# Isobaric Work and Charles Law

1. Feb 5, 2016

### UMath1

I understand if isobaric work is done on a system then its internal energy must increase, meaning that the temperature must increase as well. However, according to Charle's Law, volume is proportional to temperature which means that if volume decreases, so must temperature. How can the two happen at the same time? There seems to be a contradiction. Doing isobaric work involves decreasing volume leading to increased internal energy and temperature. But according to Charles' Law, the temperature must decrease in this situation.

2. Feb 5, 2016

### Aniruddha@94

Doing work on a system in isobaric conditions means the volume decreases, that's true.. However that doesn't necessarily mean that the internal energy increases, you also have to think about the heat transfer..
Since the volume is decreasing, so must the temperature ( from Charles' law) .. That means (delta)T is -ve, and since Q=nCp*(delta)T , that means Q is also -ve ( i.e, the system looses heat)..
So it's not certain that the internal energy increases, that depends upon the value of work done-Q

3. Feb 5, 2016

### Aniruddha@94

Actually upon further calculation I can say that in the conditions you mentioned ( isobaric and work on the system), the internal energy can never increase. That would happen only if Cp<R, and we know that's not true.

4. Feb 5, 2016

### jbriggs444

If you squeeze a volume of gas into a smaller space and you wish to keep that gas at a constant pressure while doing so then Charles Law says that you had better figure out a way to reduce its temperature.

It does not say that the temperature will magically reduce just because you muttered the word "isobaric".

5. Feb 5, 2016

### Staff: Mentor

When we use the term "isobaric work" for an irreversible adiabatic compression process, what we really mean is that the external force per unit area being applied at the piston face is being held constant during the compression. This doesn't mean that the average pressure of the gas is constant during the compression, because, in an irreversible process, the pressure within the cylinder is not uniform spatially. If we look at the final state of the gas after the compression has been completed (and equilibrium has been attained), the pressure of the gas will be higher than the initial pressure, as will the temperature. In short, we can't apply Charles' Law to a system that is not in equilibrium.

6. Feb 5, 2016

### UMath1

So if we want to increase the internal energy of gas adiabatically, work must be done such that the pressure increase is more than the decrease in volume. In other words P1V1 < P2V2 but V1>V2, so that the temperature increases?

Also, for a truly isobaric process to occur energy must be removed from the system as heat in order maintain a constant pressure correct? So does this mean an isobaric process is also isothermal?

7. Feb 5, 2016

### Staff: Mentor

I guess you could look at it that way.
No. If you are doing it reversibly (by transitioning through a continuous sequence of equilibrium states) and you want the pressure to remain constant, you need to be reducing the temperature (in proportion to the volume). So you need to remove more heat than to just keep it isothermal.

8. Feb 6, 2016

### Aniruddha@94

Is my conclusion ( that internal energy can never increase in an isobaric compression) wrong? I know that Charles' law doesn't hold for non equilibrium states but don't we assume that the real processes are done in small infinitesimal steps, each one at equilibrium?

9. Feb 6, 2016

### Staff: Mentor

No. If it is done adiabatically, the first law tells us that internal energy increases.
Absolutely not. What you describe here is what we refer to as reversible processes. For an irreversible processes, the system does not pass through a continuous sequence of thermodynamic equilibrium states.

Chet