‘change in momentum’ occurs.

9. The change in momentum occurs during the collision. But the duration of a collision is very small. Let it be Δt.
• The molecule must hit the wall and bounce back in the interval Δt.
• This is possible only if the molecule is within a distance of $\small{\left(v_{x(1)}\,\Delta t \right)}$ from the wall
(Recall that, velocity multiplied by time gives distance)
• If the molecule is at a greater distance than $\small{\left(v_{x(1)}\,\Delta t \right)}$, it will not be able to reach the wall in duration of Δt.
• This distance is marked by the magenta vertical line in fig.b
(In the fig.b, the distance $\small{\left(v_{x(1)}\,\Delta t \right)}$ is exaggerated. The magenta vertical line is in fact, very close to the left side wall. It is shown further away, only for clarity)

10. It is clear that, all the molecules (with velocity $\small{v_{x(1)}}$) between the magenta line and the left wall, have a chance of hitting the wall within the duration Δt.
• So we need the actual number of molecules (with velocity $\small{v_{x(1)}}$) between the magenta line and the left wall
• For finding that actual number, we need the volume between the magenta line and the left wall
   ♦ If A is the area of the left wall, that volume will be $\small{A v_{x(1)} \Delta t}$
• If $\small{n_{x(1)}}$ is the number of molecules (with velocity $\small{v_{x(1)}}$) in unit volume, then the number of molecules in that volume will be $\small{n_{x(1)} A v_{x(1)} \Delta t}$

11. All those molecules calculated in (10) have the chance of hitting the left wall
• But we know that, molecules move in random directions. So on the average, only half of those molecules will be hitting the wall
• So we get:
Number of molecules hitting the wall =
$\small{(1/2) n_{x(1)} A v_{x(1)} \Delta t}$

12. In (7), we saw that, the change in momentum suffered by the wall due to one molecule = $\small{2mv_{x(1)}}$
• So the change in momentum (Q) due to $\small{(1/2) n_{x(1)} A v_{x(1)} \Delta t}$ molecules can be obtained as: $\small{\left[2mv_{x(1)} \right]\left[(1/2) n_{x(1)} A v_{x(1)} \Delta t \right]}$

13. Dividing this change in momentum by Δt, we get the force $\small{F_{x(1)}}$ experienced by the wall due to the collisions
• That is., $\small{F_{x(1)}~=~\left[2mv_{x(1)} \right]\left[(1/2) n_{x(1)} A v_{x(1)} \right]}$
$\small{\Rightarrow F_{x(1)}~=~n_{x(1)} \,m\, A\, v^2_{x(1)}}$
    ♦ This force is due to the molecules traveling with velocity $\small{v_{x(1)}}$

14. Dividing this force by the area of the wall, we get the pressure P experienced by the wall
• That is., $\small{P_{x(1)}~=~n_{x(1)} \,m\,  v^2_{x(1)}}$
    ♦ This pressure is due to the molecules traveling with velocity $\small{v_{x(1)}}$

15. But there are molecules (traveling in the x-direction), with other velocities also.
• The various velocities in the x-direction can be denoted as:
$\small{v_{x(1)},~v_{x(2)}.~v_{x(3)}}$, etc.,
• So we can assume that, all molecules traveling in the x-direction, have an average velocity of $\small{\bar{v}_x }$
• This $\small{\bar{v}_x }$ is the average of $\small{v_{x(1)},~v_{x(2)}.~v_{x(3)}}$, etc.,
• Then the result in (14) becomes:
$\small{P_x=n_x\,m\,\bar{v}_x^2}$
    ♦ Where
        ✰ P_x is the pressure due to all molecules traveling in the x-direction
        ✰ n_x is the number of molecules (traveling in the x-direction) per unit volume
        ✰ $\small{\bar{v}_x }$ is the average of $\small{v_{x(1)},~v_{x(2)},~v_{x(3)}}$, etc.,

16. We derived the above result by considering the molecules traveling in the x-direction
• We can apply the same steps to the molecules which travel in the y-direction
    ♦ We will get: $\small{P_y~=~\,n_y\,m\,\bar{v}_y^2}$
• We can apply the same steps to the molecules which travel in the z-direction
    ♦ We will get: $\small{P_z~=~n_z\,m\,\bar{v}_z^2}$

17. A gas sample will be isotropic. So we cannot say that, more molecules travel in any preferred direction.

For an isotropic substance, any physical property will have same values in all directions. For example:     ♦ Strength will be same in all directions.     ♦ Density will be same in all directions.     ♦ Conductivity will be same in all directions.

• All directions are equally likely. So we can write: $\small{\bar{v}_x=\bar{v}_y=\bar{v}_z}$
• Squaring this, we get: $\small{\bar{v}_x^2=\bar{v}_y^2=\bar{v}_z^2}$
◼  Thus we can write:
In the gas sample, there are three velocities and all three are equal. Their squares are also equal.

18. Let us write the role of each velocity:
   ♦ We use $\small{\bar{v}_x^2}$ to find P_x
   ♦ We use $\small{\bar{v_y^2}}$ to find P_y
   ♦ We use $\small{\bar{v_z^2}}$ to find P_z 

19. If we can find the average of $\small{\bar{v}_x^2,~\bar{v}_y^2 ~\text{and}~ \,\bar{v}_z^2}$, we can use it to find the pressure in a single step
• So our next aim is to find that average. It can be done in 2 steps:
(i) Since they are equal, we can sum them up and divide by 3. No weightage need to be given for any of them
• So the average is: $\small{\frac{\bar{v}_x^2\;+\;\bar{v}_y^2\;+\;\bar{v}_z^2}{3}}$
(ii) We will denote ($\small{\bar{v}_x^2\;+\;\bar{v}_y^2\;+\;\bar{v}_z^2}$) as $\small{{\bar{v}^2}}$
• So the average is: $\small{{\frac{\bar{v}^2}{3}}}$

20. This average is applicable to all molecules. So we do not need to consider n_x, n_y and n_z separately. We can use n. This n is the 'number density' (number per unit volume) of molecules having the average velocity
• So based on the results in (15) and (16), we can write:
The pressure due to all the molecules is given by:

Eq.13.5: $\small{P=\frac{n\,m\,\bar{v}^2}{3}}$

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◼  Two remarks have to be written about the above derivation:
1. We chose a cube as the container. This made it easier to consider the three directions separately
• But in reality, we can choose any arbitrary shape. The reason can be written in 3 steps
(i) We considered the molecules hitting on the left wall of area A
• We need not insist on such perfect areas. A small area ΔA would be sufficient
(ii) This is because, when we find pressure, we divide force by area. Then the area in both numerator and denominator cancel each other
(iii) The time Δt also gets canceled in this way. So the distance $\small{v_{x(1)}\,\Delta t}$ is also not important. The container can be of any shape

2. Collisions between molecules
• This can be written in 5 steps
(i) We have considered the collision of molecules with the walls of the container. But we have not considered the collision between molecules
(ii) When molecules collide with each other, their velocities will change
• We considered a molecule with velocity $\small{v_{x(1)}}$
• There is no guarantee that, on it’s way towards the wall, it will not hit another molecule, thereby changing it’s velocity to another value
(iii) However, numerous collisions are taking place in the gas sample
• So velocity lost by one molecule will be gained by another molecule at some other place
(iv) Also, we are taking the average value
(v) So the collisions between molecules will not lead to much error

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Kinetic Interpretation of Temperature

• Based on Eq.13.5, we can find the relation between kinetic energy of the molecules and the temperature of the gas. It can be written in 15 steps:
1. We have Eq.13.5: $\small{P=\frac{n\,m\,\bar{v}^2}{3}}$
• Multiplying both sides by V, we get: $\small{PV=\frac{1}{3}n\,V\,m\,\bar{v}^2}$
• Multiplying the right side by $\small{\frac{2}{2}}$, we get: $\small{PV=\frac{2}{3}\times\frac{1}{2}n\,V\,m\,\bar{v}^2}$

2. Recall that, n is the 'number density' (number of molecules in unit volume)
• So nV is the total number of molecules (N) in the sample
• Thus we get: $\small{PV=\frac{2}{3}\times\frac{1}{2}Nm\bar{v}^2}$

3. $\small{\frac{1}{2}m\bar{v}^2}$ is the average kinetic energy possessed by a single molecule
• So $\small{\frac{1}{2}N\,m\,\bar{v}^2}$ is the total energy possessed by all the molecules in the sample.

4. The total kinetic energy possessed by all the molecules is the internal energy E of the gas sample
• That is., $\small{E=\frac{1}{2}N\,m\,\bar{v}^2}$
• Thus the result in (2) becomes:
Eq.13.6: $\small{PV=\frac{2}{3}E}$

5. But we have the basic equation: $\small{PV~=~\mu\,R\,T}$
• Comparing the two, we get: $\small{\mu RT=\frac{2}{3}E}$

6. In the previous section, we saw that: $\small{\mu R~=~N\,K_B}$
• So the result in (5) becomes: $\small{N\,K_B \,T=\frac{2}{3}E}$
⇒ $\small{\frac{E}{N}=\frac{3}{2}K_B T}$

7. $\small{\frac{E}{N}}$ is the kinetic energy of one molecule
• K_B is a constant
◼  So we can write:
Kinetic energy of a molecule is proportional to the absolute temperature of the gas
• It is also clear that, the internal energy of a gas depends only on it's temperature
   ♦ Pressure or volume has no effect on the internal energy

8. Next, let us consider a mixture of non-reactive gases
• We will name the gases as A, B, C, . . .
   ♦ Let the average velocity of the molecules of gas A be $\small{\bar{v}_A}$
   ♦ Let the average velocity of the molecules of gas B be $\small{\bar{v}_B}$
   ♦ Let the average velocity of the molecules of gas C be $\small{\bar{v}_C}$
so on . . .
• We will write the number densities of molecules also:
   ♦ Let the number density of the molecules of gas A be n_A
   ♦ Let the number density of the molecules of gas B be n_B
   ♦ Let the number density of the molecules of gas C be n_C
so on . . .
9. Then using Eq 13.5, we can write the individual contributions:
   ♦ Contribution of gas A towards the total pressure = $\small{\frac{n_A m_A \bar{v}_A^2}{3}}$
   ♦ Contribution of gas B towards the total pressure = $\small{\frac{n_B m_B \bar{v}_B^2}{3}}$
   ♦ Contribution of gas C towards the total pressure = $\small{\frac{n_C m_C \bar{v}_C^2}{3}}$
so on . . .

10. So the total pressure of the gas will be obtained as:
$\small{p=\frac{n_A \,m_A \,\bar{v_A}^2}{3}+\frac{n_B\, m_B \,\bar{v}_B^2}{3}+\frac{n_C \,m_C \,\bar{v}_C^2}{3}+\, .\, .\, .}$

11. Now consider the result in (6): $\small{\frac{E}{N}=\frac{3}{2}K_B T}$
• $\small{\frac{E}{N}}$ is the kinetic energy per molecule, which is equal to $\small{\frac{1}{2}m \bar{v}^2}$
• So we can write: $\small{\frac{1}{2}m \bar{v^2}=\frac{3}{2}K_B T}$

12. Applying this result to individual gases, we get:
   ♦ $\small{\frac{1}{2}m_A \bar{v}_A^2=\frac{3}{2}K_B T}$
   ♦ $\small{\frac{1}{2}m_B \bar{v}_B^2=\frac{3}{2}K_B T}$
   ♦ $\small{\frac{1}{2}m_C \bar{v}_C^2=\frac{3}{2}K_B T}$
so on . . .
• From this, we get:
   ♦ $\small{m_A \bar{v}_A^2=3K_B T}$
   ♦ $\small{m_B \bar{v}_B^2=3K_B T}$
   ♦ $\small{m_C \bar{v}_C^2=3K_B T}$

13. So the result in (10) becomes:
$\small{p=\frac{n_A 3K_B T}{3}+\frac{n_B 3K_B T}{3}+\frac{n_C3K_B T}{3}+\, .\, .\, .}$
⇒ $\small{P=(n_A+n_B+n_C+\, .\, .\, .)K_BT}$
⇒ $\small{PV=(n_AV+n_BV+n_CV+\, .\, .\, .)K_BT}$
⇒ $\small{PV=(n_AVK_B+n_BVK_B+n_CVK_B+\, .\, .\, .)T}$
⇒ $\small{PV=(N_AK_B+N_BK_B+N_CK_B+\, .\, .\, .)T}$

14. In the previous section, we saw that: $\small{\mu\,R~=~N\,K_B}$
• Applying this to individual gases, we get:
   ♦ $\small{\mu_A\,R~=~N_A\,K_B}$
   ♦ $\small{\mu_B\,R~=~N_B\,K_B}$
   ♦ $\small{\mu_C\,R~=~N_C\,K_B}$
so on . . .
15. So the result in (13) becomes:
$\small{PV=(\mu_A R+\mu_B R+\mu_C R+\, .\, .\, .)T}$
⇒ $\small{PV=(\mu_A+\mu_B+\mu_C+\, .\, .\, .)RT}$
◼  This is Dalton's law of partial pressures
• Thus we derived Dalton's law of partial pressures using kinetic theory

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• We know that, when temperature increases, the speed of the molecules increases
• For a gas,
   ♦ The internal energy is due to the kinetic energy of the molecules
   ♦ When temperature increases, kinetic energy increases
   ♦ So when temperature increases, internal energy increases
• The kinetic energy depends on speed
• Let us try to calculate the average speed of a molecule in a gas sample. It can be written in 5 steps:
1. In the step (6) above, we obtained: $\small{\frac{E}{N}=\frac{3}{2}K_B T}$
• $\small{\frac{E}{N}}$ is the kinetic energy of one molecule
• So we can write: $\small{\frac{1}{2}m\bar{v}^2=\frac{3}{2}K_B T}$
⇒ $\small{m\bar{v}^2 =3K_BT}$
⇒ $\small{\bar{v}=\left(\frac{3K_BT}{m}\right)^{\frac{1}{2}}}$
    ♦ Where m is the mass of one molecule

2. Consider a sample of nitrogen gas
    ♦ One mole of nitrogen gas will have a mass of 28 grams
    ♦ One mole of nitrogen gas will have 6.023 × 10²³ molecules
• So one molecule of nitrogen gas will have a mass of $\small{\left( \frac{28 \times 10^{-3}}{6.023 \times 10^{23}}\right)~=~4.65~\times~10^{-26}~\text{Kg}}$

3. So we have the mass of one molecule
    ♦ Let the sample be at a temperature of 300 K
    ♦ We know the value of K_B. It is: 1.38 × 10^-23 J K^-1
• Substituting these known values in (1), we get:
$\small{\bar{v}=\left( \frac{3 \times 1.38 \times 10^{-23}(J\,K^{-1})\times 300\, (K)}{4.65 \times 10^{-26}(kg)}\right)^{\frac{1}{2}}=517\;(J\;kg^{-1})^{\frac{1}{2}}}$
⇒ $\small{\bar{v}=517\;(kg\;m\;s^{-2}\;m\;kg^{-1})^{\frac{1}{2}}=517 \; m\;s^{-1}}$
◼ This speed is comparable to the speed of sound in air

4. We take the root of $\small{\bar{v}^2}$ to find $\small{\bar{v}}$
• So $\small{\bar{v}}$ is called root mean square velocity
    ♦ It's symbol is: v_rms
• So the result in (1) can be modified as:
Eq.13.7: $\small{v_{rms}=\left( \frac{3K_BT}{m}\right)^{\frac{1}{2}}}$

5. Note that, in Eq.13.7, m is in the denominator
• So we can write:
In a mixture of gases, the lighter molecules will have greater v_rms

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Link to two solved examples based on the above discussion is given below:

Solved example 13.12 and 13.13

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In the next section, we will see the Law of equipartition of energy

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Chapter 13 - Kinetic Theory

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

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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

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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

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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_B nT
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: $\mathbf\small{\rm{P=\frac{\rho RT}{M_0}}}$

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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

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


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