Question

ChemActivity T10 Gibbs Energy as a Function of Temperature and Pressure Focus Question: The most stable phase of a substance at a fixed T,P is that phase with the lowest molar Gibbs energy. If temperature is increased what happens to the Gibbs energy? What happens to the Gibbs energy if pressure is increased? Model 1: The Gibbs Energy. G=U + PV-TS Critical Thinking Questions 1. Describe the meaning of equation 1 using grammatically correct English sentences. 2. Based on equation 1 and the First Law, derive equation 2 below for a reversible process in which the only work done is pressure-volume work: dG = VdP - SAT (2) Clearly state the assumptions and conditions which are necessary for this derivation. 119

Answer 1

Answer to the focus qustion:

As the physical quantity G is a function of temperature and pressure, either of them being changed, changes the whole Gibbs free energy magniude along with its sign. Now we know that dG = Vdp -SdT, so if we increase the temperature keeping the pressure to be constant, the expression reduces to dG = -SdT, which suggests the Gibbs free energy also decreases only if S is positive. If S is negative, then with increase in temperature dG becomes more and more positive and hence the reaction becomes more non-spontaneous.

If pressure is increased but temperature is kept constant, then our expression reduces to dG = Vdp, where p=partial pressure of the component in the system. With increase in pressure dG increases and becomes more and more positive, hence deviates from spontaneity.

1. G = U + PV - TS

This expression is a measure of the fundamental quantity, the Gibb's free energy. U= internal energy; P = pressure of the system; V = volume of the system; T= temperature of the system; S = entropy of the system. Al these variables point to a particular temperature and pressure pair. The quantities like S, U, V depent on this pair of parameters only, i.e., T and P.

We know H = U + PV. So we can say that G is equal to the enthalpy of the system( heat released or taken up by the sysem to undergo a reaction) which is completely free of the entropy and the temperature of the system.

2. From the first law, we know U =Q + W

=> dU = dQ + dW (assuming reversible work done by the system, so we te infinitesimal small changes in each quantites)

Again dQ = SdT, using this we get,

dU = SdT - PdV, as dW = -PdV.

Putting this above equation in

G= U +PV -ST,

we get,

dG = dU + d(PV) -d(ST)

=> dG = SdT - PdV + (PdV + VdP) - (SdT +TdS)

=> dG = VdP - SdT

At constant temperature, we obtain Gibb's free energy is equal to the change in work due to volume and pressure, i.e., dG​​​​​T = VdP.