Page 54 - Modern physical chemistry

3.5 Gaseous Solutions 43

In practice, this limit is determined at standard temperatures for such gases as nitro-
gen, hydrogen, helium, and neon. It is found that

To =273.15 K [3.30]
and that
R = 0.0820578 I atm K- I mol- l [3.31 ]
or
R = 8.31451 J K- I mOrl. [3.32]

Example 3.4
Change the energy unit in R ( a) to the cubic centimeter bar, (b) to the calorie.
(a) Since the cubic centimeter bar equals 0.1 J, we have

R = 8.31451 cm3 bar K-I mor l = 83.1451 cm 3 bar K-I mol-I.
0.1
(b) A calorie is the energy needed to raise the temperature of 1 g water from 14S C
to 15.5° C. In conversions, it is taken as 4.18400 J. So from (3.32), we get

I
l
R = 8.31451 J K- mor = 1.98722 cal K-I mol-I.
4.18400 J cal- l
Example 3.5

Calculate the root-mean-square velocity of a helium gas molecule at 20° C.
Solve equation (3.13) for the mean-square velocity:

- 3PV
u 2 = __ .
Nm
Introduce formula (3.24) for PV and the molecular mass M for Nmln:

- 3RT
u 2 = __ .
M
Introduce the value of R in joules per degree per mole, of T in kelvins, and of M in kilo-
grams per mole:

- 3(8.31451 J K- mOI- )(293.15 K)
I
I
u 2 = = 1.827 x 10 6 m 2 S-2.
0.0040026 kg mol- l
Finally, take the square root
-1/2
l
3
2
(u ) = 1.35 x 10 ms- .
3.5 Gaseous Solutions
Two or more gases can be mixed in any proportions to form a uniform solution. In
the solution, the molecules generally interact as in a pure gas.