PHYSICAL CHEMISTRY IN BRIEF

CHAP. 3: FUNDAMENTALS OF THERMODYNAMICS [CONTENTS] 70

Solution

For an isothermal process we obtain from (3.12)

∆F= ∆U−∆(T S) = ∆U−T∆S, [T].

For a non-isothermal process we have

∆F= ∆U−∆(T S) = ∆U−T 2 S 2 +T 1 S 1.

Since the values of entropy in the initial and final states,S 1 , S 2 , are not specified and we only

know that∆S=S 2 −S 1 , the change in the Helmholtz energy cannot be calculated.

3.2.3 Gibbs energy

The Gibbs energy (or the Gibbs function)Gis a function of state defined by the relation

G=H−T S. (3.14)

U Main unit:J.

Note:The Gibbs energy is defined up to the additive constant, just like internal energy.

Example

During a certain thermodynamic process, a system passed from its initial state defined by the

values of volumeV 1 and pressurep 1 to its final state defined by the valuesp 2 andV 2. The change

in the Helmholtz energy was∆F. Calculate the change in the Gibbs energy.

Solution

From the definitions (3.9), (3.12) and (3.14) forH,F, andGwe obtain

G=H−T S=U+pV−T S=F+pV ,

from which it follows:

∆G= ∆F+ ∆(pV) = ∆F+p 2 V 2 −p 1 V 1.