# Enthalpy change of an isothermal expansion at constant pressure.

1. Oct 17, 2011

### LogicX

For a process at constant pressure, ΔH=q.

My textbook clearly says that the only way that enthalpy can change is with a change in temperature. So ΔH=0. But q≠0 for an isothermal process.

I know that ΔU=0 for an isothermal process. So ΔH=0+Δ(PV)=Δ(nRT)=0

It really seems like ΔH should be zero for an isothermal process, but I can also make ΔH=q, which is not zero. What is the issue here?

2. Oct 17, 2011

### Bill_K

What kind of process can take place at both constant pressure and constant temperature? Answer: a first-order phase transition such as melting. In general you need a system which is heterogeneous, and a process that gradually converts one phase into the other. The enthalpy will be the sum of two parts, one for each phase.

Furthermore, enthalpy is extensive, i.e. proportional to the mass present. And it's really the enthalpy per unit mass which is constant. For a melting process, the enthalpy per mass h1 of the liquid is greater than the enthalpy per mass h2 of the solid. And so if you're converting one to the other the total enthalpy will change (= heat being added) even though h1 and h2 are constant.

3. Oct 17, 2011

### LogicX

Sorry I'm having a little trouble following you. So, it is possible for ΔH to not equal zero for an isothermal process?

If a question just asked you "what is ΔH for an isothermal expansion at constant pressure" what would you say? Are you saying that the overall total enthalpy ≠0? I'm just a little confused because in class we normally don't think of things in terms of actual processes that we could experimentally do, my professor just says "isothermal expansion at constant pressure" and expects us to be able to do the math. But I will give a couple examples that help illustrate my point:

1) A sample of 1mol of a perfect gas is expanded isothermally at 0°C from 22.4L to 44.8L. What is ΔH?

Textbook answer: It gives the proof that I did above and then says "Hence, ΔH=0 at constant temperature for all processes of a perfect gas."

2) 3 mol of a gas at 230K and 150kPa are compressed isothermally until the entropy decreases by 25J/K. Calculate the final pressure, and the Gibbs free energy change.

To calculate Gibbs, you need ΔH. Would ΔH be 0 in this case?

*begin AHA moment*
So, the reason ΔH is able to not be 0 for an isothermal process is because the ONLY way an isothermal process can be at constant pressure is for it to be a phase change, in which case Δ(nRT)≠0. So Δ(nRT)=-q? (Δn the change in mols of gas)?

VdP cannot equal 0 for an isothermal process of a gas, i.e. you can't expand a gas at constant temp and expect the pressure to not change?

I know I'm kind of rambling, let me know if I am not being clear about anything.

Last edited: Oct 17, 2011
4. Oct 17, 2011

### Andrew Mason

Yes, but not for a "perfect" or ideal gas.

You cannot expand an ideal gas isothermally and at constant pressure. That just follows from PV=nRT. If n, T and P are constant, V must be constant.

Since for an ideal gas, dH = dU + d(PV) = dU + d(nRT), and since dU = nCvdT, we have dH = n(Cv + R)dT = nCpdT. So if dT = 0, dH = 0.

Is this an ideal gas?

The "aha" is that it is not an ideal gas. Water, for example, undergoes a change in enthalpy when it condenses to liquid water at 100C and 1 atm pressure. This is an isothermal process. Water vapour is not an ideal gas.

AM

5. Oct 17, 2011

### LogicX

But if you did have an ideal gas condensing into a liquid, ΔH equals 0, right? Does q=-ΔnRT?

Sometimes I get confused in p chem by the differences in solving problems for ideal gases or non ideal gases. Like you said above, sometimes the reasoning for the answer is that it is not an ideal gas. But still, when I am given a question it seems like most of the time you could either say "perfect gas" or "water vapor" and the math that I would be required to use is still the same. ΔH still equals ΔU + Δ(PV) whether you specify this or not.

How would your answer change if I said no instead of yes? (My answer is yes, the problem specifies that it is a perfect gas)

6. Oct 17, 2011

### Andrew Mason

An ideal gas remains a gas down to absolute 0. It never condenses.

For an ideal gas, you can calculate the change in pressure from the change in volume. Change in volume is determined from the work done on the gas, which is determined from the heat flow (ie. apply the first law), which is determined from the change in entropy.

So, from the change in entropy, determine the heat flow out of the gas between the beginning and end states (ie. in a reversible process). Since it is isothermal, what is the relationship between the work done on the gas and the heat flow out of the gas in this process? Once you have calculated the work done on the gas, you can determine its change in volume from W = ∫PdV and P = nRT/V. Once you have the change in volume you can work out the change in pressure.

AM

7. Oct 17, 2011

### LogicX

Ok I think I get how to do that question.

A little more about this enthalpy thing though. So if someone said, calculate ΔH during the vaporization of water, would this be ambiguous? They would have to say, calculate ΔHvap of water, or calculate ΔH of the surroundings (of the heater). These two values would be nonzero, but would sum to 0? Since the heat provided by the heater would be the negative value of the heat absorbed by water.

8. Oct 17, 2011

### LogicX

Ok that makes sense, I was just confused because my professor made us a "cheat sheet" for the different types of changes and how to calculate different values.

There is section labeled:

Expansion against constant pressure:

Isothermal; w=-pdV, q=pdV, U=0, ΔT=0
Adiabatic; w=-pdV, q=0, U=-pdV, ΔT= -pΔV/Cv

Is this wrong or is he just referring to an irreversible expansion?

9. Oct 17, 2011

### Andrew Mason

Correct.

The measurements can be done at constant pressure and temperature using conventional calorimeters. ΔQ = ΔU + PΔV = ΔH - VΔP = ΔH (since there is no change in pressure). So you just measure the heat flow from the calorimeter into the water until it is all vapourized.

AM

10. Oct 17, 2011

### Andrew Mason

This must refer to a non-ideal gas eg. water vapour.
It is easier just to remember the first law: dQ = dU - dW where dW is the work done ON the gas = -PdV (or: dQ = dU + dW where dW = PdV, the work done BY the gas).

An adiabatic constant pressure change must involve a non-quasi-static process. So it is not reversible.

Adiabatic means dQ = 0 so dU - dW = 0 (dW = work done ON the gas); If external pressure is constant, the work done on the gas is -PextΔV. The change in internal energy is nCvΔT. This must be equal to the work done on the gas: -PextΔV, so ΔT = -PextΔV/nCv

AM

Last edited: Oct 17, 2011