Questions about thermodynamic relations

1. Jun 12, 2014

kelvin490

We know that for constant pressure thermodynamic processes which only expansion work is possible, dH=dQp. My question is, is it necessary that both work W and Qv be reversible to arrive this relation? What if the heat transfer is irreversible?

Similar question for dU=dQv, does the heat transfer need to be reversible?

i.e. Besides the work of the system do we also need to consider the reversibility of δQp and δQv in order to determine whether dU=δQv and dH=δQp ?

Another question is, do the above relations have anything to do with whether or not the system is an ideal gas? I have heard from a lecture that du≠dqv in Joule's free expansion for non-ideal gas. Consider reversibility and whether it's ideal gas there are four combinations of situations (ideal gas reversible,ideal gas irreversible, non-ideal gas reversible, non-ideal gas irreversible). I get confused with how these factors affect the thermodynamic relations.

Is it true that reversible process can only occur in the form of infinitesimal expansion/compression?

In addition, in a process is it possible that the work is irreversible but the heat transfer is reversible? Or the work is reversible but the heat transfer is irreversible?

2. Jun 12, 2014

Adithyan

Conservation of energy always holds irrespective of a process' reversibility.

du≠dqv holds in the case of non ideal gas because of intermolecular attractions. But conservation of energy still holds. The heat given to the system gets used up in internal energy, work done, energy lost by the molecules due to the collision with the walls of the beaker etc in a real life situation. Only problem we have is, since real gases don't obey PV=NRT the formulas for work done and internal energy become invalid.

Yes.

See, reversibility is a property of the process and not the energy exchanges in a particular process. So, if a process is carried out in a reversible manner, that means the work done, heat etc all will be reversible.

3. Jun 12, 2014

kelvin490

Does it mean that in order to get dU=δQv and dH=δQp, the only assumption is work can only be done in the form of expansion/compression so that it can be represented by dW=-PdV ?

Also, is it true that since both reversible and irreversible expansion work can be represented by dW=-PdV, dU=δQv and dH=δQp are also true for irreversible process?

4. Jun 12, 2014

Adithyan

Work is always represented by

dW= PressureextdV (Reversible and Irreversible)

But, when the process is reversible, dW = (Pgas$\pm$ dP )dV
Ignoring dPxdV (since it's too small), we get dW =PgasdV

So there is no "assumption" here.

The answer to your second question is 'yes'.

5. Jun 12, 2014

Andrew Mason

Careful there Adithyan. The first law always holds for any substance. So if no work is done, Q = ΔU for any substance. The existence of intermolecular forces simply means that the ΔU term includes molecular potential energy as well as molecular kinetic energy.

AM

6. Jun 12, 2014

Adithyan

You are right. Actually, I wanted to write something like this:
du≠dqv holds in the case of non ideal gas because of intermolecular attractions which causes inelastic collisions between the molecules and the walls of the container.

But then I thought I'll let the OP figure out why intermolecular attraction/repulsion is the reason behind du≠dqv.

7. Jun 12, 2014

Andrew Mason

I am still not sure why you are still saying du≠dqv. Net heat flow into a system has to be equal to the change in internal energy if no mechanical work is done, regardless of intermolecular forces. The presence of intermolecular forces just means that some of that internal energy change relates to a change in intermolecular potential energy as well as molecular kinetic energy.

Also, I don't quite understand what you mean when you say that intermolecular attractions cause inelastic collisions between molecules and the walls of the container. It seems to me that if a gas molecule collides with a wall molecule the kinetic energy of the gas molecule is converted to kinetic vibrational and rotational energy of the gas/wall molecules and may possibly result in some change in potential energy. Are you saying that the conversion of some kinetic to potential energy makes the collision inelastic?

AM

8. Jun 12, 2014

kelvin490

It seems that in some textbooks works in other forms are mentioned, e.g. electrical work, friction etc. That's why I suggested compression/expansion work is an assumption to get dU=dQv and dH=dQp. Do you mean all sorts of works can be expressed in the form of expansion work?

Sorry that I have wrongly written "du≠dqv in Joule's free expansion for non-ideal gas". It should be "dU≠dW in Joule's free expansion". It is from the lecture: , stated at 36:49.
What I would like to ask is why in this irreversible adiabatic process, dU≠dW? Is it because the W here doesn't include other sort of work?
Sorry again for the confusion.

Last edited by a moderator: Sep 25, 2014
9. Jun 12, 2014

Staff: Mentor

For more perspective on all this, check out my Physics Forums Blog. It addresses these issues in detail.

Chet

10. Jun 12, 2014

Adithyan

I was talking about a real life situation. Net heat flow gets used up in changing internal energy and other losses like inelastic collision with the wall. Real gas molecules don't collide elastically with the wall since kinetic theory doesn't apply here.

11. Jun 13, 2014

kelvin490

In this case and in the case of Joule's free expansion mentioned before, how the equation U=Q+W be modified (or re-define the terms) to better represent the first law?

12. Jun 13, 2014

Andrew Mason

No modification is needed. The first law always applies: ΔU = Q + W where W is the work done ON the gas. Since no work is done on the gas and no heat flows into the gas in a free expansion, ΔU = 0.

For a non-ideal gas in which there are intermolecular forces, the increase in volume necessarily means that molecular potential energy (PE) has increased. Since ΔU = 0, this increase in PE must be done at the expense of internal kinetic energy (KE) so the temperature of the gas must decrease. This free-expansion cooling is used in refrigeration.

AM

13. Jun 13, 2014

Andrew Mason

Why are molecular collisions - between gas molecules and wall molecules - inelastic? In inelastic molecular collisions, where does the initial kinetic energy go after collision?

AM

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