In the above, it's mentioned that any quantity other than MOLAR VOLUME can be written that way.

But how can we say generalize it with certainty when molar volume itself doesn't fit in or to say any property which might be having a relationship with Pressure (like entropy) can satisfy the above equation for ideal gas mixture.

I.E

$$V_i'(T,P) \neq V_i(T,P_i)$$

but

$$S_i'(T,P) = S_i(T,P_i)$$

where $P_i$ is partial pressure.

## Best Answer

## Foundations

Given a state property $P$, the definition of its partial molar value in a mixture at a given temperature $T$ and total pressure $p$ is below.

$$ \check{P}_i \equiv \left( \frac{\partial P}{\partial n_j} \right)_{T, p, n_j \neq n_i} $$

The definition measures the change in the total intensive property of the system when the number of moles of a component are varied holding all other number of moles constant.

An ideal solution is one where the particles have zero volume, no interactions, and collide elastically. An ideal gas, as derived through kinetic theory, is an example of an ideal solution.

The total molar property $\bar{P}$ of an ideal solution always depends on partial molar values and mole fractions $z_i$ identically as below.

$$ \bar{P} = \sum\ z_i \check{P}_i $$

## Partial Molar Volume

When a number of moles of only one component $i$ are added to an ideal solution as $\Delta n_i$, the total volume increases in proportion.

$$\Delta V = \frac{\Delta n_i R T}{p} $$

This is a direct consequence of the mechanical equation of state of the ideal mixture (the ideal gas law). Convert this to a partial derivative to obtain the following:

$$ \left(\frac{\partial V}{\partial n_i}\right)_{T, p, n_j \neq n_i} = \frac{\partial n_i}{\partial n_i} \frac{R T}{p} = \bar{V} $$

Recall that, in an ideal solution, the total molar volume $\bar{V}$ is the same as the molar volume of any component $\bar{V}_i$. Therefore, the partial molar volume of an ideal solution is the molar volume of the component at the given conditions of $T, p$ (total pressure).

$$ \check{V}_i(T, p) = \bar{V}_i(T, p)$$

This formulation stands for all cases of ideal solutions. It requires no theorem, rather it is a consequence of ideal behavior as stated by first principles in the ideal mechanical equation of state for the system.

## The Gibbs Theorem

The Gibbs theorem is a hypothetical framework. It is used to express the partial molar values of the thermodynamic state properties $U, H, S, A$, and $G$ for ideal mixtures. While we can obtain the total intensive values of a mixture using a mechanical equation of state, we have no rigorous way to define the partial molar properties without making an additional assumption. In other words, unlike the partial molar volume, the behavior of the partial molar value of any thermodynamic state function cannot be derived by first principles using only a mechanical equation of state for the total mixture. The Gibbs theorem is based on the assumptions that all particles in an ideal mixture are self-contained and that the intensive thermodynamic state property of any component $i$ in the mixture is defined solely by the system temperature $T$ and the partial pressure of the component $p_i$ in the mixture.

## Conclusion

The partial molar volume of a component $i$ in an ideal mixture is identically defined and its expression is directly derived when we make the statement that the system behaves ideally. No other assumptions or theorems are required. By comparison, the partial molar value of any thermodynamic state function for a component $i$ in a mixture cannot be defined solely by the mechanical equation of state. An additional statement is required, and that statement is the Gibbs theorem.

## Further Discussions

The remaining discussion here is left as a supplement. It is not needed to finalize the conclusive answer given above.

Two of the fundamental assumptions in the kinetic theory used to derive ideal gas behavior are: particles have zero volume and particles do not interact. The second statement is the more important one to determine ideal versus non-ideal behavior in thermodynamic functions. The first statement is the more important one to determine ideal versus non-ideal mixing behavior of volumes. Both must be true to apply the Gibbs theorem.

When particles interact, their thermodynamic energies in a mixture will depend on the composition of the mixture. When they do not interact, their thermodynamic energies in a mixture are the same as though they are at the $T, p$ that they would have to occupy the entire volume of the container entirely by themselves. By this statement, the particles in the ideal mixture have no identifiable partial volume. Instead they behave in energy functions as though they occupy the total volume of the container. That is permitted because all other particles have zero volume.

Note that we do not set the conditions at the total pressure. Even though the ideal particles occupy the total container, they contribute only their part of the collisions in the container. So, in a 50/50 mixture of an ideal gas, when we conceptually remove half of the gas particles, the remaining half still hit the walls of the container with the same frequency as they did before. They do not get to make up the loss of collisions on the wall from the missing particles. By kinetic theory, even when half the particles are removed, the remaining half still contribute only half to the pressure.

In a different view, consider that in any closed system, only two of the three parameters $T, p, V$ are independent. Allow that we choose $T$ as a constant. Subsequently we can pick constant $p$ and allow $V$ to be defined for us or we can pick constant $V$ and allow $p$ to be defined for us.

In summary, the Gibbs theorem is constructed on the basis of a defined $T, p$ state. It is a theorem only about thermodynamic (energy) functions in a mixture. At a defined $T, p$ state in a closed system, the

totalmolar energy functions $\bar{U}, \bar{H} \ldots$ for the mixture are defined (because they are state functions). The Gibbs theorem establishes how these total molar energy functions are distributed among the components. For an ideal mixture, they are distributed in accord with the mole fraction of the components. The theorem is not about molar volume because we cannot independently divide molar volume AND partial pressures in a closed system at a given $T, p$. We can divide one or the other.We could construct a hypothetical framework using partial volume at total pressure as the basis for a new theorem of partial molar energy functions. This choice of $T, V$ as our basis would not be the Gibbs theorem.

Perhaps a direct answer will serve as well. The Gibbs theorem does not state for partial molar volume that is does not equal molar volume.

$$\check{V}_j(T, p) \neq \bar{V}_j(T, p)$$

This is false on two accounts. First, for an ideal gas, the absolute truth is

$$\check{V}_i(T, p) \equiv \bar{V}_i(T, p)$$

Secondly, this is absolute truth not because we need a theorem. This is absolute truth because in a closed system only two of the three parameters $T, p, V$ are independent.