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
• 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 18 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 |
• 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 vx(1) is the initial velocity before collision, the velocity after collision will be -vx(1)
5. Let m be the mass of the molecule. Then the momentum can be calculated as:
♦ Initial momentum = mvx(1)
♦ Final momentum = -mvx(1)
6. Then change in momentum suffered by the molecule
= (final momentum – initial momentum)
= (-mvx(1) - mvx(1)) = -2mvx(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 2mvx(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 vx(1) × Δt from the wall
(Recall that, velocity multiplied by time gives distance)
• This distance is marked by the magenta vertical line in fig.b
(In the fig.b, the distance vx(1) × Δt 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 vx(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 vx(1)) between the magenta line and the left wall
• For 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 A × vx(1) × Δt
• If nx(1) is the number of molecules (with velocity vx(1)) in unit volume, then the number of molecules in that volume will be nx(1) × A × vx(1) × Δt
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 = 1⁄2 × nx(1) × A × vx(1) × Δt
12. In (7), we saw that, the change in momentum suffered by the wall due to one molecule = 2mvx(1)
• So the change in momentum (Q) due to 1⁄2nx(1)AvxΔt molecules can be obtained as: (2mvx(1))×(1⁄2nx(1)AvxΔt)
13. Dividing this change in momentum by Δt, we get the force Fx(1) experienced by the wall due to the collisions
• That is., Fx(1) = (2mvx(1))×(1⁄2nx(1)Avx(1)) = nx(1)mAv2x(1)
♦ This force is due to the molecules traveling with velocity vx(1)
14. Dividing this force by the area of the wall, we get the pressure P experienced by the wall
• That is., Px(1) = nx(1)mv2x(1)
♦ This pressure is due to the molecules traveling with velocity vx(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:
vx(1), vx(2), vx(3), etc.,
• So we can assume that, all molecules traveling in the x-direction, have an average velocity of $\mathbf\small{\rm{\bar{v}_x }}$
• This $\mathbf\small{\rm{\bar{v}_x }}$ is the average of vx(1), vx(2), vx(3), etc.,
• Then the result in (14) becomes: $\mathbf\small{\rm{P_x=n_xm\bar{v_x^2}}}$
♦ Where
✰ Px is the pressure due to all molecules traveling in the x-direction
✰ nx is the number of molecules (traveling in the x-direction) per unit volume
✰ $\mathbf\small{\rm{\bar{v}_x }}$ is the average of vx(1), vx(2), vx(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: $\mathbf\small{\rm{P_y=n_xm\bar{v_y^2}}}$
• We can apply the same steps to the molecules which travel in the z-direction
♦ We will get: $\mathbf\small{\rm{P_z=n_xm\bar{v_z^2}}}$
17. A gas sample will be isotropic. So we cannot say that, more molecules travel in any preferred direction
• All directions are equally likely. So we can write: $\mathbf\small{\rm{\bar{v}_x=\bar{v}_y=\bar{v}_z }}$
• Squaring this, we get: $\mathbf\small{\rm{\bar{v_x^2}=\bar{v_y^2}=\bar{v_z^2}}}$
◼ 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 $\mathbf\small{\rm{\bar{v_x^2}}}$ to find Px
♦ We use $\mathbf\small{\rm{\bar{v_y^2}}}$ to find Py
♦ We use $\mathbf\small{\rm{\bar{v_z^2}}}$ to find Pz
19. If we can find the average of $\mathbf\small{\rm{\bar{v_x^2},\bar{v_y^2} \,and \,\bar{v_z^2} }}$, 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: $\mathbf\small{{\frac{\bar{v_x^2}\;+\;\bar{v_y^2}\;+\;\bar{v_z^2}}{3}}}$
(ii) We will denote ($\mathbf\small{{\bar{v_x^2}\;+\;\bar{v_y^2}\;+\;\bar{v_z^2}}}$) as $\mathbf\small{{\bar{v^2}}}$
• So the average is: $\mathbf\small{{\frac{\bar{v^2}}{3}}}$
18. This average is applicable to all molecules. So we do not need to consider nx, ny and nz separately. We can use n. This n is the 'number density' 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: $\mathbf\small{\rm{P=\frac{nm \bar{v^2}}{3}}}$
◼ 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 vx(1)Δt 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 vx(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
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: $\mathbf\small{\rm{P=\frac{nm \bar{v^2}}{3}}}$
• Multiplying both sides by V, we get: $\mathbf\small{\rm{PV=\frac{1}{3}nVm\bar{v^2}}}$
• Multiplying the right side by 2⁄2, we get: $\mathbf\small{\rm{PV=\frac{2}{3}\times\frac{1}{2}nVm\bar{v^2}}}$
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: $\mathbf\small{\rm{PV=\frac{2}{3}\times\frac{1}{2}Nm\bar{v^2}}}$
3. $\mathbf\small{\rm{\frac{1}{2}m\bar{v^2}}}$ is the average kinetic energy possessed by a single molecule
• So $\mathbf\small{\rm{\frac{1}{2}Nm\bar{v^2}}}$ 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., $\mathbf\small{\rm{E=\frac{1}{2}Nm\bar{v^2}}}$
• Thus the result in (2) becomes Eq.13.6: $\mathbf\small{\rm{PV=\frac{2}{3}E}}$
5. But we have the basic equation: PV = 𝝁RT
• Comparing the two, we get: $\mathbf\small{\rm{\mu RT=\frac{2}{3}E}}$
6. In the previous section, we saw that: 𝝁R = NKB
• So the result in (5) becomes: $\mathbf\small{\rm{NK_B T=\frac{2}{3}E}}$
⇒ $\mathbf\small{\rm{\frac{E}{N}=\frac{3}{2}K_B T}}$
7. $\mathbf\small{\rm{\frac{E}{N}}}$ is the kinetic energy of one molecule
• KB 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 $\mathbf\small{\rm{\bar{v}_A}}$
♦ Let the average velocity of the molecules of gas B be $\mathbf\small{\rm{\bar{v}_B}}$
♦ Let the average velocity of the molecules of gas C be $\mathbf\small{\rm{\bar{v}_C}}$
so on . . .
• We will write the number densities of molecules also:
♦ Let the number density of the molecules of gas A be nA
♦ Let the number density of the molecules of gas B be nB
♦ Let the number density of the molecules of gas C be nC
so on . . .
9. Then using Eq 13.5, we can write the individual contributions:
♦ Contribution of gas A towards the total pressure = $\mathbf\small{\rm{\frac{n_A m_A \bar{v_A^2}}{3}}}$
♦ Contribution of gas B towards the total pressure = $\mathbf\small{\rm{\frac{n_B m_B \bar{v_B^2}}{3}}}$
♦ Contribution of gas C towards the total pressure = $\mathbf\small{\rm{\frac{n_C m_C \bar{v_C^2}}{3}}}$
so on . . .
10. So the total pressure of the gas will be obtained as:
$\mathbf\small{\rm{p=\frac{n_A m_A \bar{v_A^2}}{3}+\frac{n_B m_B \bar{v_B^2}}{3}+\frac{n_C m_C \bar{v_C^2}}{3}+\, .\, .\, .}}$
11. Now consider the result in (6): $\mathbf\small{\rm{\frac{E}{N}=\frac{3}{2}K_B T}}$
• $\mathbf\small{\rm{\frac{E}{N}}}$ is the kinetic energy per molecule, which is equal to $\mathbf\small{\rm{\frac{1}{2}m \bar{v^2}}}$
• So we can write: $\mathbf\small{\rm{\frac{1}{2}m \bar{v^2}=\frac{3}{2}K_B T}}$
12. Applying this result to individual gases, we get:
♦ $\mathbf\small{\rm{\frac{1}{2}m_A \bar{v_A^2}=\frac{3}{2}K_B T}}$
♦ $\mathbf\small{\rm{\frac{1}{2}m_B \bar{v_B^2}=\frac{3}{2}K_B T}}$
♦ $\mathbf\small{\rm{\frac{1}{2}m_C \bar{v_C^2}=\frac{3}{2}K_B T}}$
so on . . .
• From this, we get:
♦ $\mathbf\small{\rm{m_A \bar{v_A^2}=3K_B T}}$
♦ $\mathbf\small{\rm{m_B \bar{v_B^2}=3K_B T}}$
♦ $\mathbf\small{\rm{m_C \bar{v_C^2}=3K_B T}}$
13. So the result in (10) becomes:
$\mathbf\small{\rm{p=\frac{n_A 3K_B T}{3}+\frac{n_B 3K_B T}{3}+\frac{n_C3K_B T}{3}+\, .\, .\, .}}$
⇒ $\mathbf\small{\rm{P=(n_A+n_B+n_C+\, .\, .\, .)K_BT}}$
⇒ $\mathbf\small{\rm{PV=(n_AV+n_BV+n_CV+\, .\, .\, .)K_BT}}$
⇒ $\mathbf\small{\rm{PV=(n_AVK_B+n_BVK_B+n_CVK_B+\, .\, .\, .)T}}$
⇒ $\mathbf\small{\rm{PV=(N_AK_B+N_BK_B+N_CK_B+\, .\, .\, .)T}}$
14. In the previous section, we saw that: 𝝁R = NKB
• Applying this to individual gases, we get:
♦ 𝝁AR = NAKB
♦ 𝝁BR = NBKB
♦ 𝝁CR = NCKB
so on . . .
15. So the result in (13) becomes:
$\mathbf\small{\rm{PV=(\mu_A R+\mu_B R+\mu_C R+\, .\, .\, .)T}}$
⇒ PV = (𝝁A + 𝝁B + 𝝁C + . . .)RT
◼ This is Dalton's law of partial pressures
• Thus we derived Dalton's law of partial pressures using kinetic theory
• 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: $\mathbf\small{\rm{\frac{E}{N}=\frac{3}{2}K_B T}}$
• $\mathbf\small{\rm{\frac{E}{N}}}$ is the kinetic energy of one molecule
• So we can write: $\mathbf\small{\rm{\frac{1}{2}m\bar{v^2}=\frac{3}{2}K_B T}}$
⇒ $\mathbf\small{\rm{m\bar{v^2} =3K_BT}}$
⇒ $\mathbf\small{\rm{\bar{v}=\left( \frac{3K_BT}{m}\right)^{\frac{1}{2}}}}$
♦ 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 $\mathbf\small{\rm{\left( \frac{28 \times 10^{-3}}{6.023 \times 10^{23}}\right)}}$ = 4.65 × 10-26 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 KB. It is: 1.38 × 10-23 J K-1
• Substituting these known values in (1), we get:
$\mathbf\small{\rm{\bar{v}=\left( \frac{3 \times 1.38 \times 10^{-23}(J\,K^{-1})\times 300\, (K)}{4.65 \times 10^{-26}(kg)}\right)^{\frac{1}{2}}=517\;(J\;kg^{-1})^{\frac{1}{2}}}}$
⇒ $\mathbf\small{\rm{\bar{v}=517\;(kg\;m\;s^{-2}\;m\;kg^{-1})^{\frac{1}{2}}=517 \; m\;s^{-1}}}$
◼ This speed is comparable to the speed of sound in air
4. We take the root of $\mathbf\small{\rm{\bar{v^2}}}$ to find $\mathbf\small{\rm{\bar{v}}}$
• So $\mathbf\small{\rm{\bar{v}}}$ is called root mean square velocity
♦ It's symbol is: vrms
• So the result in (1) can be modified as:
Eq.13.7: $\mathbf\small{\rm{v_{rms}=\left( \frac{3K_BT}{m}\right)^{\frac{1}{2}}}}$
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 vrms
Link to two solved examples based on the above discussion is given below:
Solved example 13.12 and 13.13
In the next
section, we will see the Law of equipartition of energy
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