Register to reply

Temperature changes and degrees of freedom and intermolecular forces

Share this thread:
sgstudent
#1
Jun16-13, 04:07 AM
P: 645
Hi I have a couple of questions relating to these topic hope you guys can help :)

Comparing ethanol and water, ethanol has weaker intermolecular forces than water. However, it has a greater specific heat capacity by kJ/mol.K than water. Meaning more energy would have to be absorbed per mole of ethanol to raise the temperature by 1K than for water.

1)I read that having a greater degree of freedom would mean the specific heat capacity is greater but I don't quite understand why this is so. Can someone explain this? cos I can't seem to find an explanation for this. I would think that the reason for that is that when ethanol gains 1J of energy, it is distributed to translational, rotational and vibrational kinetic energy as well as potential energy. So when we compare per mole of ethanol and water, a larger portion of that 1J goes to the translational energy component causing it to have a greater increase in temperature in water than in ethanol?

2)However reading this link "http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/inteng.html#c2" they said that the total KE of 1gram of copper is the same for 1gram in water. However I don't quite understand this. Since total KE is the average KE multiplied by the number of molecules. In 1g of water and 1g of copper there are different numbers of molecules. So at the same temperature, the total KE shouldn't be the same right?

3)However, how would the specific heat capacity correlate to molecular bond strength too? Or is there no link in the 2 concepts besides that once it reaches a certain amount of kinetic energy, then it would have enough energy to vapourize? So enthapy of vaporization is greater for water than for ethanol.

4)Then again, the ethanol boils at a lower temperature than water. So it would have to have a greater vapour pressure at any given temperature - meaning more ethanol evaporates at a given temperature than water. So would we explain that because the enthalpy of vaporization is less for water than ethanol due to the weaker intermolecular forces so more ethanol would evaporate at a given temperature?

But I'm not quite sure about that. Because even though the enthalpy of vaporization is greater for water than ethanol, due to the higher molar heat capacity I'm not too sure why it would evaporate at faster rate actually. I'm thinking these 2 factors contradict each other somehow. Where would my misconception be?

Sorry for the long post. Hope you guys can help :)
Phys.Org News Partner Physics news on Phys.org
UCI team is first to capture motion of single molecule in real time
And so they beat on, flagella against the cantilever
Tandem microwave destroys hazmat, disinfects
Andrew Mason
#2
Jun16-13, 11:07 AM
Sci Advisor
HW Helper
P: 6,684
Quote Quote by sgstudent View Post
r.

1)I read that having a greater degree of freedom would mean the specific heat capacity is greater but I don't quite understand why this is so. Can someone explain this?
cos I can't seem to find an explanation for this. I would think that the reason for that is that when ethanol gains 1J of energy, it is distributed to translational, rotational and vibrational kinetic energy as well as potential energy. So when we compare per mole of ethanol and water, a larger portion of that 1J goes to the translational energy component causing it to have a greater increase in temperature in water than in ethanol?
The difference does not appear to be due to potential energy since the intermolecular forces in water appear to be higher than for ethanol (ethanol has a lower boiling point). So the difference must be due to a higher number of degrees of freedom of the ethanol molecule that are available. The numbers suggest that it has 9 degrees of freedom so there must be some active vibrational modes.

2)However reading this link "http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/inteng.html#c2" they said that the total KE of 1gram of copper is the same for 1gram in water. However I don't quite understand this. Since total KE is the average KE multiplied by the number of molecules. In 1g of water and 1g of copper there are different numbers of molecules. So at the same temperature, the total KE shouldn't be the same right?
You are right. It appears to be a mistake. They should be comparing samples with equal numbers of molecules.

If the two samples are at the same temperature, the average translational KE per molecule is the same. So if they have the same total translational KE they must have the same number of molecules. The molar specific heat of water is only about 3 times that of copper. That difference is due to the inter-molecular forces (potential energy), not the 10 times as indicated in this example.

You should point out the mistake to Hyperphysics.

AM
Andrew Mason
#3
Jun17-13, 10:34 AM
Sci Advisor
HW Helper
P: 6,684
I have noticed another error in the thermodynamics section on the Hyperphysics site. This page: http://hyperphysics.phy-astr.gsu.edu.../eqpar.html#c3
suggests that the potential energy of steam is 0! In fact, the internal energy of steam is much greater than that of water at the same temperature because it takes energy to evaporate water (latent heat of vaporization of water = 2260 J/g). All of this goes into potential energy of steam.

AM

sgstudent
#4
Jun20-13, 05:46 AM
P: 645
Temperature changes and degrees of freedom and intermolecular forces

Quote Quote by Andrew Mason View Post
I have noticed another error in the thermodynamics section on the Hyperphysics site. This page: http://hyperphysics.phy-astr.gsu.edu.../eqpar.html#c3
suggests that the potential energy of steam is 0! In fact, the internal energy of steam is much greater than that of water at the same temperature because it takes energy to evaporate water (latent heat of vaporization of water = 2260 J/g). All of this goes into potential energy of steam.

AM
Hi Andrew Mason sorry for the late reply I just got back from a camp so I just got the chance to read this.

Thanks for clarifying 1 and 2 for me. For question 1 so does it mean per joule of energy for ethanol, some of it is converted to the other modes of kinetic energy so less energy is converted to the translational kinetic energy component? Also what do you mean by potential energy not being a factor here? Shouldn't the potential energy component also increase as heat is introduced to it?

I still have some doubts for the evaporation of water versus ethanol. Generally the explain given to me was that "the enthalpy of vaporization of ethanol is less compared to water. Thus at any given temperature, more ethanol can vaporize by evaporating." But I was thinking that ethanol requires more energy to raise the temperature so shouldn't that contradict that statement? I'm not sure why it contradicts just that I have a feeling that it does. What's wrong with my concept here?

Thanks so much for the help :)
Andrew Mason
#5
Jun20-13, 10:09 AM
Sci Advisor
HW Helper
P: 6,684
Quote Quote by sgstudent View Post
Hi Andrew Mason sorry for the late reply I just got back from a camp so I just got the chance to read this.

Thanks for clarifying 1 and 2 for me. For question 1 so does it mean per joule of energy for ethanol, some of it is converted to the other modes of kinetic energy so less energy is converted to the translational kinetic energy component?

Also what do you mean by potential energy not being a factor here? Shouldn't the potential energy component also increase as heat is introduced to it?
Since the latent heat of vaporization of ethanol is less than that of water, the intermolecular forces in ethanol must be less for ethanol (heat of vaporization is the energy required to break the intermolecular bonds). Therefore, for the same increase in temperature, the increase in the potential energy component of internal energy in a mole of ethanol is less than that of a mole of water at the same temperature.

For each unit of heat flow (Q) the temperature change of the ethanol is less than that of water. This means that, for the same increase in temperature, the increase in internal energy component of a mole of ethanol is greater than that of water. Since the increase in potential energy is not greater, this means that the increase in kinetic energy must be greater for ethanol than for water. Since the increase in translational KE of the mole of ethanol is the same as that for water (same T), the additional internal KE must consist of non-translational KE. This means that ethanol molecules must have more degrees of freedom and molecules of H2O.

I still have some doubts for the evaporation of water versus ethanol. Generally the explain given to me was that "the enthalpy of vaporization of ethanol is less compared to water. Thus at any given temperature, more ethanol can vaporize by evaporating." But I was thinking that ethanol requires more energy to raise the temperature so shouldn't that contradict that statement? I'm not sure why it contradicts just that I have a feeling that it does. What's wrong with my concept here?

Thanks so much for the help :)
See above.

[tex]U = PE + \sum_{i=1}^n KE_i[/tex].

Uwater < Ueth at same T (because ethanol has higher heat capacity). If PEwater > PEeth and since KE1 is the same for both (same T), n must be greater for ethanol than for water.

AM
sgstudent
#6
Jun21-13, 12:06 AM
P: 645
Quote Quote by Andrew Mason View Post
Since the latent heat of vaporization of ethanol is less than that of water, the intermolecular forces in ethanol must be less for ethanol (heat of vaporization is the energy required to break the intermolecular bonds). Therefore, for the same increase in temperature, the increase in the potential energy component of internal energy in a mole of ethanol is less than that of a mole of water at the same temperature.

For each unit of heat flow (Q) the temperature change of the ethanol is less than that of water. This means that, for the same increase in temperature, the increase in internal energy component of a mole of ethanol is greater than that of water. Since the increase in potential energy is not greater, this means that the increase in kinetic energy must be greater for ethanol than for water. Since the increase in translational KE of the mole of ethanol is the same as that for water (same T), the additional internal KE must consist of non-translational KE. This means that ethanol molecules must have more degrees of freedom and molecules of H2O.

See above.

[tex]U = PE + \sum_{i=1}^n KE_i[/tex].

Uwater < Ueth at same T (because ethanol has higher heat capacity). If PEwater > PEeth and since KE1 is the same for both (same T), n must be greater for ethanol than for water.

AM
Oh i better understand it now. So is there a link between water having a lower internal energy at the same temperature than ethanol and how much ethanol would vaporize compared to water? Because I feel that even though the total amount of energy to raise the temperature of ethanol to its boiling point and boiling it is less than for water to do the same thing, having different internal energies would have an effect on this.

Thanks :)
Andrew Mason
#7
Jun21-13, 12:57 AM
Sci Advisor
HW Helper
P: 6,684
Quote Quote by sgstudent View Post
Oh i better understand it now. So is there a link between water having a lower internal energy at the same temperature than ethanol and how much ethanol would vaporize compared to water? Because I feel that even though the total amount of energy to raise the temperature of ethanol to its boiling point and boiling it is less than for water to do the same thing, having different internal energies would have an effect on this.

Thanks :)
No. From the higher heat capacity for ethanol, you can say is that the internal energy change for ethanol is greater than that of water for a given ΔT.

This is just the first law of thermodynamics: Q = ΔU + W. Since W = 0 (no appreciable PdV work being done), Q = ΔU. Since Q is higher for ethanol than water for a given ΔT, ΔU is higher for ethanol than for water for a given ΔT. Since ΔU consists of change in PE and change in total KE, and since the KE associated with each degree of freedom is equal (equipartition theorem), we have:

[tex]\Delta U = \Delta{PE} + \sum_{i=1}^{n} \Delta KE_i = \Delta PE + n\Delta KE_1[/tex]

This does not tell you how much is in PE and how much is in KE. The relative strengths of the intermolecular forces determines the relative PE.

AM
sgstudent
#8
Jun21-13, 03:55 AM
P: 645
Quote Quote by Andrew Mason View Post
No. From the higher heat capacity for ethanol, you can say is that the internal energy change for ethanol is greater than that of water for a given ΔT.

This is just the first law of thermodynamics: Q = ΔU + W. Since W = 0 (no appreciable PdV work being done), Q = ΔU. Since Q is higher for ethanol than water for a given ΔT, ΔU is higher for ethanol than for water for a given ΔT. Since ΔU consists of change in PE and change in total KE, and since the KE associated with each degree of freedom is equal (equipartition theorem), we have:

[tex]\Delta U = \Delta{PE} + \sum_{i=1}^{n} \Delta KE_i = \Delta PE + n\Delta KE_1[/tex]

This does not tell you how much is in PE and how much is in KE. The relative strengths of the intermolecular forces determines the relative PE.

AM
Ohh but what about the vaporizing parts? Would the only explanation be for that the energy required to turn it into a liquid be less?

Also, even if it doesn't tell where the energy goes (like either KE or PE), how does it relate to this actually?

Sorry for not understanding this too well. Thanks so much :)
sgstudent
#9
Jun21-13, 10:53 PM
P: 645
Quote Quote by Andrew Mason View Post
No. From the higher heat capacity for ethanol, you can say is that the internal energy change for ethanol is greater than that of water for a given ΔT.

This is just the first law of thermodynamics: Q = ΔU + W. Since W = 0 (no appreciable PdV work being done), Q = ΔU. Since Q is higher for ethanol than water for a given ΔT, ΔU is higher for ethanol than for water for a given ΔT. Since ΔU consists of change in PE and change in total KE, and since the KE associated with each degree of freedom is equal (equipartition theorem), we have:

[tex]\Delta U = \Delta{PE} + \sum_{i=1}^{n} \Delta KE_i = \Delta PE + n\Delta KE_1[/tex]

This does not tell you how much is in PE and how much is in KE. The relative strengths of the intermolecular forces determines the relative PE.

AM
Because I'm thinking since ethanol has a greater internal energy at any given temperature would it be able to vapourize more too?

Also I'm not too sure about the explanations regarding the KE and PE. Even of we are sure about where the energy goes, how does it explain whether ethanol would vaporize more easily?
Andrew Mason
#10
Jun21-13, 11:57 PM
Sci Advisor
HW Helper
P: 6,684
Quote Quote by sgstudent View Post
Because I'm thinking since ethanol has a greater internal energy at any given temperature would it be able to vapourize more too?

Also I'm not too sure about the explanations regarding the KE and PE. Even of we are sure about where the energy goes, how does it explain whether ethanol would vaporize more easily?
The specific heat does NOT tell you whether ethanol will vaporize at a lower temperature. That is determined by the intermolecular forces. The stronger the intermolecular forces are, the higher the vaporization temperature will be and the higher the potential energy component of total energy will be.

AM
sgstudent
#11
Jun22-13, 01:44 AM
P: 645
Quote Quote by Andrew Mason View Post
The specific heat does NOT tell you whether ethanol will vaporize at a lower temperature. That is determined by the intermolecular forces. The stronger the intermolecular forces are, the higher the vaporization temperature will be and the higher the potential energy component of total energy will be.

AM
Ohh! But actually if the intermolecular forces are greater why would it mean that the temperature required to boil is greater? Because now each degree change in temperature for different compounds would require different amounts of energy, so I'm not too sure why having a greater intermolecular force would mean greater temperature when boiling occurs actually.

Also, why would a compound which has a low boiling point evaporate more easily? Because I thought if a high boiling point substance has a high amount of internal energy it should be easier to evaporate too?

Thanks Andrew Mason :)
Andrew Mason
#12
Jun22-13, 08:06 AM
Sci Advisor
HW Helper
P: 6,684
Quote Quote by sgstudent View Post
Ohh! But actually if the intermolecular forces are greater why would it mean that the temperature required to boil is greater? Because now each degree change in temperature for different compounds would require different amounts of energy, so I'm not too sure why having a greater intermolecular force would mean greater temperature when boiling occurs actually.

Also, why would a compound which has a low boiling point evaporate more easily? Because I thought if a high boiling point substance has a high amount of internal energy it should be easier to evaporate too?

Thanks Andrew Mason :)
This may help explain the relationship between the intermolecular forces and boiling point:

http://butane.chem.uiuc.edu/cyerkes/...e%2018-102.htm

AM
sgstudent
#13
Jun24-13, 03:37 AM
P: 645
Quote Quote by Andrew Mason View Post
This may help explain the relationship between the intermolecular forces and boiling point:

http://butane.chem.uiuc.edu/cyerkes/...e%2018-102.htm

AM
Hi Andrew Mason, thanks for the great link.

From the link they explained that temperature rise is caused by the rise in kinetic energy and boiling is when the energy is used for the change in potential energy. They also mentioned about having stronger IMF to be related to having a greater boiling point.

However, why is this so? I still can't seem to get why the greater the IMF the greater the BP because isn't the energy required to raise the temperature for both the KE component and the PE component. So each degree change would have different raises in PE as well as KE contributing to the different heat capacities. So I don't see how there's a link in them actually.

Thanks for the help :)
Andrew Mason
#14
Jun25-13, 07:52 AM
Sci Advisor
HW Helper
P: 6,684
Quote Quote by sgstudent View Post
Hi Andrew Mason, thanks for the great link.

From the link they explained that temperature rise is caused by the rise in kinetic energy and boiling is when the energy is used for the change in potential energy. They also mentioned about having stronger IMF to be related to having a greater boiling point.

However, why is this so? I still can't seem to get why the greater the IMF the greater the BP because isn't the energy required to raise the temperature for both the KE component and the PE component. So each degree change would have different raises in PE as well as KE contributing to the different heat capacities. So I don't see how there's a link in them actually.

Thanks for the help :)
The relationship between IMF and BP is not about heat capacity: higher IMF -> higher BP

AM
sgstudent
#15
Jun25-13, 09:11 AM
P: 645
Quote Quote by Andrew Mason View Post
The relationship between IMF and BP is not about heat capacity: higher IMF -> higher BP

AM
Ohh but when a substance has a greater BP, won't it also mean that more kinetic energy is transferred too? So in the ethanol and water case, ethanol has weaker IMF so it would have a lower BP.

But why is it that the greater the IMF the greater the BP? Because I don't see the relationship clearly. When I heat it up, the heat goes to both the PE and the KE so if the IMF is weaker why is it that the boiling occurs earlier?

Thanks Andrew Mason :)
Andrew Mason
#16
Jun25-13, 11:48 AM
Sci Advisor
HW Helper
P: 6,684
Quote Quote by sgstudent View Post
Ohh but when a substance has a greater BP, won't it also mean that more kinetic energy is transferred too? So in the ethanol and water case, ethanol has weaker IMF so it would have a lower BP.

But why is it that the greater the IMF the greater the BP? Because I don't see the relationship clearly. When I heat it up, the heat goes to both the PE and the KE so if the IMF is weaker why is it that the boiling occurs earlier?

Thanks Andrew Mason :)
The intermolecular forces do not change with temperature. Those depend on the characteristics of the molecules.

I don't know what else to say to you except that you should just try to understand what I have posted.

Think of the atoms stuck together as having negative energy. Positive heat flow causes the kinetic energy of these molecules to increase. If it increases enough so that the total energy is positive, the molecules overcome these bonds and are free to move. The greater the IMF, the more negative is the binding energy. The more negative the binding energy, the greater the temperature (translational KE) required to make the total energy positive (free molecules moving with 0 PE and positive KE).

AM
sgstudent
#17
Jun27-13, 04:29 AM
P: 645
Quote Quote by Andrew Mason View Post
The intermolecular forces do not change with temperature. Those depend on the characteristics of the molecules.

I don't know what else to say to you except that you should just try to understand what I have posted.

Think of the atoms stuck together as having negative energy. Positive heat flow causes the kinetic energy of these molecules to increase. If it increases enough so that the total energy is positive, the molecules overcome these bonds and are free to move. The greater the IMF, the more negative is the binding energy. The more negative the binding energy, the greater the temperature (translational KE) required to make the total energy positive (free molecules moving with 0 PE and positive KE).

AM
Ohh I think i get it now. That was a great explanation :)

So substances with high IMF would be more negative.

But the explanation seems to suggest that per joule of energy absorbed by the 2 substances would have an equal distribution to both the translational KE and the PE but isn't the transfer of energy between the 2 substances different?

Thanks so much for the help Andrew Mason :)


Register to reply

Related Discussions
Degrees of Freedom Set Theory, Logic, Probability, Statistics 1
Capillary forces, intermolecular forces and surface tension questions Classical Physics 2
Intermolecular forces >>> trends in London Dispersion Forces Chemistry 2