Thermodynamics and ideal gas law concepts

In summary, the conversation discusses the struggle to understand the relationship between pressure, volume, and temperature in thermodynamics and the ideal gas law. The expert explains that it is important to consider the specific process being used and clarifies that compression of a gas does not typically turn it into a liquid. They also explain that when a gas changes into a liquid through a decrease in temperature, the internal energy of the gas actually decreases.
  • #1
ryley
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I'm having trouble wrapping my head around some thermodynamics and ideal gas law concepts. I don't have a specific textbook question but Just a concept I'm having trouble with.

What I'm struggling with is understanding some of the relations between pressure, volume and temperature. Specifically, if a gas is compressed it heats up because the molecules are colliding more often causing an increase in internal energy. But if that process adds heat to the gas and increase in temperature how is it then that compression of a gas turns in into a liquid? Correct me if I'm wrong but if state change occurs from gas to liquid through a lowering of temperature, therefore the molecules come closer, which according the the ideal gas law will increase heat, how then does a liquid form if the internal energy is greater than before? Thanks for the help!

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  • #2
ryley said:
What I'm struggling with is understanding some of the relations between pressure, volume and temperature.

A common pitfall in thermodynamics is to fail to consider the specific process by which various changes are made. You seem to have done this, as will be noted below.

ryley said:
if a gas is compressed it heats up because the molecules are colliding more often causing an increase in internal energy

There is an increase in internal energy (equal to the work done on the gas by compression), but it's not best thought of as molecules "colliding more often". In fact, it's important to keep distinct the internal energy of the gas (which, as just noted, increases by the amount of compression work done) and the temperature of the gas. The latter is the average kinetic energy of the molecules of the gas; this increases during the compression process because that process pushes on the boundary of the gas (for example, a piston) and thereby transfers momentum and hence kinetic energy to the gas molecules. The internal energy of the gas increases because the average energy of the gas molecules increases, and the number of molecules stays the same, so internal energy = average energy per molecule times number of molecules must increase. So it isn't how often the molecules are colliding that has to increase; it's how much energy they have.

ryley said:
if that process adds heat to the gas and increase in temperature how is it then that compression of a gas turns in into a liquid?

Adiabatic compression of a gas, which is what you are describing (no heat transfer into or out of the gas, just work done on it) won't turn it into a liquid. (Technically, it's possible in principle that it could, depending on what the phase diagram of the gas looks like, but I think it's safe to say it's extremely unlikely, because the rise in temperature of the gas would be faster than the rise in its liquefaction temperature with pressure). It's true that if the pressure is higher, the temperature at which a gas will liquefy is higher, but that's not the same thing: the comparison is not between the same sample of gas before and after compression, but between two separate samples of gas that have different pressures but the same temperature (which would be above the liquefaction temperature at the lower pressure but below it at the higher pressure).

ryley said:
if state change occurs from gas to liquid through a lowering of temperature, therefore the molecules come closer, which according the the ideal gas law will increase heat, how then does a liquid form if the internal energy is greater than before?

Now you're talking about a different process, one in which a sample of gas has its temperature lowered while keeping its pressure constant. In such a process, the internal energy of the gas will decrease, not increase. (The temperature of the gas will remain constant while the phase is changing, but that's because the change in internal energy is not due to a change in the average kinetic energy of the molecules; it's due to the phase change, in which molecules are changing from freely moving states in the gas to states in which they are bound to neighboring molecules in the liquid.)
 
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  • #3
PeterDonis said:
A common pitfall in thermodynamics is to fail to consider the specific process by which various changes are made. You seem to have done this, as will be noted below.
There is an increase in internal energy (equal to the work done on the gas by compression), but it's not best thought of as molecules "colliding more often". In fact, it's important to keep distinct the internal energy of the gas (which, as just noted, increases by the amount of compression work done) and the temperature of the gas. The latter is the average kinetic energy of the molecules of the gas; this increases during the compression process because that process pushes on the boundary of the gas (for example, a piston) and thereby transfers momentum and hence kinetic energy to the gas molecules. The internal energy of the gas increases because the average energy of the gas molecules increases, and the number of molecules stays the same, so internal energy = average energy per molecule times number of molecules must increase. So it isn't how often the molecules are colliding that has to increase; it's how much energy they have.
Adiabatic compression of a gas, which is what you are describing (no heat transfer into or out of the gas, just work done on it) won't turn it into a liquid. (Technically, it's possible in principle that it could, depending on what the phase diagram of the gas looks like, but I think it's safe to say it's extremely unlikely, because the rise in temperature of the gas would be faster than the rise in its liquefaction temperature with pressure). It's true that if the pressure is higher, the temperature at which a gas will liquefy is higher, but that's not the same thing: the comparison is not between the same sample of gas before and after compression, but between two separate samples of gas that have different pressures but the same temperature (which would be above the liquefaction temperature at the lower pressure but below it at the higher pressure).
Now you're talking about a different process, one in which a sample of gas has its temperature lowered while keeping its pressure constant. In such a process, the internal energy of the gas will decrease, not increase. (The temperature of the gas will remain constant while the phase is changing, but that's because the change in internal energy is not due to a change in the average kinetic energy of the molecules; it's due to the phase change, in which molecules are changing from freely moving states in the gas to states in which they are bound to neighboring molecules in the liquid.)
Thank you very much I read you post and was re reading the textbook, I worded my question way too simply and that caused my understanding to get muddled. So again please correct me where I go wrong but my textbook says in these exact words that the internal energy of an ideal gas is a function gas temperature only; It does not depend on any other variable. So my understanding is then that the internal energy when increased because of heat Q (In this case I won't talk about work just heat) is just an increase in kinetic energy, not necessarily that the molecules will collide more, but that leaves me with some fuzziness around heat and temperature. Heat is then the total amount of energy (joules) either added or removed depending on the process and temperature is the average kinetic energy of the gas. So this is going to sound dumb but the total energy is all the kinetic energy of all the molecules added up, and the temperature is the average kinetic energy of the whole group? So an increase in temperature is an increase in the average kinetic energy which results in a greater total heat Q, because the range of speeds/ kinetic energy of the gas has increased?

Thanks!
 
  • #4
ryley said:
my textbook says in these exact words that the internal energy of an ideal gas is a function gas temperature only; It does not depend on any other variable

That's correct in the sense I explained, that the internal energy is just the average energy per molecule (the temperature) times the number of molecules (which is assumed to be constant). (There is also a qualification to this with regard to kinetic energy of the molecules being the only source of internal energy; I go into this further below.)

ryley said:
my understanding is then that the internal energy when increased because of heat Q (In this case I won't talk about work just heat)

You can't not talk about work if you are talking about a process in which no heat is transferred to and from the gas, but work is done on it. When you talk about compressing a gas and heating it up, that is what you are talking about: adiabatic compression, in which work is done on the gas, and the internal energy of the gas increases by the amount of work that is done. There is no heat transfer Q at all.

If you are talking about a process in which heat is added to the gas but no work is done, that's a different process than compressing the gas and heating it, and you would need to specify what that process is in order to say anything definite about it.

ryley said:
the total energy is all the kinetic energy of all the molecules added up, and the temperature is the average kinetic energy of the whole group?

Yes. At least, that's true on the assumption that there is no other source of internal energy, which is often a valid assumption but is not always the case. For example, if a chemical reaction is taking place, internal energy can change even if temperature does not change, so internal energy must have some component in such cases other than just adding up the kinetic energy of all the molecules.

ryley said:
an increase in temperature is an increase in the average kinetic energy which results in a greater total heat Q

A greater total internal energy (assuming the number of molecules is constant, as above). But that is not the same as heat Q; heat Q is heat transferred to or from the gas, not the internal energy of the gas. They're different things.
 
  • #5
Yes sorry I was meaning to say I just wanted to focus on one part of the equation not to disregard work. I follow most of what youre saying its just the final statement I'm still struggling with. In my textbook I see for the sections on molar specific heat at constant volume as well as molar specific heat for constant pressure a formula that has Q= n*C*delta T as the formula for both processes. The "C" is the constant depending on the process (constant volume or pressure). So my question is in both those equations temperature is affecting the final heat Q, so when you say an increase in temperature increases the total internal energy if the value of heat Q has deltaT in its calculation wouldn't it stand to reason that an increase in temperature increases the amount of energy transferred as heat, in turn increasing the total internal energy? Thanks again for the help.
 
  • #6
ryley said:
in both those equations temperature is affecting the final heat Q

No, the change in temperature is affecting the final heat Q. In other words, you are talking about a process where some amount of heat Q is being transferred, and nothing else is happening; in any such process, there will be a linear relationship between the heat transferred, Q, and the change in temperature, delta T. The specific constant in this linear relationship depends on what kind of process--whether it's at constant volume or constant pressure.

But none of this has anything to do with the temperature T itself; it has to do with the change in temperature. They're not the same thing. For example, suppose delta T is 10 degrees; that equates to the same total heat Q transferred whether the temperature T itself starts out at 10 degrees or 1000 degrees.
 
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  • #7
PeterDonis said:
No, the change in temperature is affecting the final heat Q. In other words, you are talking about a process where some amount of heat Q is being transferred, and nothing else is happening; in any such process, there will be a linear relationship between the heat transferred, Q, and the change in temperature, delta T. The specific constant in this linear relationship depends on what kind of process--whether it's at constant volume or constant pressure.

But none of this has anything to do with the temperature T itself; it has to do with the change in temperature. They're not the same thing. For example, suppose delta T is 10 degrees; that equates to the same total heat Q transferred whether the temperature T itself starts out at 10 degrees or 1000 degrees.
Oh okay I see that now, so yes if delta T went from Tinitial=10 to TFinal = 20, that is the same amount of heat transferred as Tinitial=50 to Tfinal=60. The amount they change by is the same. If you could answer one more question I would appreciate it. Its regarding expansion again, I can't wrap my head around how during a free expansion the temperature stays the same. If the volume of the gas expands freely with no opposing pressure my understanding then is that the temperature should decrease because the volume increased. But my textbook describes an example of a gas freely expanding and in order to calculate the change in entropy a reversible process must be calculated which will be equal to the irreversible change in entropy. The example states that a gas freely expands and that it is kept at a constant temperature, how is that possible to keep the temperature the same if the gas is expanding freely? I understand in free expansion change in internal energy is zero, but how does the expansion not change the temperature, or is it that you are only able to take measurements at equilibrium points because you can't calculate all the different things going on during the expansion?
 
  • #8
ryley said:
if delta T went from Tinitial=10 to TFinal = 20, that is the same amount of heat transferred as Tinitial=50 to Tfinal=60. The amount they change by is the same

Yes.

ryley said:
in order to calculate the change in entropy a reversible process must be calculated which will be equal to the irreversible change in entropy

I don't understand what you're saying here. The change in entropy of a reversible process is zero.

ryley said:
The example states that a gas freely expands and that it is kept at a constant temperature, how is that possible to keep the temperature the same

The example should say so: for example it might say the gas is in thermal contact with a heat bath. Or it might be relying on the reasoning given below.

ryley said:
I understand in free expansion change in internal energy is zero, but how does the expansion not change the temperature

If there is no heat transferred, and no work done (because the expansion is free), then there is no way for the internal energy to change. If the internal energy doesn't change, and the number of molecules remains the same, then the temperature must remain the same (since the temperature is just the average energy per molecule).
 
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  • #9
PeterDonis said:
Yes.
I don't understand what you're saying here. The change in entropy of a reversible process is zero.
The example should say so: for example it might say the gas is in thermal contact with a heat bath. Or it might be relying on the reasoning given below.
If there is no heat transferred, and no work done (because the expansion is free), then there is no way for the internal energy to change. If the internal energy doesn't change, and the number of molecules remains the same, then the temperature must remain the same (since the temperature is just the average energy per molecule).
Yes okay I see that now, I re read that part of the chapter, I'll keep plugging away with the entropy section but I think I'm getting it now, thanks for the help! Its been much appreciated! have yourself a good weekend.
 
  • #10
Compressing a vapor adiabatically and reversibly does not cause a vapor to condense to a liquid. In fact, it moves further away from the saturation line.
 
  • #11
PeterDonis said:
The change in entropy of a reversible process is zero.
This is not quite correct. The change in entropy for the combination of system and surroundings of a reversible process is zero. But each individually can experience a change in entropy (unless the reversible process is adiabatic).
 
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  • #12
ryley said:
thanks for the help! Its been much appreciated!

You're welcome!
 
  • #13
Chestermiller said:
Compressing a vapor adiabatically and reversibly does not cause a vapor to condense to a liquid. In fact, it moves further away from the saturation line.

I suspected this was the case. Is there a general proof for why this is always true?
 
  • #14
Chestermiller said:
unless the reversible process is adiabatic

That's actually the kind of process that was under discussion. I should have made clear that my specific statement only applied to that case.
 
  • #15
PeterDonis said:
I suspected this was the case. Is there a general proof for why this is always true?
Well, one can compare the slope of an adiabat at saturation with the slope of the Clausius-Clapeyron equation, and I think it would always lead to this result.
 

Related to Thermodynamics and ideal gas law concepts

1. What is the ideal gas law?

The ideal gas law is a mathematical equation that describes the relationship between the pressure, volume, temperature, and number of moles of an ideal gas. It is written as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature in Kelvin.

2. How does the ideal gas law relate to thermodynamics?

The ideal gas law is a fundamental concept in thermodynamics, which is the study of energy and its transformations. It helps to explain the behavior of gases under different conditions and is used to calculate properties such as work, heat, and internal energy.

3. What is the difference between an ideal gas and a real gas?

An ideal gas is a theoretical concept that follows the ideal gas law exactly, meaning that its particles have no volume and do not interact with each other. A real gas, on the other hand, does have volume and experiences intermolecular forces, so it may deviate from the ideal gas law at high pressures or low temperatures.

4. How does temperature affect the behavior of gases?

According to the ideal gas law, temperature and volume are directly proportional, while temperature and pressure are directly proportional. This means that as temperature increases, the volume and pressure of a gas will also increase. Additionally, temperature affects the average kinetic energy of gas particles, which determines their speed and collisions with the container walls.

5. Can the ideal gas law be applied to all gases?

The ideal gas law is a simplified model that is most accurate for low pressures and high temperatures. It is a good approximation for most gases, but it may not apply to gases with strong intermolecular forces or at extreme conditions. In these cases, more complex equations, such as the van der Waals equation, may be used.

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