DG is for isothermal thus dH =0?

[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?
Please help clear my confusion, thanks.

 

[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.

 

[Dickfore]

The fundamental thermodynamic relation is:
<br /> dU = T \, dS - P \, dV<br />
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:
<br /> T = \left(\frac{\partial U}{\partial S}\right)_{V}, \; P = -\left(\frac{\partial U}{\partial V}\right)_{S}<br />
Because the mixed second partial derivatives have to be equal, we have the following identity:
<br /> \left(\frac{\partial T}{\partial V}\right)_{S} = -\left(\frac{\partial P}{\partial S}\right)_{V}<br />
which is one of the Maxwell relations.

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

 

[sparkle123]

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.

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

 

[danago]

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

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.