Find the change in the Kinetic energy of an Ideal Gas

In summary, the answer to question 1 is that when the volume doubled at constant temperature, the kinetic energy per molecule became greater by 0.4. When the volume was doubled without any transfer of heat, the kinetic energy per molecule became greater by 1.4.
  • #1
Helly123
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Homework Statement


Let 3/2kT be the kinetic energy of ideal gas per molecules. T the absolute temperature and N the avogadro number. Answer the following questions :

1) when the volume doubled at constant temperature. How many times the kinetic energy per molecule become greater than before?

2) when the volume doubled without transfer of heat. But the pressure 0.3 lower than before. How many times the kinetic energy per molecule become greater than before?

Homework Equations

The Attempt at a Solution


1) since the temperature is constant. I think that Ek doesn't change. So 0 times greater ? However, my answer is wrong.

2) T2 = P2*V2/P1*V1 = 0.7P1*2V1/P1*V1
T2 = 1.4T1

EK2 = 1.4EK1
EK2 0.4 greater than EK1. But again, my answer wrong. Why?
 
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  • #2
It looks like you are not clear about what is meant by "how many times" and "lower than"
If you say quantity a is 10 times quantity b, that means a = 10 b.
If you say a is 5 lower than b, that means a = b - 5

So for example, according to the statement of the problem, P2 = P1 - 0.3 (units?). That does not mean P2 = 0.7*P1
 
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  • #3
What you are looking for are ratios not differences. Keep in mind these are absolute temperatures.
 
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  • #4
Chandra Prayaga said:
It looks like you are not clear about what is meant by "how many times" and "lower than"
If you say quantity a is 10 times quantity b, that means a = 10 b.
If you say a is 5 lower than b, that means a = b - 5

So for example, according to the statement of the problem, P2 = P1 - 0.3 (units?). That does not mean P2 = 0.7*P1

I see..
Now, T2 = ##\frac{(P1-0.3)(2V1)}{P1V1} T1##
Then, what should i do?
Ek2 = 3/2kT2 or
Ek2 = 3/2k(T2-T1) ?
Btw. Is it U = 3/2nRT or ##\Delta##U = 3/2nR(T2-T1) ?
 
  • #5
Dr Dr news said:
What you are looking for are ratios not differences. Keep in mind these are absolute temperatures.
Internet says absolute temperature is temperature measured from absolute zero kelvins.
But, how does that relate to the question?
 
  • #6
Suppose T(initial) increases from 100 K to 140 K. Then T(final)/T(initial) = 1.40 = 140 % and [T(final) - T(initial)]/T(initial) = ΔT/T(initial) = 40/100 = 0.4 = 40% increase. A matter of semantics. You are mixing pV = NkT, p(n/m^2), V(m^3), N( no. of molecules), k = Boltzmann's constant = 1.381x10^-23 J/K, T(K) and pV = nRT, n(no. of moles) = N/NA, NA = Avogadro's no. = 6.022x10^23 molecules/g-mol, R = gas constant = 8.314 J/K-mol, p, V, T the same units as before..
 
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  • #7
Helly123 said:
Now, T2 = (P1−0.3)(2V1)P1V1T1(P1−0.3)(2V1)P1V1T1\frac{(P1-0.3)(2V1)}{P1V1} T1
Then, what should i do?
Ek2 = 3/2kT2 or
Ek2 = 3/2k(T2-T1) ?
Btw. Is it U = 3/2nRT or ΔΔ\DeltaU = 3/2nR(T2-T1) ?

Let us take care of some important things first.
In your first equation, P1 - 0.3 does not make sense unless you give the proper units for the quantity 0.3
You used two different symbols in the next three relations, Ek and U. What are they, and is there a difference? This relates to the remarks by Dr Dr news.
 
  • #8
Chandra Prayaga said:
Let us take care of some important things first.
In your first equation, P1 - 0.3 does not make sense unless you give the proper units for the quantity 0.3
You used two different symbols in the next three relations, Ek and U. What are they, and is there a difference? This relates to the remarks by Dr Dr news.
Yes. It kind of makes sense.. i will try
 
  • #9
Dr Dr news said:
Suppose T(initial) increases from 100 K to 140 K. Then T(final)/T(initial) = 1.40 = 140 % and [T(final) - T(initial)]/T(initial) = ΔT/T(initial) = 40/100 = 0.4 = 40% increase. A matter of semantics. You are mixing pV = NkT, p(n/m^2), V(m^3), N( no. of molecules), k = Boltzmann's constant = 1.381x10^-23 J/K, T(K) and pV = nRT, n(no. of moles) = N/NA, NA = Avogadro's no. = 6.022x10^23 molecules/g-mol, R = gas constant = 8.314 J/K-mol, p, V, T the same units as before..
Do we calculate Ek at certain temperature or we calculate Ek based on the temperature change?

If there is no heat transfer/adiabatic (number 2) then the Q is the same, does it imply that temperature is the same?
 
  • #10
Dr Dr news said:
What you are looking for are ratios not differences. Keep in mind these are absolute temperatures.
So, the question says "how many times Ek become greater", in the sense that we are to find the ratio? The ratio A to B is the same as how many times A bigger than B?
 
  • #11
It is not a question of how we calculate, but a question of how we define ot. What is Ek?
 
  • #12
Chandra Prayaga said:
It is not a question of how we calculate, but a question of how we define ot. What is Ek?
Kinetic energy. Which formula is 3/2kT
Kinetic energy defined based on average velocity of molecules. Is it?
 
  • #13
Helly123 said:
Kinetic energy. Which formula is 3/2kT
Kinetic energy defined based on average velocity of molecules. Is it?
Kinetic energy of what?
 
  • #14
Chandra Prayaga said:
Kinetic energy of what?
Kinetic energy of one mol ideal gas
 
  • #15
Then the formula is not (3/2) kT. That formula is for the average energy per molecule. This is part of what Dr Dr news was pointing out. There are a lot of "formulas" and one has to be careful about what one means. The following are different quantities, which are related to each other, and all of them are proportional to the absolute temperature:
Average kinetic energy per molecule
Average kinetic energy per mole
Average kinetic energy of N molecules
Average kinetic energy of n moles
The new symbol U that you used
These are all related, all proportional to T.

Incidentally, the kinetic energy is not related to the average velocity. The average velocity of a molecule in a gas in equilibrium is zero.
 
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  • #16
Your original question says (3/2 kT) is the kinetic energy per molecule. Let us call it Ek. This is the average kinetic energy per molecule.

So Ek = (3/2) kT. That means the average kinetic energy per molecule is proportional to the absolute temperature, not to the temperature difference. It also means that this quantity does not depend on the volume, so the answer to question 1 should be obvious.
 
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  • #17
Chandra Prayaga said:
Your original question says (3/2 kT) is the kinetic energy per molecule. Let us call it Ek. This is the average kinetic energy per molecule.

So Ek = (3/2) kT. That means the average kinetic energy per molecule is proportional to the absolute temperature, not to the temperature difference. It also means that this quantity does not depend on the volume, so the answer to question 1 should be obvious.
Hmm.. i see. Absolute temperature no depend on volume? If temperature not absolute? Will still not depend on volume?
 
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  • #18
Chandra Prayaga said:
Then the formula is not (3/2) kT. That formula is for the average energy per molecule. This is part of what Dr Dr news was pointing out. There are a lot of "formulas" and one has to be careful about what one means. The following are different quantities, which are related to each other, and all of them are proportional to the absolute temperature:
Average kinetic energy per molecule
Average kinetic energy per mole
Average kinetic energy of N molecules
Average kinetic energy of n moles
The new symbol U that you used
These are all related, all proportional to T.

Incidentally, the kinetic energy is not related to the average velocity. The average velocity of a molecule in a gas in equilibrium is zero.
What does it mean when gas is in equilibrium? Adiabatic?
 
  • #19
Certainly, everything depends on everything else. But, for an ideal gas, if the temperature is kept constant, the average energy per molecule, and therefore the thermal energy U is a constant, independent of the volume.
Equilibrium means the properties of the gas do not change with time. That is when the gas is said to be in an equilibrium state. The word adiabatic refers to a process, a change of state, done in such a way that the system does not exchange heat with its surroundings.
 
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  • #20
Chandra Prayaga said:
Certainly, everything depends on everything else. But, for an ideal gas, if the temperature is kept constant, the average energy per molecule, and therefore the thermal energy U is a constant, independent of the volume.
Equilibrium means the properties of the gas do not change with time. That is when the gas is said to be in an equilibrium state. The word adiabatic refers to a process, a change of state, done in such a way that the system does not exchange heat with its surroundings.
I see.. U independent of volume, how about the case when volume change and temperature change? So U depend on volume?
 
  • #21
If the temperature changes, U changes. U = (3/2)nRT = (3/2)NkT.
 
  • #22
Chandra Prayaga said:
Your original question says (3/2 kT) is the kinetic energy per molecule. Let us call it Ek. This is the average kinetic energy per molecule.

So Ek = (3/2) kT. That means the average kinetic energy per molecule is proportional to the absolute temperature, not to the temperature difference. It also means that this quantity does not depend on the volume, so the answer to question 1 should be obvious.
So, the answer for number 1 is 1?
 
  • #23
So now you tell us. What do you think, and why?
 
  • #24
case 2) Without going into the gory detail, for an adiabatic process, p(V)^γ = con, where γ = c(p)/c(V) = 1.67 for a monatomic gas. which means that
p1(V1)^γ = p2(V2)^γ = (p1/1.3)(2 V1)^γ; 1.3 ≠ (2)^1.67. This is some unknown process - not adiabatic.
 
  • #25
Chandra Prayaga said:
So now you tell us. What do you think, and why?
Temperature no changin. Hence, Ek2 the same as Ek1. The ratio Ek2/Ek1 = 1
Is it?
 
  • #26
Dr Dr news said:
case 2) Without going into the gory detail, for an adiabatic process, p(V)^γ = con, where γ = c(p)/c(V) = 1.67 for a monatomic gas. which means that
p1(V1)^γ = p2(V2)^γ = (p1/1.3)(2 V1)^γ; 1.3 ≠ (2)^1.67. This is some unknown process - not adiabatic.
Then the question wrong?
 
  • #27
Chandra Prayaga said:
Then the formula is not (3/2) kT. That formula is for the average energy per molecule. This is part of what Dr Dr news was pointing out. There are a lot of "formulas" and one has to be careful about what one means. The following are different quantities, which are related to each other, and all of them are proportional to the absolute temperature:
Average kinetic energy per molecule
Average kinetic energy per mole
Average kinetic energy of N molecules
Average kinetic energy of n moles
The new symbol U that you used
These are all related, all proportional to T.

Incidentally, the kinetic energy is not related to the average velocity. The average velocity of a molecule in a gas in equilibrium is zero.
Then, what is formula for Ek per n mole? 3/2nRT?
When Ek for N molecule = 3/2NkT.
Why is it the same formula as ##\Delta##U?
##\Delta##U is the total of Ek and Ep of molecules. If we asked to find Ek is the same as we find ##\Delta##U? Or what?
 
  • #28
The kinetic theory of gases yields the relation that the kinetic energy equals kT/2 for each degree of freedom of each molecule. The kinetic theory considers each molecule as a particle which undergoes elastic collisions with the container walls and other molecules. The potential energy is not considered. If the molecules under consideration are one atom / molecule, called a monatomic gas, it has three degrees of freedom, i.e. it can move in three directions and thus has a velocity component in each of three orthogonal directions. The thermal energy of a gas, also called the internal energy, is just the kinetic energy.
 
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  • #29
Dr Dr news said:
The kinetic theory of gases yields the relation that the kinetic energy equals kT/2 for each degree of freedom of each molecule. The kinetic theory considers each molecule as a particle which undergoes elastic collisions with the container walls and other molecules. The potential energy is not considered. If the molecules under consideration are one atom / molecule, called a monatomic gas, it has three degrees of freedom, i.e. it can move in three directions and thus has a velocity component in each of three orthogonal directions. The thermal energy of a gas, also called the internal energy, is just the kinetic energy.
I see. Thanks
 
  • #30
What am i supposed to do with number 2 btw?
 
  • #31
You need to clarify the problem statement. How do you get from the initial state to the final state?
 
  • #32
Dr Dr news said:
You need to clarify the problem statement. How do you get from the initial state to the final state?
Hmm..
The volume doubled and pressure 0.3 lower than before. Gases do work. But in adiabatic condition Q = 0
##\Delta##U = W
So Ek = W?
 
  • #33
Dr Dr news said:
case 2) Without going into the gory detail, for an adiabatic process, p(V)^γ = con, where γ = c(p)/c(V) = 1.67 for a monatomic gas. which means that
p1(V1)^γ = p2(V2)^γ = (p1/1.3)(2 V1)^γ; 1.3 ≠ (2)^1.67. This is some unknown process - not adiabatic.
Is this related to number 2? Or what?
 
  • #34
The volume doubling as the pressure decreases by 30$ is not an adiabatic process. Ref. #24.
 
  • #35
Dr Dr news said:
You need to clarify the problem statement. How do you get from the initial state to the final state?
20180508_125607.jpg


This is the full questions. Hope it still readable
 

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<h2>1. What is the formula for finding the change in kinetic energy of an ideal gas?</h2><p>The formula for finding the change in kinetic energy of an ideal gas is ΔKE = 1/2 * m * (Δv)^2, where ΔKE is the change in kinetic energy, m is the mass of the gas, and Δv is the change in velocity.</p><h2>2. How is the change in kinetic energy of an ideal gas related to its temperature?</h2><p>The change in kinetic energy of an ideal gas is directly proportional to its temperature. As the temperature of the gas increases, the average kinetic energy of its molecules also increases, resulting in a larger change in kinetic energy.</p><h2>3. What factors affect the change in kinetic energy of an ideal gas?</h2><p>The change in kinetic energy of an ideal gas is affected by the mass of the gas, the change in velocity, and the temperature of the gas. Additionally, the type of gas and its specific heat capacity can also impact the change in kinetic energy.</p><h2>4. Can the change in kinetic energy of an ideal gas be negative?</h2><p>Yes, the change in kinetic energy of an ideal gas can be negative if the gas is losing kinetic energy. This can happen if the gas is undergoing a decrease in temperature or if it is experiencing a decrease in velocity.</p><h2>5. How is the change in kinetic energy of an ideal gas calculated in real-life situations?</h2><p>In real-life situations, the change in kinetic energy of an ideal gas can be calculated by using experimental data and applying it to the formula ΔKE = 1/2 * m * (Δv)^2. This can be done by measuring the mass and change in velocity of the gas and using the ideal gas law to determine the temperature.</p>

1. What is the formula for finding the change in kinetic energy of an ideal gas?

The formula for finding the change in kinetic energy of an ideal gas is ΔKE = 1/2 * m * (Δv)^2, where ΔKE is the change in kinetic energy, m is the mass of the gas, and Δv is the change in velocity.

2. How is the change in kinetic energy of an ideal gas related to its temperature?

The change in kinetic energy of an ideal gas is directly proportional to its temperature. As the temperature of the gas increases, the average kinetic energy of its molecules also increases, resulting in a larger change in kinetic energy.

3. What factors affect the change in kinetic energy of an ideal gas?

The change in kinetic energy of an ideal gas is affected by the mass of the gas, the change in velocity, and the temperature of the gas. Additionally, the type of gas and its specific heat capacity can also impact the change in kinetic energy.

4. Can the change in kinetic energy of an ideal gas be negative?

Yes, the change in kinetic energy of an ideal gas can be negative if the gas is losing kinetic energy. This can happen if the gas is undergoing a decrease in temperature or if it is experiencing a decrease in velocity.

5. How is the change in kinetic energy of an ideal gas calculated in real-life situations?

In real-life situations, the change in kinetic energy of an ideal gas can be calculated by using experimental data and applying it to the formula ΔKE = 1/2 * m * (Δv)^2. This can be done by measuring the mass and change in velocity of the gas and using the ideal gas law to determine the temperature.

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