Physical Chemistry Third Edition

6.1 Ideal Solutions 239

quantity with the value ofRTln(P∗ixi/P◦), assuming thatxi 0 .500.

Vm∗

46 .069 g mol−^1

0 .7885 g cm−^3

 58 .437 cm^3 mol−^1

∆μVm1∗(P 1 ∗−P)

(58. 437 × 10 −^6 m^3 mol−^1 )(760 torr− 40 .0 torr)

(

101325 Pa

760 torr

)

 5 .609 J mol−^1

RTln

(

P∗ixi

P◦

)

(8.3145 J K−^1 mol−^1 )(292.15 K) ln

(

(40.0 torr)(0.500)

750 torr

)

−8804 J mol−^1

where we have used the fact that 1.000 bar is equal to 750 torr.∆μis 0.064% as large as

RTln(Pi∗xi/P◦).

At equilibrium the chemical potential of pure componentiis equal to the chemical

potential of gaseous componentiat pressureP∗i, so that

μ∗i(T,P)≈μ∗i(T,Pi∗)μ◦i(g)+RTln

(

P∗i

P◦

)

(6.1-4)

When Eq. (6.1-4) is substituted into Eq. (6.1-3), we obtain

μ

◦(g)

i +RTln(P

∗

i/P

◦)+RTln(xi)μ◦

i(g)+RTln(Pi/P

◦) (6.1-5)

After canceling and combining terms,

RTln(Pi∗xi/P◦)RTln(Pi/P◦) (6.1-6)

We divide byRTand take antilogarithms:

PiPi∗xi (6.1-7)

which is Raoult’s law, Eq. (6.1-2). An ideal solution is sometimes defined as a solution

in which every substance in the solution obeys Raoult’s law for all compositions. If

this definition is taken, it can be shown that Eq. (6.1-1) follows as a consequence.

Exercise 6.1

Assuming that Raoult’s law holds for componentifor all compositions of an ideal solution, show

that the chemical potential of this component is given by Equation (6.1-1).