# DG is for isothermal thus dH =0?

• sparkle123
In summary: This is because enthalpy is an extensive property, meaning it depends on the amount of substance present. Therefore, if you increase the number of moles of gas, the enthalpy will also increase.
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?

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.

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.

danago said:
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?

sparkle123 said:
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.

## 1. What does it mean for DG to be isothermal?

DG stands for Gibbs free energy, which is a measure of the amount of energy available to do work in a thermodynamic system. When DG is isothermal, it means that there is no change in the temperature of the system, and therefore, no exchange of heat energy with the surroundings.

## 2. How does isothermal DG affect the enthalpy of a system?

Since enthalpy (H) is a measure of the total energy of a system, when DG is isothermal, it means that there is no change in the total energy of the system. Therefore, the enthalpy (H) remains constant, and dH = 0.

## 3. Is isothermal DG always equal to zero?

No, isothermal DG does not always equal zero. It only equals zero when there is no change in temperature (T) and pressure (P) of the system, and the process is reversible. In other cases, DG may have a non-zero value.

## 4. What is the significance of dH = 0 in an isothermal process?

dH = 0 in an isothermal process means that there is no change in the enthalpy of the system. This can be useful in determining the energy requirements for a process or reaction, as the enthalpy remains constant.

## 5. How is isothermal DG related to the spontaneity of a reaction?

When DG is isothermal, it means that the change in free energy is equal to zero. This can be used to determine the spontaneity of a reaction, as a negative DG indicates a spontaneous reaction, while a positive DG indicates a non-spontaneous reaction. Therefore, an isothermal process with dH = 0 can be used to determine the spontaneity of a reaction.

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