# Homework Help: DG is for isothermal thus dH =0?

1. Jul 4, 2011

### sparkle123

If U, H are functions of T only, then ΔU and ΔH should be zero for isothermal processes
ΔG and ΔA are only defined at constant T (thus define isothermal processes)
ΔG = ΔH – TΔS thus equals -TΔS always (since ΔH=0)?
this makes no sense since from previous calculations I have done in many problems ΔH has nonzero value
similarly, ΔA = ΔU - TΔS= -TΔS?

2. Jul 5, 2011

### danago

U and H are functions of only temperature if the gas is ideal (in which case ΔH=ΔU=0 for an isothermal process). For a non-ideal system, temperature is not the only relevant variable.

3. Jul 5, 2011

### Dickfore

The fundamental thermodynamic relation is:
$$dU = T \, dS - P \, dV$$
which would imply that the internal energy U is a function of entropy S and volume V as natural variables. The partial derivatives can be read off from this differential form:
$$T = \left(\frac{\partial U}{\partial S}\right)_{V}, \; P = -\left(\frac{\partial U}{\partial V}\right)_{S}$$
Because the mixed second partial derivatives have to be equal, we have the following identity:
$$\left(\frac{\partial T}{\partial V}\right)_{S} = -\left(\frac{\partial P}{\partial S}\right)_{V}$$
which is one of the Maxwell relations.

The other therodynamic potentials are Legendre transfroms:
$$\begin{array}{lcr} H = U + P \, V & dH = T \, dS + V \, dP & H = H(S, P) \\ A = U - T \, S & dA = -S \, dT - P \, dV & A = A(T, V) \\ G = U - T \, S + P \, V & dG = -S \, dT + V \, dP & G = G(T, P) \end{array}$$
You should be able to deduce the other three Maxwell relations from these expressions.

4. Jul 5, 2011

### sparkle123

Wow is this really it?
perhaps changes in moles of gas is also a factor?

5. Jul 5, 2011

### danago

If we are talking about the total enthalpy (i.e. in units of energy), then yes it will change with the number of moles of gas; however the specific enthalpy (i.e. energy/mole) will remain unchanged.