Download Average Spacing of Gas Molecules

ME12001 Thermodynamics T4
Average Spacing of Gas Molecules
Consider an ideal gas at 27.0 degrees Celsius and 1.00 atmosphere pressure. Imagine the molecules to be
uniformly spaced, with each molecule at the center of a small cube.
Part A
What is the length L of an edge of each small cube if adjacent cubes touch but don't overlap?
Express your answer numerically in meters.
Hint 1. How to approach the problem
Calculate the volume of one mole of the gas, being careful about the units, and then use this result to
find the volume of one molecule, or rather, the volume of the imaginary cube that is assumed to
surround each molecule. Then use the volume per molecule to calculate the length of a side of the cube.
Hint 2. Calculate the volume per mole
What is the volume Vmole of one mole of the gas?
Express your answer numerically in cubic meters using three significant figures.
Hint 1. Ideal gas equation
The ideal gas equation states that pV = nRT . For this question, since you want to find the
volume of one mole of the gas, n = 1 . Also, be careful of the units that you use for the gas
constant R.
ANSWER:
Vmole
= 2.46×10−2 m3 Hint 3. Calculate the volume per molecule
Calculate the volume per molecule Vmolecule of the gas.
Express your answer in cubic meters using three significant figures.
Hint 1. How to approach the problem
Recall that one mole of a gas has Avogadro's number, i.e., NA , molecules. So above you found
the volume explored by NA molecules. Use this information to find the volume surrounding one
molecule.
ANSWER:
Vmolecule
= 4.09×10−26 m3 1/15
ME12001 Thermodynamics T4
Hint 4. The edge length of a cube
Recall that the volume Vc of a cube is related to its edge length l by Vc
the correct expression for l in terms of Vc ?
= l
3
. Which of the following is
ANSWER:
Vc
Vc
3
V
3
c
3Vc
1/3
Vc
ANSWER:
L
= 3.45×10−9 m Correct
For carbon dioxide gas, the physical volume of a molecule (based on the van der Waals equation constant 5
3
−29
3
4.27 × 10
m /mol ) is approximately 7.09 × 10
m . This implies that the linear dimension of the
molecule is only about 4.14 × 10 −10 m .
Comparing this number to the result of your calculation above, you can see that the size of the molecule is
less than one eighth of the length of the cube edge that surrounds the molecule, which is also the average
distance separating one molecule from the next. A small molecular size, as compared to the distance
between molecules, is a necessary assumption in the kinetic­molecular model of an ideal gas.
Exercise 18.27
Part A
How many moles are there in a 1.80 kg bottle of water?
ANSWER:
M
= 100 mol Correct
Part B
How many molecules are there in the bottle?
ANSWER:
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ME12001 Thermodynamics T4
N
= 6.02×1025
Correct
Exercise 18.38
Part A
Calculate the mean free path of air molecules at a pressure of 3.50×10−13 atm and a temperature of 305 K .
(This pressure is readily attainable in the laboratory.) Model the air molecules as spheres with a radius of
2.00×10−10 m .
ANSWER:
λ
= 1.67×105 m Correct
Speed and Energy Scaling
Helium atoms have a mass of 4u and oxygen molecules have a mass of 32u , where u is defined as an atomic
mass unit (u = 1.660540 × 10 −27 kg). Compare a gas of helium atoms to a gas of oxygen molecules.
Part A
At what gas temperature TE would the average translational kinetic energy of a helium atom be equal to that of
an oxygen molecule in a gas of temperature 300 K?
Express the temperature numerically in kelvins.
Hint 1. Find the energy associated with one degree of freedom
The Equipartition Theorem states that the average energy associated with each degree of freedom is (1/2)k B T . Therefore, what is the average contribution of motion in the x direction to the kinetic
energy?
Express the average energy due to the x motion in terms of k B and T .
Hint 1. Equipartition Theorem
For the translational kinetic energy of gas particles, the Equipartition Theorem states
1
2
⟨
2
m vx
⟩=
1
2
⟨
2
m vy
where m is the mass of each gas particle.
⟩=
1
2
⟨
2
m vz
⟩=
1
2
kB T ,
ANSWER:
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ME12001 Thermodynamics T4
1
2
⟨
2
m vx
⟩ = 1
2
kB T
Hint 2. Total translational kinetic energy
Add the contributions from the x, y, and z degrees of freedom to find the total translational kinetic
energy Etrans .
Express the total translational kinetic energy in terms of k B and T .
Hint 1. Total translational energy in terms of velocities
We can write the total translational energy as follows:
E trans =
1
2
⟨
2
m( v x
⟩ + ⟨v ⟩ + ⟨v ⟩).
2
y
2
z
ANSWER:
Etrans
= 3
2
kB T
ANSWER:
TE
= 300 K Correct
Part B
At what gas temperature Trms would the root­mean­square (rms) speed of a helium atom be equal to that of an
oxygen molecule in a gas at 300 K?
State your answer numerically, in kelvins, to the nearest integer.
Hint 1. Find the rms speed
From the Equipartition Theorem, you can determine that
1
2
⟨
m v
2
⟩ = 3(
1
2
kB T ) ,
where the factor of 3 results from the fact that the particles can move in three possible directions. From
−
−
−
−
this equation, find v rms = √ ⟨v 2 ⟩ .
Express your answer in terms of k B , T , and m.
ANSWER:
−
−
−
−
2
v rms = √ v
⟨
⟩ = −
−
−
−
−
√
3kB T
m
ANSWER:
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ME12001 Thermodynamics T4
Trms
= 38 K Correct
Note that 38 K is is a very low temperature. At this temperature, helium still behaves approximately like an
ideal gas, as do hydrogen and neon, but all other gasses would be frozen solid!
Particle Gas Review
A particle gas consists of N monatomic particles each of mass m all contained in a volume V at temperature T .
Your answers should be written in terms of the Boltzmann constant k B and Avagadro's number NA rather than R = NA k B .
Part A
Find ⟨v 2 ⟩ , the average speed squared for each particle.
Express the average speed squared in terms of the gas temperature T and any other given quantities.
Hint 1. how to approach the problem
Since v
−
−
−−
−−
−−
−−
2
2
2
= √vx + vy + vz
, it follows that ⟨v 2 ⟩
=
⟨v ⟩ + ⟨v ⟩ + ⟨v ⟩ where ⟨v ⟩ is the average
2
x
2
y
2
z
2
i
velocity squared for each particle in the i direction. Find the average velocity squared for each particle
along each direction, then find the algebraic sum.
Hint 2. Find ⟨v 2x ⟩ for each particle
Find ⟨v 2x ⟩ , the average x velocity squared for each particle.
Express the average x velocity squared in terms of the gas temperature T and any other given
quantities.
Hint 1. How to approach the problem
Consider a single gas particle moving only in the x direction. Find one formula for its kinetic
energy in terms of its velocity and another formula in terms of the gas temperature. Set these
equations equal to each other and solve for v 2x .
Hint 2. Kinetic energy in terms of temperature
A monatomic particle has three degrees of freedom, one for translation in each of three
dimensions. The kinetic energy associated with one degree of freedom (for example, motion in
the x direction) is K = (1/2)k B T .
Hint 3. Kinetic energy in terms of velocity
Another expression for a particle's kinetic energy associated with motion in the x direction is 2
K = (1/2)m⟨v x ⟩ .
ANSWER:
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ME12001 Thermodynamics T4
⟨v ⟩ = 2
x
kB T
m
Hint 3. Relating the x, y , and z velocities
Particles of an ideal gas move randomly, so in a contained volume no direction is preferred. Therefore
⟨v 2x ⟩ = ⟨v 2y ⟩ = ⟨v 2z ⟩ .
ANSWER:
⟨v ⟩ = 2
3kB
T
m
Correct
Part B
Find U , the internal energy of the gas.
Express the internal energy in terms of the gas temperature T and any other given quantities.
Hint 1. How to approach the problem
In an ideal gas, intermolecular forces are negligible. The internal energy of the whole gas is a sum of the
kinetic energies of each gas particle.
Hint 2. Kinetic energy of a single gas particle
The kinetic energy of a single gas particle is K = (1/2)m⟨v 2 ⟩ . As stated in the problem introduction,
there are a total of N particles comprising the gas.
ANSWER:
U
= 3
2
kB T N
Correct
Part C
Find CV , the molar heat capacity (heat capacity per mole) of the gas at constant volume.
Express the molar heat capacity in terms of NA and k B .
Hint 1. Definition of heat capacity
Heat capacity is generally defined to be dQ/dT . From the first law of thermodynamics,
At constant volume dW
= 0
dQ = dU + dW .
. Therefore, the heat capacity at constant volume is given by
C
= dU /dT
.
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ME12001 Thermodynamics T4
CV = dU /dT
.
This question asks you to find the molar heat capacity, or heat capacity per mole of gas particles. The
easiest way to do this is to compute heat capacity assuming that there is exactly one mole of gas (i.e., N = NA ).
ANSWER:
CV
= 3
2
N A kB
Correct
You can use the molar heat capacity CV to find the total heat capacity of the gas at constant volume cV .
Since cV is the total heat capacity of the gas, it is just the heat capacity per mole multiplied by the
number of moles of gas particles n = N /NA . Hence, cV = (3/2)N k B .
Part D
Express the pressure p of the gas in terms of its energy density U /V .
Enter the numerical factor that multiplies U /V in your expression for p to at least three significant
figures.
Hint 1. How to approach the problem
Find an expression that relates the pressure p to the volume V of the gas. Then use your expressions
for U from Part C to write p in terms of U /V .
Hint 2. Find an expression for pressure
Find p, the pressure of the gas using the Ideal Gas Law.
Express your answer in terms of N , T , V , and k B .
Hint 1. Ideal gas law
The ideal gas law is pV
= N kB T
.
ANSWER:
p
= N kB
T
V
Correct
ANSWER:
p
= 0.667 U /V
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ME12001 Thermodynamics T4
Correct
As an interesting aside, note that for an ultra­relativistic particle gas, such as a gas of photons,
p=
1
U
3
V
.
Now imagine that the mass of each gas particle is increased by a factor of 3. All other information given in the
problem introduction remains the same.
Part E
What will be the ratio of the new molar mass M ′ to the old molar mass M ?
ANSWER:
M
′
M
= 3
Correct
Part F
What will be the ratio of the new rms speed v ′rms to the old rms speed v rms ?
Hint 1. Definition of rms speed
−
−
−
−
By definition v rms ≡ √ ⟨v 2 ⟩ . A monatomic particle has three degrees of freedom, one for translation in
each of three dimensions. The kinetic energy associated with one degree of freedom (for example,
motion in the x direction) is K = (1/2)k B T . Another expression for a particle's kinetic energy
associated with motion in the x direction is K = (1/2)m⟨v 2x ⟩ .
ANSWER:
′
v rms
v rms
= 0.577
Correct
So if the internal energy (temperature) remains the same, when the mass of the particles increases, the
velocity correspondingly decreases. Observe that the pressure and so the average force that the particles
exert on the walls also remains the same.
Part G
′
What will be the ratio of the new molar heat capacity CV to the old molar heat capacity CV ?
Hint 1. How to approach the problem
The total heat capacity of the gas can be determined by multiplying the heat capacity per mole by the
number of moles of gas particles n = N /NA . How does the heat capacity per mole depend on the
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ME12001 Thermodynamics T4
mass of the particles? How does the number of moles of gas particles depend on the mass of the
particles?
ANSWER:
′
CV
CV
= 1
Correct
In fact, the molar heat capacity is (almost) the same for all monoatomic gases, as the general expression
derived earlier shows. For example, helium, neon, and argon all have almost the same molar heat
capacity. The tiny differences in the values tell us that the ideal gas equation describes gases pretty well,
but it is not perfect.
± Gas Scaling
When doing numerical calculations involving temperature, you need to pay particular attention to the temperature
scale you are using. In general, you should use the Kelvin scale (for which T = 0 represents absolute zero) in such
calculations. This is because the standard thermodynamic equations (i.e., the ideal gas law and the formula for
energy of a gas in terms of temperature) assume that zero degrees represents absolute zero.
If you are given temperatures measured in units other than kelvins, convert them to kelvins before plugging them
into these equations. (You may then want to convert back into the initial temperature unit to give your answer.)
Part A
The average kinetic energy of the molecules of an ideal gas at 10 ∘ C has the value K10 . At what temperature T1 (in degrees Celsius) will the average kinetic energy of the same gas be twice this value, 2K10 ?
Express the temperature to the nearest integer.
Hint 1. Formula for energy in terms of temperature
From the Equipartition Theorem we know that each translational degree of freedom of particles in a gas
contributes (1/2)k B T to the average internal energy, where the temperature T is measured in kelvins.
Recalling that a gas molecule is free to move in three perpendicular directions, give an expression for
the kinetic energy Ktr of a molecule due to its translational degrees of freedom.
Express your answer in terms of k B and T .
ANSWER:
Ktr
= 3
2
kB T
Hint 2. Convert from Celsius to Kelvin scale
Don't forget to convert the degrees Celsius into kelvins before plugging in for the temperature: TCelsius = TKelvin − 273 .
ANSWER:
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ME12001 Thermodynamics T4
T1
= 293 ∘ C Correct
Part B
The molecules in an ideal gas at 10 ∘ C have a root­mean­square (rms) speed v rms . At what temperature T2 (in
degrees Celsius) will the molecules have twice the rms speed, 2v rms ?
Express the temperature to the nearest integer.
Hint 1. What is rms speed?
The root­mean­square or rms speed of molecules in a gas is equal to the square root of the average
velocity squared of the molecules: v rms
−
−−−−
2
= √ (v )
av
. The rms speed is related to the translational
kinetic energy Ktr of the gas via the following equation:
1
2
mv rms = K tr ,
2
where m is the mass of the gas molecules.
Hint 2. Find the change in translational kinetic energy
Assume that, at 10 ∘ C , the molecules have translational kinetic energy K1 . When the rms speed of the
molecules is doubled, what is their new translational kinetic energy K2 ?
Express your answer in terms of K1 .
ANSWER:
K2
= 4K1
ANSWER:
T2
= 859 ∘ C Correct
Exercise 18.52
A physics lecture room has a volume of 207 m3 .
Part A
For a pressure of 1.00 atm and a temperature of 26.0 ∘ C , use the ideal­gas law to estimate the number of air
molecules in the room. Assume all the air is N2 .
Express your answer to three significant figures and include the appropriate units.
ANSWER:
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ME12001 Thermodynamics T4
N
= 5.08×1027 molecules Correct
Part B
Calculate the particle density­that is, the number of N2 molecules per cubic centimeter.
Express your answer to three significant figures and include the appropriate units.
ANSWER:
particle density = 2.46×1019 molecules/cm3 Correct
Part C
Calculate the mass of the air in the room.
Express your answer to three significant figures and include the appropriate units.
ANSWER:
m
= 236 kg
Correct
Problem 18.68
Part A
The size (radius) of an oxygen molecule is about 2.0 ×10 −10 m . Make a rough estimate of the pressure at
which the finite volume of the molecules should cause noticeable deviations from ideal­gas behavior at ordinary
temperatures (T = 300K ). Assume that deviatons would be noticeable when volume of the gas per molecule
equals the volume of the molecule itself.
Express your answer using one significant figure.
ANSWER:
P
= 1×108 Pa Correct
Problem 18.76
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ME12001 Thermodynamics T4
The surface of the sun has a temperature of about 5800K and consists largely of hydrogen atoms.
Part A
Find the rms speed of a hydrogen atom at this temperature. (The mass of a single hydrogen atom is −27
1.67 × 10
kg .)
ANSWER:
V
= 1.20×104 m/s Correct
Part B
1/2
The escape speed for a particle to leave the gravitational influence of the sun is given by (2GM /R)
,
where M is the sun's mass, R its radius, and G the gravitational constant. The sun`s mass is 30
8
−11
2
2
M = 1.99 × 10
kg , its radius R = 6.96 × 10 m and G = 6.673 × 10
N ⋅ m /kg . Calculate the
escape speed for the sun.
ANSWER:
v
= 6.18×105 m/s Correct
Part C
Can appreciable quantities of hydrogen escape from the sun?
ANSWER:
Yes
No
Correct
Part D
Can any hydrogen escape?
ANSWER:
Yes
No
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ME12001 Thermodynamics T4
Correct
Exercise 18.37
Part A
Oxygen O 2 has a molar mass of 32.0 g/mol. What is the average translational kinetic energy of an oxygen
molecule at a temperature of 296 K ?
ANSWER:
K
= 6.13×10−21 J Correct
Part B
What is the average value of the square of its speed?
ANSWER:
2
(v )
av
= 2.31×105 m2 /s 2
Correct
Part C
What is the root­mean­square speed?
ANSWER:
v rms
= 480 m/s Correct
Part D
What is the momentum of an oxygen molecule traveling at this speed?
ANSWER:
p
= 2.55×10−23 kg ⋅ m/s Correct
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ME12001 Thermodynamics T4
Part E
Suppose an oxygen molecule traveling at this speed bounces back and forth between opposite sides of a
cubical vessel 0.10 m on a side. What is the average force the molecule exerts on one of the walls of the
container? (Assume that the molecule's velocity is perpendicular to the two sides that it strikes.)
Express your answer using two significant figures.
ANSWER:
F av
= 1.2×10−19 N Correct
Part F
What is the average force per unit area?
Express your answer using two significant figures.
ANSWER:
P av
= 1.2×10−17 Pa Correct
Part G
How many oxygen molecules traveling at this speed are necessary to produce an average pressure of 1 atm?
Express your answer using two significant figures.
ANSWER:
N
= 8.3×1021 molecules Correct
Part H
Compute the number of oxygen molecules that are actually contained in a vessel of this size at 296 K and
atmospheric pressure.
Express your answer using two significant figures.
ANSWER:
N
= 2.5×1022 molecules Correct
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ME12001 Thermodynamics T4
Part I
Your answer for part H should be three times as large as the answer for part G. Where does this discrepancy
arise?
ANSWER:
3710 Character(s) remaining
At part G we calculated pressure exerted on area while in part h we calculated with volume
Submitted, grade pending
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