# Enthalpy and isothermal, where is my logic wrong

1. Apr 9, 2010

### EvilKermit

Assume ideal gas and isothermal :

$$\Delta H = \int_{T_{i}}^{T_{f}}\! C_{p} dT = 0$$
(no change in temperature, no change in enthalpy)

$$\Delta H = \Delta U + W, U = 0$$

There is no change in internal energy but there is change in work done. How do these two contradicting statements work.

Last edited: Apr 9, 2010
2. Apr 9, 2010

### SpectraCat

What makes you think that?

3. Apr 9, 2010

### nonequilibrium

Well there is no change in internal energy because you're talking about an ideal gas.

But I don't get how "isothermal => no change in enthalpy" follows... Your integral makes no sense. There is no dT. And you can only use Cp if you're dealing with constant pressure.

4. Apr 9, 2010

### SpectraCat

What do you mean? The internal energy of an ideal gas is inversely proportional to the volume. So, if there is an expansion (which I am guessing there is since he is talking about "work done"), then the internal energy must be changing, right?

5. Apr 9, 2010

### kanato

H = U + PV is the definition of enthalpy.

A monoatomic ideal gas has U = 3/2 NkT and PV = NkT, so H = 5/2 NkT and you are right that the enthalpy does not change when pressure is applied at constant temperature.

The differential of H is $$dH = TdS + V dP$$.
To get your integral equation for delta H you need to divide by dT at constant P, but you are talking about a situation where T is held constant, not P. So your integral is wrong, as it can only be used at constant pressure.

Since we know that dH is zero for an ideal gas at constant temperature, then we must have the relation $$TdS = -VdP$$ where the left hand side is the differential heat added to the system. If dP is positive so pressure is increasing, then the differential heat is negative, which means that as work is done on the system, heat must be removed in order for it to remain at constant temperature. If heat does not get removed, then you will not have an isothermal process (instead it will be adiabatic).

Last edited: Apr 9, 2010
6. Apr 9, 2010

### EvilKermit

Alright so

$$dU = TdS - pdV$$

$$dH = TdS + VdP$$

Set dU and dH to 0. Then

$$-pdV = Vdp$$

This correct?

Also, if there is no change in enthalpy for a valve, does that mean that valves also work as an isothermal process for an ideal gas?

Last edited: Apr 9, 2010
7. Apr 9, 2010

### nonequilibrium

Yes, which means PV is constant, which you could've gotten out of the gas-law, as PV = nRT = constant

8. Apr 9, 2010

### kanato

I'm not really sure what you mean but the answer is probably no. Simply having no change in enthalpy means there is some specific balance between the heat and work. At constant enthalpy, you can say

$$T \left( \frac{\partial S}{\partial P} \right)_H = -V$$

which will depend on the details of the system being studied, and will definitely not be ideal gas like unless you really have something which is an ideal gas.