Page 89 - Physical chemistry understanding our chemical world

56 INTRODUCING INTERACTIONS AND BONDS

Table 2.4 Van der Waals constants for various gases
−3 −1
−3 −2
Gas a/(mol dm ) bar b/(mol dm )
Monatomic gases
Helium 0.034 589 0.023 733
Neon 0.216 66 0.017 383
Argon 1.3483 0.031 830
Krypton 2.2836 0.038 650
Diatomic gases
Hydrogen 0.246 46 0.026 665
Nitrogen 1.3661 0.038 577
Oxygen 1.3820 0.031 860
Carbon monoxide 1.4734 0.039 523
Polyatomic gases
Ammonia 4.3044 0.037 847
Methane 2.3026 0.043 067
Ethane 5.5818 0.065 144
Propane 9.3919 0.090 494
Butane 13.888 0.116 41
Benzene 18.876 0.119 74

Worked Example 2.1 0.04 mol of methane gas is enclosed within
The value of p calcu- a flask of volume 0.25 dm . The temperature is 0 C. From Table 2.4,
3
◦
lated with the ideal-gas 6 −2 3 −1
a = 2.3026 dm bar mol and b = 0.043 067 dm mol . What is the
equation is 3.63 × pressure p exerted?
5
10 Pa, or 3.63 bar.
We first rearrange Equation (2.2), starting by dividing both sides by
the term (V − nb), to yield
2
n a nRT
p + =
V 2 (V − nb)
2
2
We then subtract (n a) ÷ V from both sides:
nRT n 2
p = − a
(V − nb) V
Calculations with the ◦
Next, we insert values and convert to SI units, i.e. 0 C is expressed
van der Waals equation as 273.15 K.
are complicated be-
3
cause of the need to 0.04 mol×(8.314×10 −2 dm bar K −1 mol )×273.15 K
−1
convert the units to p = 3 3 −1
accommodate the SI (0.25 dm − 0.04 mol×0.043 067 dm mol )
system. The value of R 0.04 mol 2
−1 2
3
comes from Table 2.3. − 3 × 2.3026 (dm mol ) bar
0.25 dm