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:
$\small{C\,=\,\frac{3 R\,T}{T}\,=\,3 R}$
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.
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:
$\small{C\,=\,\frac{9 R\,T}{T}\,=\,9 R}$
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.
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.
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.
In the next
chapter, we will see oscillations.
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