Faculty of Engineering
International Credit Hours Engineering Programs
Communication Systems Engineering Program
Academic Year 2023/2022 – Spring 2022
PHM 123
Thermal and Statistical Physics
Major Task Phase (2)
Name
ID
Youssef Hany Youssef
21P0398
Omar Mohamed Kamal
20P7817
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Table of content:
1. Question I…………………………………………….... 3
2. Question II……………………………….…….………. 5
3. Question III…………………………………………………. 7
4. Question IV…………………………………………... 11
5. Question V……………………………………..…….. 13
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Question I:
𝑮𝒊𝒗𝒆𝒏𝒔:
𝑅 = 8.3144621
𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 = 100 ℃/373 𝐾
𝑎 = 0.55
𝑏 = 3.0 × 10−5
𝑛 = 1 𝑚𝑜𝑙𝑒
𝑉𝑜𝑙𝑢𝑚𝑒 𝑟𝑎𝑛𝑔𝑒 ∶ 0.5 – 1 𝑚3
Firstly, we set 100 steps for the volume. Then, we set up equations
for the real P and the ideal P. We then plot both of them, with the
pressure being the y-axis and the volume being the x-axis. Then,
we identify a function at V for the real P, so that we can get the
exact integration answer (from 0.5 to 1). As for the numerical
method, we use the trapezoidal method to get the approximated
answer. To calculate the error, we use the rule:
𝒆𝒓𝒓𝒐𝒓 = √|𝒓𝒆𝒂𝒍𝟐 − 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒆𝒅𝟐 |
As most of the numerical methods word, if we increase the
number of steps, the error decreases, and vice versa.
This is how the plot should look like:
Figure (1): PV diagram for a gas at 373 K at a pressure from 0.5 to 1 m 3
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In this range, the differences will not be clear. However, if we
zoom in on a smaller range, it will be clearer.
Figure (2): PV diagram for a gas at 373 K at a pressure from 0 to 5 dm 3
It is clear that with really high pressure and low temperature,
the ideal starts deviating from the real. As we can see, when the
pressure decreases there are almost no differences again. This can
be explained by the fact that ideal gases have no definite volume,
which can be negligible, unlike real gases. This is why the ideal
gases have more pressure than the real gases, under the same
volume.
𝑹𝒆𝒔𝒖𝒍𝒕𝒔:
𝑇ℎ𝑒 𝒂𝒑𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒕𝒆 𝑤𝑜𝑟𝑘 (𝑐𝑎𝑙𝑢𝑐𝑙𝑎𝑡𝑒𝑑 𝑏𝑦 𝑛𝑢𝑚𝑒𝑟𝑖𝑐𝑎𝑙 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝟐𝟏𝟒𝟗. 𝟐𝟏𝟔𝟐
𝑇ℎ𝑒 𝒆𝒙𝒂𝒄𝒕 𝑤𝑜𝑟𝑘 (𝑐𝑎𝑙𝑢𝑐𝑙𝑎𝑡𝑒𝑑 𝑏𝑦 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝟐𝟏𝟒𝟗. 𝟏𝟗𝟔𝟓
𝑇ℎ𝑒 𝒆𝒓𝒓𝒐𝒓 𝑖𝑠 𝟎. 𝟎𝟎𝟎𝟗𝟏𝟗𝟓𝟒 %
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Question II:
𝐺𝑖𝑣𝑒𝑛𝑠:
𝑇 = 20℃/293 𝐾
𝑚 = 28.78 × 1.66 × 10−27
𝑘𝑏 = 1.38 × 10−23
Here we start by putting the velocity from 0 to 1500, with a
step of 0.1. We then set up our function, as a function of v,
and plot it. Then, we got the most probable speed by using
the equation:
2𝐾𝑇
𝑀
We multiply the answer we got by 1.5 to know where the
point we want is (let’s call it v1). We also plot that point.
Here is the plot of the function and the point.
𝑉𝑚𝑝 = √
Figure (3): Maxwell-Boltzmann distribution at 293K with the specified point in red.
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Then we iterate until we find the velocity at which the
graph after it saturates to approximately zero and call it the
case max speed. The numerical answer is done by
numerically integrating from the start speed to the case max
speed and then we compare it with the exact answer given
by integrating the function itself from the start speed to
infinity and we calculate the error.
𝑹𝒆𝒔𝒖𝒍𝒕𝒔
𝑇ℎ𝑒 𝒏𝒖𝒎𝒆𝒓𝒊𝒄𝒂𝒍 𝒑𝒆𝒓𝒆𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝑖𝑠 𝟐𝟏. 𝟐𝟏𝟗𝟓 %
𝑇ℎ𝑒 𝒂𝒏𝒂𝒍𝒚𝒕𝒊𝒄𝒂𝒍 𝒂𝒏𝒔𝒘𝒆𝒓 𝑖𝑠 𝟐𝟏. 𝟐𝟐𝟗 %
𝑎𝑛𝑑 𝑡ℎ𝑒 𝒆𝒓𝒓𝒐𝒓 𝑖𝑠 𝟎. 𝟎𝟎𝟎𝟒𝟒𝟕𝟔𝟑%
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Question III:
A)
A theory of the specific heat capacity of solids put
forward by Peter Debye in 1912, in which it was assumed
that the specific heat is a consequence of the vibrations of the
atoms of the lattice of the solid. In contrast to the Einstein
theory of specific heat, which assumes that each atom has the
same vibrational frequency, Debye postulated that there is a
continuous range of frequencies that cuts off at a maximum
frequency νD, which is characteristic of a particular solid.
Debye treated the solid as a continuous elastic body in which
the vibrations of the atoms generate stationary waves.
According to Debye, the atoms in a solid do not vibrate
independently with the same frequency. The oscillations
belong to the entire solid and frequencies of various modes of
oscillation vary from 0 to a certain maximum value which is
characteristic of a substance and one can express it in terms
of elastic constants.
Figure (4): Comparison of Debye and Einstein Models
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The theory leads to the conclusion that the specific heat
capacity of solids is proportional to T3, where T is the
thermodynamic temperature. A key quantity in this theory is
the Debye temperature, θD, defined by θD = hνDk, where h is
the Planck constant and k is the Boltzmann constant. At very
low temperatures, the T dependence of specific heat
agrees for nonmetals. And for metals, the specific heat
of the atoms at high temperatures is defined by the
Einstein model. The temperature dependence of the
Einstein model is just on T. Both Einstein model and
Debye model gives major contribution to the high
temperature. To explain the low-temperatures specific
heats of metals, they include electron contribution to
the specific heat.
Figure (5): Summary of Debye vs Einstein Model
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B)
𝑮𝒊𝒗𝒆𝒏𝒔:
𝑇𝑑 = 2230𝐾
𝑛=1
Here we put different number of steps for each temperature,
where the approximate has 100 steps while the exact has
1000 steps. Then we put our entropy functions, again one for
the approximate and another for the exact, and we plot both
of them after.
Figure (5): Entropy – temperature graph for diamond from 4k to 40k
We then identify a function at T so we can integrate it. We
use trapezoidal method to integrate it numerically then we
integrate it normally from 4 to 40 to get the exact answer.
This is how our results looked like:
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𝑹𝒆𝒔𝒖𝒍𝒕𝒔:
𝑵𝒖𝒎𝒆𝒓𝒊𝒄𝒂𝒍 𝒂𝒏𝒔𝒘𝒆𝒓 𝑖𝑛 𝐽/𝐾 𝑖𝑠 𝟎. 𝟎𝟎𝟑𝟔𝟏𝟑𝟏
𝑬𝒙𝒂𝒄𝒕 𝒂𝒏𝒔𝒘𝒆𝒓 𝑖𝑛 𝐽/𝐾 𝑖𝑠 𝟎. 𝟎𝟎𝟑𝟔𝟏𝟑
C) We already calculated the exact answer and error in B, and
they were:
𝑹𝒆𝒔𝒖𝒍𝒕𝒔:
𝑬𝒓𝒓𝒐𝒓 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑖𝑠 𝟎. 𝟎𝟎𝟑𝟕𝟐𝟐𝟕%
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Question IV:
𝑈𝑠𝑒𝑑 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 = 500𝐾
𝐶ℎ𝑒𝑚𝑖𝑐𝑎𝑙 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 = +0.005 𝑒𝑉
𝐾𝑏 = 1.38 × 1023
𝑒 = 1.6 × 10−19
𝐾𝑇 = 𝑇 ∗ 𝐾𝑏
First, we put the equations for the three types of distributions:
Boltzman_dist = 1./exp((E+0.005*e)./KT) ;
Fermi_dist = 1./((exp((E+0.005*e)./KT))+1) ;
Boson_dist = 1./((exp((E+0.005*e)./KT))-1) ;
We then plot the three equations. Also, we put out y limit at 1
because boson distribution wouldn’t fit the plot at low values.
The plot is as following:
Figure (6): the energy distribution for the 3 energy distributions at 500K and chemical potential of 0.005 ev
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The distribution of particles in a system at a particular temperature
is described by the Maxwell-Boltzmann, Bose-Einstein, and
Fermi-Dirac statistics.
Systems of non-interacting, non-identical particle systems, such as
gases made up of atoms or molecules, are covered by the
Maxwell-Boltzmann statistics. It claims that the temperature
divided by the negative exponential of the energy is inversely
proportional to the likelihood of a particle possessing a particular
energy. This can be concluded from the graph because the
energy decreases as the energy increases until it saturates at a
zero probability at about 4KT.
Systems of identical, non-interacting bosons, such as photons or
atoms in a Bose-Einstein condensate, are subject to the BoseEinstein statistics. It says that the exponential of the negative of
the energy divided by the temperature, minus one, is proportional
to the likelihood of a particle having a given energy. This minus
one is the reason it approaches infinity at zero energy because
at this point the dominator will equal zero. This may mean that
these particles cannot have zero energy.
Systems containing identical, non-interacting fermions, such as
the electrons in a metal or the atoms in a Fermi gas, are covered
by the Fermi-Dirac statistics. According to this formula, the
chance of a particle possessing a specific amount of energy is
equal to the negative exponential of the energy divided by the
temperature, plus one. From the graph, this can be understood
because the fermi is just like the Maxwell-Boltzmann except
that it’s always less than it.
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Question V
Part 1:
𝑈𝑠𝑒𝑑 𝐺𝑎𝑠 = 𝑜𝑥𝑦𝑔𝑒𝑛
𝑀 = 5.312 × 1026
𝐾𝑏 = 1.38 × 10−23
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 = 106
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑟𝑎𝑛𝑔𝑒 = 0: 2000
First, we declare the main variables and constants. Our first job is
to calculate the temperature, so we need to put a range. We can
assume that the temperature is between 1 and 1000 K, so we will
iterate through range with a step 0.01 K and calculate the value of
the number of molecules in the three specified range and compare
it with the real values we have in the problem. The error was
calculated using the following formula:
𝒆𝒓𝒓𝒐𝒓 = √|𝒓𝒆𝒂𝒍𝟐 − 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒆𝒅𝟐 |
Then we add the error for the three ranges and compare it with the
minimum error (its initial value was very big to be changed in the
Figure (7): Maxwell-Boltzmann distribution at the specified temperature with the given ranges specified using dotted lines.
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iteration). If the error is less than the minimum error, then this
temperature is the new best temperature, and this error is the new
minimum error. The process continues until the range is finished.
After we made this Process, the best temperature found was
499.96 K which is approximately 500 K. For this temperature we
have plotted the velocity distribution and calculated the needed
number.
Answer:
𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 = 𝟒𝟗𝟗. 𝟗𝟔 𝑲 ≃ 𝟓𝟎𝟎 𝑲
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 ℎ𝑎𝑣𝑒 𝑠𝑝𝑒𝑒𝑑 𝑣𝑎𝑙𝑢𝑒𝑠 𝑓𝑟𝑜𝑚 400 𝑚/𝑠 𝑡𝑜 450 𝑚/𝑠
= 𝟕𝟔𝟔𝟔𝟗 𝒑𝒂𝒕𝒓𝒊𝒄𝒍𝒆𝒔
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 ℎ𝑎𝑣𝑒 𝑠𝑝𝑒𝑒𝑑 𝑣𝑎𝑙𝑢𝑒𝑠 𝑓𝑟𝑜𝑚 950 𝑚/𝑠 𝑡𝑜 1000 𝑚/𝑠
= 𝟐𝟎𝟗𝟎𝟐 𝑷𝒂𝒓𝒕𝒊𝒄𝒍𝒆𝒔
Part 2:
First, we started by declaring the main variables and constants.
Then we started to take the input from the user and ensure that the
input is logical (like that the number of particles is less than 108).
Then we start by normalizing the portion of the graph that the
user gives as an input.
This normalizing process is done by integrating the function in the
specified range and divide all the values in the range by this
factor, so that the total area of this part is exactly 1.
Then, we start by iterating through each velocities range. For
each range we approximate the number of particles that have a
velocity in this range. Then we randomly generate the needed
number of particles’ speeds in this range by the following code
𝑛𝑒𝑤_𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
= (𝑟𝑎𝑛𝑑(1, 𝑟𝑜𝑢𝑛𝑑(𝑟𝑎𝑡𝑖𝑜_𝑖𝑛_𝑟𝑎𝑛𝑔𝑒 ∗ 𝑛𝑢𝑚_𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠)) ∗ 𝑠𝑡𝑒𝑝_𝑠𝑖𝑧𝑒)
+ 𝑖;
Then we add the velocities for each range to the main velocities
array.
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Figure (8): Flowchart describing the model of part 2 of question 5.
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After this process, we ensure the velocities graph contains exactly
the number of speeds needed. We may eliminate or add small
number of velocities to regulate the array.
Then, we iterate through the speed to find the number of velocities
that range between 200 and 900 m/s. Then, we calculate the exact
number from the integration of the main function and the error is
calculated.
The root mean square velocity Is calculated by squaring the array
of velocities, find its mean and then calculate its square root. The
exact value is calculated using the theoretical formula.
The average velocity is calculated by just finding the mean of the
array of velocities.
Then the mode is calculated by rounding the values in the array
and finding the mode by a direct function. Allt he numerical
values were compared with the theoretical formulas that are
summarized in Figure (9).
Figure (9): Maxwell-Boltzmann velocity distribution and speeds formulas
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To test the output of this code, we will show a test case:
𝑰𝒏𝒑𝒖𝒕𝒔
𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 4 𝑎𝑣𝑖𝑙𝑎𝑏𝑙𝑒 𝑔𝑎𝑠𝑒𝑠: 1 − 𝑜𝑥𝑦𝑔𝑒𝑛 2 − 𝑁𝑖𝑡𝑟𝑜𝑔𝑒𝑛 3
− 𝐻𝑦𝑑𝑟𝑜𝑔𝑒𝑛 4 − 𝐻𝑒𝑙𝑖𝑢𝑚
𝐸𝑛𝑡𝑒𝑟 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑎𝑠 𝑦𝑜𝑢 𝑤𝑎𝑛𝑡 , 𝑖𝑓 𝑛𝑜 𝑜𝑛𝑒 𝑖𝑠 𝑢𝑠𝑒𝑑 𝑂𝑥𝑦𝑔𝑒𝑛 𝑖
1
𝐸𝑛𝑡𝑒𝑟 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 , 𝑖𝑡 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑡 10^8
1𝑒5
𝐸𝑛𝑡𝑒𝑟 𝑡ℎ𝑒 𝑡𝑒𝑚𝑝𝑟𝑒𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
250
𝐸𝑛𝑡𝑒𝑟 𝑡ℎ𝑒 𝑏𝑒𝑔𝑒𝑛𝑛𝑖𝑛𝑔 𝑡ℎ𝑒 𝑠𝑝𝑒𝑒𝑑 𝑟𝑎𝑛𝑔𝑒
0
𝐸𝑛𝑡𝑒𝑟 𝑡ℎ𝑒 𝑒𝑛𝑑 𝑡ℎ𝑒 𝑠𝑝𝑒𝑒𝑑 𝑟𝑎𝑛𝑔𝑒
1500
𝑶𝒖𝒕𝒑𝒖𝒕
𝑇ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑐𝑎𝑙𝑢𝑙𝑐𝑎𝑡𝑒𝑑 𝑓𝑟𝑜𝑚 200 𝑡𝑜 900
𝑖𝑠 𝟖𝟖𝟔𝟔𝟒
𝑇ℎ𝑒 𝑒𝑥𝑎𝑐𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠 𝟖𝟖𝟔𝟖𝟔 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑒𝑟𝑟𝑜𝑟 𝑖𝑠 𝟎. 𝟎𝟐𝟒𝟖𝟎𝟕 %
𝑇ℎ𝑒 𝑐𝑎𝑙𝑢𝑙𝑐𝑎𝑡𝑒𝑑 𝑉𝑟𝑚𝑠 𝑖𝑠 𝟒𝟒𝟏. 𝟒𝟕𝟔𝟑
𝑇ℎ𝑒 𝑒𝑥𝑎𝑐𝑡 𝑉𝑟𝑚𝑠 𝑖𝑠 𝟒𝟒𝟏. 𝟒𝟎𝟗 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑒𝑟𝑟𝑜𝑟 𝑖𝑠 𝟎. 𝟎𝟏𝟓𝟐𝟓𝟏 %
𝑇ℎ𝑒 𝑐𝑎𝑙𝑢𝑙𝑐𝑎𝑡𝑒𝑑 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑠 406.6994
𝑇ℎ𝑒 𝑒𝑥𝑎𝑐𝑡 𝑉𝑟𝑚𝑠 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 406.6779 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑒𝑟𝑟𝑜𝑟
𝑖𝑠 𝟎. 𝟎𝟎𝟓𝟐𝟗𝟖𝟔 %
𝑇ℎ𝑒 𝑐𝑎𝑙𝑢𝑙𝑐𝑎𝑡𝑒𝑑 𝑚𝑜𝑠𝑡 𝑝𝑟𝑜𝑏𝑎𝑏𝑙𝑒 𝑠𝑝𝑒𝑒𝑑 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑠 𝟑𝟔𝟗
𝑇ℎ𝑒 𝑒𝑥𝑎𝑐𝑡 𝑚𝑜𝑠𝑡 𝑝𝑟𝑜𝑏𝑎𝑏𝑙𝑒 𝑠𝑝𝑒𝑒𝑑 𝑖𝑠 𝟑𝟔𝟎. 𝟒𝟎𝟖𝟗 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑒𝑟𝑟𝑜𝑟
𝑖𝑠 𝟐. 𝟑𝟖𝟑𝟕 %
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The Histogram for the velocities range is as follows:
Figure (10): Histogram of the generated array of velocities
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