PHYSICAL CHEMISTRY IN BRIEF

CHAP. 2: STATE BEHAVIOUR [CONTENTS] 47

Solution

From (2.15) it follows thatVm=RT /pand

αp=

p

RT

R

p

=

1

T

2.1.7 Coefficient of isothermal compressibilityβT.

Definition^1

βT=−

1

Vm

(

∂Vm

∂p

)

T

=−

(

∂lnVm

∂p

)

T

. (2.10)

U Main unit: Pa−^1.

Example

Derive the relation for the isothermal compressibility coefficient from the equation of state of an

ideal gas (2.15).

Solution

From (2.15) we haveVm=RT /p, and

βT=−

p

RT

(

−

RT

p^2

)

=

1

p

2.1.8 Partial pressurepi

The partial pressurepiof componentiin a mixture of gases is defined by the relation

pi=xip , (2.11)

wherexiis the mole fraction of componentiandpis the pressure of a mixture of gases.
U Main unit:Pa.

(^1) Sometimes the symbolκTis used.