# I thought enthalpy meant total internal energy?

1. Feb 2, 2014

### kq6up

Then I see this line in a Wiki article:

For an ideal gas, the heat capacity is constant with temperature. Accordingly we can express the enthalpy as H = n C_P T and the internal energy as U = n C_V T, where n is the amount of substance in moles. Thus, it can also be said that the heat capacity ratio is the ratio between the enthalpy to the internal energy:
\gamma = \frac{H}{U}

Is this statement false? If not, some clarification would be greatly appreciated.

Thanks,
Chris Maness

2. Feb 2, 2014

### Andrew Mason

I am not sure where you got the impression that H and U were the same. Enthalpy, H, is a state variable, like U, but it is defined as: H = U + PV. If P is constant, then ΔH = ΔU + PΔV, in which case ΔH = Q i.e. the total heat flow during the process (first law: Q = ΔU + W where W is the work done by the gas).

AM

3. Feb 3, 2014

### Sunfire

This would be true:

$\gamma= \frac{H_s}{U}$,

where $H_s$ is the static enthalpy of the gas. If the gas is in motion with velocity v, then the total enthalpy of the gas is

$H=H_s+\frac{v^2}{2}=c_pT$, where T is the total temperature of the gas; while the static enthalpy is
$H_s=c_pT_s$. Since $U=c_vT_s$, one can write $\gamma= \frac{H_s}{U}$.

But $\gamma= \frac{H}{U}$, is wrong , if H denotes the total enthalpy. Total enthalpy contains both static enthalpy and the kinetic energy of the gas.

4. Feb 3, 2014

### Staff: Mentor

Actually, for an ideal gas, the heat capacity is not constant with temperature. I think you are thinking of what physicists call a perfect gas, rather than an ideal gas. At low pressures where real gases approach ideal gas behavior, the heat capacity does vary with temperature.

For ideal gases, the enthalpy H and the internal energy U are independent of pressure, but are functions of temperature, with

dH = Cp(T)dT

dU = Cv(T)dT

So, when you integrate to get H and U, you need to take into account the temperature dependence of the heat capacities. However, you can write:
$$\frac{dH}{dU}=γ(T)$$
Hope this helps.

5. Feb 3, 2014

### kq6up

I was deriving P/V and T/V relationships for an adiabatic process for a refresher. I was successful, but I assumed Cp=Cv+R. I would like to derive that too, but there was some holes in my understanding (obviously). When I get home, I will take a closer look so that I can glean more insight. Then take a shot at Cp=Cv+R. I would imagine these "constants" are approximately constant over a range. At least for understanding a non real gas.

Thanks,
Chris Maness

6. Feb 3, 2014

### Staff: Mentor

Try this:

dH = CpdT=dU+d(PV)=CvdT+d(RT)

This only applies to an ideal gas.

Chet

7. Feb 4, 2014

### kq6up

I guess the whole time I thought U contained all forms of potential and kinetic energy. I didn't know it excluded this term PdeltaV. I guess it is useful in chemistry where a reaction can/cannot outgas and expand into free space. The change in energy from the gas leaving the system would no longer be internal. I am thinking you would no longer consider the gas part of the system, but part of the environment.

Did I get the touchy feely explanation bit down? Or did I bodge it?

Thanks,
Chris Maness

8. Feb 4, 2014

### Staff: Mentor

I can't really follow what you are saying. The form of the changes in internal energy and enthalpy I wrote down for an ideal gas apply to a closed system (no exchange of mass with the surroundings).