PHYSICAL CHEMISTRY IN BRIEF

CHAP. 3: FUNDAMENTALS OF THERMODYNAMICS [CONTENTS] 69

Solution

It follows from the definition (3.9) that

∆H= ∆U+ ∆(pV),

where∆(pV) =p 2 V 2 −p 1 V 1. For an ideal gaspV =nRT. Then

∆H= ∆U+nR∆T= 800 + 5× 8. 314 ×(400−300) = 4957 J.

3.2.2 Helmholtz energy

The Helmholtz energyFis a function of state defined by the relation

F=U−T S. (3.12)

U Main unit:J.

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

energy.

The change in the Helmholtz energy ∆Fduring a reversible isothermal process is equal to

the work supplied to the system

∆F=W , [T, reversible process]. (3.13)

Example

During a certain isothermal process, internal energy changed by∆Uand entropy by∆S. Derive

the relation for the change in the Helmholtz energy. Is it possible to calculate the change in the

Helmholtz energy during a non-isothermal process if in addition to∆U and∆Swe also know

the initial and final temperatures?