Page 91 - Physical chemistry understanding our chemical world

58 INTRODUCING INTERACTIONS AND BONDS

Inserting numbers (and taking care how we cite the value of R) yields

3
83.145 cm bar mol −1 K −1 × 273 K
V m =
1bar
× (1 − 4.626 × 10 −2 bar −1 × 1bar)
The value of V m cal-
culated with the ideal- V m = 22 697 × 0.954 cm 3
gas equation (Equation 3
(1.13)) is 4.4 per cent V m = 21 647 cm
higher. 1cm represents a volume of 1 × 10 −6 m , so expressing this value
3
3
3
of V m in SI units yields 21.6 × 10 −3 m .

SAQ 2.3 Calculate the temperature at which the molar volume of oxygen
3
is 24 dm . [Hint: you will need some of the data from Worked Example 2.2.
Assume that B has not changed, and be careful with the units, i.e.

3
V m = 24 000 cm .]
The relationship bet- An alternative form of the virial equation is expressed in terms

ween B and B is B = of molar volume V m rather than pressure:
(B ÷ RT).
B C

PV m = RT 1 + + + ... (2.4)
A positive virial coef-
V m V m
ficient indicates repul-
sive interactions Note that the constants in Equation (2.4) are distinguishable from
between the particles. those in Equation (2.3) because they lack the prime symbol. For
The magnitude of B both Equations (2.3) and (2.4), the terms in brackets represents
indicates the strength the molar compressibility Z. Table 2.5 lists a few virial coeffi-
of these interaction. cients.

SAQ 2.4 Calculate the pressure of 1 mol of gaseous argon housed within
3
2.3dm 3 at 600 K. Take B = 11.9cm mol −1 , and ignore the third virial
term, C . [Hint: take care with all units; e.g. remember to convert the

3
volume to m .]
Table 2.5 Virial coefficients B for real gases as a function
3
of temperature, and expressed in units of cm mol −1
Gas 100 K 273 K 373 K
Argon −187.0 −21.7 −4.2
Hydrogen −2.0 13.7 15.6
Helium 11.4 12.0 11.3
Nitrogen −160.0 −10.5 6.2
Neon −6.0 10.4 12.3
Oxygen −197.5 −22.0 −3.7