Resolution 1

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Molecular Description
Felipe do Nascimento Crissanto
June 2024
1 Resolution of proposed problems
1.1 Verify that PV = vRgT is equivalent to P = nKbT , P = p
m
KbT
Before we can even reach this equivalence, we need to be aware of the following relationships and
variables :
v =
N
NA
=
M
M
(1)
Where:
v is the gas amount.
N is the amount of matter (number of moles).
NA is Avogadro’s number.
M is the mass of the substance.
M is the molar mass of the substance.
Also, we have to:
Kb =
Rg
Na
(2)
On what:
Kbis the Boltzmann constant.
Rgis the ideal gas constant.
Even more, with (m) being the mass of a single particle, it is possible to verify that:
Rg
M
=
Kb
m
(3)
Having this in mind, we can write the equation of state in terms of numerical density(n), implying:
PV = vRgT (4)
Knowing that:
P is the pressure.
V is the volume.
T is the temperature.
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Replacing with eq (1), we get:
PV =
N
Na
RgT (5)
We can then use eq (2), obtaining then:
PV = NKbT (6)
So:
P =
N
V
KbT (7)
reaching the following equation:
P = nKbT (8)
We can, with the same development, rewrite eq (4) in terms of mass density, using eq (1) , we
have:
PV =
m
M
RgT (9)
Isolating Rg in eq (3) and substituting in eq (9), demonstrating that:
PV =
M
M
Kb
m
MT (10)
Developing, we obtain:
PV =
M
m
KbT (11)
Then:
P =
M
V ·m
KbT (12)
Finally:
P =
ρ
m
KbT (13)
1.2 Calculate the mass of one atom of helium and xenon
We know that:
m =
M
Na
(14)
Given that the molecular mass of Helium is equal to 0.0040025Kg/mol and Avogadro’s constant
is equivalent to 6.022× 1023 mol−1. Substituting into eq (14), we can observe that:
m =
0.0040025
6.022× 1023
= 6.6466× 10−27Kg (15)
Likewise, we have that the molecular mass of xenon is approximately 0.131293 Kg, substituting
again into eq (14), we obtain:
m =
0.131293
6.022× 1023
= 2.180× 10−25Kg (16)
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1.3 Estimate the EFP of helium and xenon under standard conditions,
that is, T = 273.15K and p = 1atm
We know the following relationship:
λ =
µVm
P
(17)
On what:
λ equivalent free path
Vm is the molecular speed
P is the pressure
For Helium we have that (µ = 19µPas, Vm = 1.1× 103m/s, P = 1.01× 105Pa ), replacing in
eq (17), we get:
λ =
19× 1.1× 103
1.01× 105
= 0.206267µm (18)
For Xenon we have that (µ = 21µPas, Vm = 1.9× 102m/s, P = 1.01× 105Pa ), then:
λ =
21× 1.9× 102
1.01× 105
= 0.0393782µm (19)
1.4 Estimate the EFP for helium under typical conditions of rough vac-
uum P = 1Pa and high vacuum P = 107Pa. Assume T = 273.15K.
In this case, the helium is under the following rough vacuum conditions(µ = 19µPas, Vm =
1, 1× 103m/s , P = 1Pa), applying to eq (17):
λ =
19× 1.1× 103
1
= 0.021m (20)
for high vacuum, he have (µ = 19µPas, Vm = 1, 1 × 103m/s, P = 10−7Pa), adding the data
in eq (17), we obtain:
λ =
19× 1.1× 103
10−7
= 2.1× 105m (21)
1.5 Calculate the Loschmidt constant, that is, the number of particles
per cubic put under standard conditions
Rearranging eq (8), isolating the numerical density (n) and substituting for the standard conditions(P =
1.01× 105, T = 273.5K), we get:
n =
P
KbT
(22)
Under the conditions:
n =
1.01× 105
1.38× 1023 × 273.15
= 2.68804× 1025m−3 (23)
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1.6 Calculate the number of particles contained in 1 m3 under the ul-
trahigh vacuum condition P = 1010Pa at T = 300K.
Substituting the values in eq (22), we achieve:
n =
10−10
1.38× 1023 × 300
= 2.4154× 1010m−3 (24)
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