# Proving RTln($a_i$) equal to non PV work done by system,where $a_i$ is activity

We know that chemical potential is defined as ${\mu}_i={\mu}_i^{standard}+RT\ln(a_i)$. Here $a_i$ is the activity of $i^\text{th}$ component of solution. In case of gases instead of $a_i$ it is $f_i$ (fugacity). Activity is the activity coefficient multiplied by concentration or in case of fugacity it is the product of the fugacity coefficient multiplied by partial pressure of the gas.

But now this is not the only definition of Chemical potential. Chemical potential also means the partial molar Gibbs free energy when pressure and temperature of the system are constant. We know that change in Gibbs free energy of a system at constant temperature and pressure also represents Non-PV work done by system.

How can the term $RTln(a_i)$ can be correlated to Non-PV work done by system in order for the two definitions to be equivalent?

Philipp, Ok after reading all your answers I got now modified question that still remained.First of all we know K and Kc are related but kc can be modified by adding activity coefficients and standard concentrations to give equilibrium constant that should be equal to K.This requires the activity coefficients (That we used with Kc) to always follow relation with fugacity coefficients and RT terms which is not proved in answers there and that is what actually my question is asking.Sorry but I am reluctant here to write maths but I just say refer to the second equation from last in your This Answer

Actually earlier my question has title ways of expressing concentration in activity term

• No one even trying to give answers,I am disappointed.Just try to join the discussion – Vishvajeet Patil Apr 22 '14 at 14:36
• I'm not sure if I understand you correctly. Are you asking why I use fugacity coefficients rather than activity coefficients? – Philipp Apr 23 '14 at 9:13
• @Philipp No I understand that using fugacity for gases we can get it's relation with non-pv work but not for solutions where we use concentrations and there is no partial pressure to use. – Vishvajeet Patil Apr 23 '14 at 9:41
• Ah, ok. There is actually the concept of a partial pressure in solutions: the vapor pressure. In equilibrium the vapor pressure of the component in solution is equal to the partial pressure of the same component in the gas phase directly above the solution. And there are Raoult's Law and Henry's Law linking the pressure to the mole fraction/concentration in the solution. But there is some uncertainty in my mind concerning the relation between fugacity coefficients and activity coefficients... – Philipp Apr 23 '14 at 10:56
• ... Maybe this Wikipedia article about activity is of interest to you. – Philipp Apr 23 '14 at 10:57

We know that chemical potential is defined as ${\mu}_i={\mu}_i^{standard}+RT\ln(a_i)$.

Be a bit careful with your wording. This equation is not part of the definition of the chemical potential but rather it is derived from it. The derivation of this equation can be found in this answer of mine. The definition of the chemical potential is

$$\mu_{i} = \left( \frac{\partial G}{\partial n_{i}} \right)_{p,T} \ ,$$

where $n_{i}$ is the amount of the $i^{\text{th}}$ component in the system.

We know that Gibbs free energy of a system at constant temperature and pressure also represents Non-PV work done by system.

Again this is not quite correct: It is the change in Gibbs free energy of a system during some process, say a chemical reaction or a phase transition, at constant temperature and pressure that can be identified with non-PV work done by the system.

Now considering a general chemical reaction

$$\ce{\nu_{1}A + \nu_{2}B + ... + \nu_{x}C <=> \nu_{x+1}D + \nu_{x+2}E + ... +\nu_{n}F}$$

the differential change in Gibbs free energy is given by

$$\text{d} G = -S \text{d} T + V \text{d} p + \sum_{i} \mu_{i} \text{d} n_{i}$$

or introducing the extend of reaction, $\text{d} \xi = \frac{\text{d} n_{i}}{\nu_{i}}$

$$\text{d} G = -S \text{d} T + V \text{d} p + \sum_{i} \nu_{i} \mu_{i} \text{d} \xi \ .$$

From this it follows that the amount of Gibbs free energy that is produced per formula conversion (meaning that it is defined for an extend of reaction $\xi = 1 \, \text{mol}$) at constant pressure and temperature is given by

$$\Delta G = \left( \frac{\partial G}{\partial \xi} \right)_{p,T} = \sum \nu_{i} \mu_{i} \ .$$

Here you can see that the non-PV work done by the system originates from the differences in the chemical potentials of the reactants. Substituting the equation for the chemical potential

$$\mu_{i} = \mu_{i}^{0} + RT \ln a_{i}$$

the equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, leads to

$$\Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ ,$$

where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced. This is the relation between the non-PV work, represented by $\Delta G$, and the activity terms. Admittedly, it is not a very graphic explanation but I fear that it simply is a pretty mathematical matter.