EMPIRICAL EQUATIONS
2.7 DETERMINING THE MOLAR MASS OF A NONIDEAL GAS
A nonideal gas, even if it is pure, will generally not occupy a molar volume of 24.789 dm3 at p=1.000 bar and T =298.15 K; hence any molar mass computed on the basis of this molar volume will be wrong. Depending on the temperature, the error may be 50% or more. If, however, the weight and the volume of a pure gas sample are known along with the pressure and temperature, the (incorrect) molar mass, often called the effective molecular weight (EMW), can be calculated from the ideal gas law. Historically, EMWs were measured at several different pressures, treating a real gas as though it were ideal. The EMWs were then extrapolated as a function of p to p=0 to obtain the true molecular weight. It is fruitless to ask about the meaning of the properties of a gas at zero pressure; one has simply “extrapolated out” the error due to nonideality. Extrapolating out is a common applied mathematical device. A number of more accurate methods for determining molar mass now exist.
PROBLEMS AND EXERCISES Exercise 2.1 The van der Waals Cubic Show that the van der Waals equation is a cubic.
PROBLEMS AND EXERCISES 29 Solution 2.1 Expanding the van der Waals equation and collecting terms, we get
p+ a
V2
(V−b)= RT pV −pb+ a
V −ba V2 = RT pV3−pbV2+aV−ba= RT V2 pV3−( pb+RT) V2+aV−ba=0
Comment: Nonideal Behavior—Another View We know that molecules are not dimensionless points. Figure 2.8 shows that Boyle’s law, the lower curve, is not obeyed above about 200 bars pressure for nitrogen. Positive deviation from Boyle’s law is a high-pressure phenomenon shown by all real gases because all real molecules occupy a nonzero volume. The real sample volume is larger than it “ought to be” on the basis of Boyle’s law. Molecules, which may be thought of roughly as hard spheres, are bumping into each other and refusing to invade each other’s space. The volume in which the particles can move is the total volume minus the volume actually taken up by the particles. As higher pressures are imposed, the particle volume becomes a larger proportion of the total volume; hence the deviation from the Boyle’s law curve is larger. The aggregate of molecular volume, as distinct from total volume, for all molecules in the sample is the excluded volume.
V (dm3)
0.5 0.4
0.3 0.2
0.1 0.0
p (bar)
0 100 200 300 400 500 600
T = 273.15 K
FIGURE 2.8 Boyle’s law plot for an ideal gas (lower curve) and for nitrogen (upper curve).
TABLE 2.1 Observed Real Gas Behavior from 10 to 100 bar Expressed as (p, pVm).
p (bar) pVm(dm3bar)
10.0000 24.6940
20.0000 24.6100
30.0000 24.5400
40.0000 24.4820
50.0000 24.4380
60.0000 24.4070
70.0000 24.3880
80.0000 24.3830
90.0000 24.3910
100.0000 24.4120
Exercise 2.2
Given the problem that experimental values of pVmbetween 10 and 100 bar have been measured at intervals of 10 bar with the results in Table 2.1, find the analytical equation that expresses these results (Fig. 2.9).
p (bar)
100 80
60 40
20 0
pV (dm3 bar)
24.3 24.4 24.5 24.6 24.7 24.8
FIGURE 2.9 Experimental values of pVmvs. p for one mole of a real gas.
PROBLEMS AND EXERCISES 31 TABLE 2.2 Observed Real Gas Behavior
Expressed as (p, pVm).
p (bar) pVm (dm3bar)
10.0000 24.6940
20.0000 24.6100
30.0000 24.5400
40.0000 24.4820
50.0000 24.4380
60.0000 24.4070
70.0000 24.3980
80.0000 24.3990
90.0000 24.4180
100.0000 24.4600
Solution 2.2 This problem and its solution are expressed using SigmaPlot 11.0C
plotting software. Load the data set in the form of Table 2.2. Click on statistics→ nonlinear regression→regression wizard→quadratic,→next, specify columns as x and y variables, and click finish.
The graph in Fig. 2.10 is shown with its fitted quadratic curve. The curve param- eters are, as they should be, the parameters we started with in the previous problem:
y0 24.7906
a −0.0103
b 6.5341E-005
The parameters are expressed in the form y=y0+ax+bx2+ cx3+ · · ·. Please notice the slight change in notation: The y intercept is called y0, the slope is a, and the quadratic coefficient is b. The general independent variable is x, which is the pressure p in our case. Translated to the terms of the problem, we have
pV = RT+B [T ] p+C [T ] p2+D [T ] p3+ · · ·
=24.7906−0.0103 p+6.5341×10−5p2 where we have truncated the series at the quadratic term.
These two exercises make the important point that any data set can be expressed as a collection of numbers (table of observations), a graph, or an analytical equation.
Tables, graphs, and empirical equations are merely different ways of saying the same thing. Difficulties may be encountered (imaginaries, singular points, multiple real roots, discontinuities, etc.), but they often point to interesting phenomena and their explanation may lead to new concepts in science. Larger data sets and more complicated functional behavior can be treated in the same way as this quadratic, except that the curve fit may be cubic, quartic, and so on.
p, bar
100 80
60 40
20 0
pV , dm barm3
24.3 24.4 24.5 24.6 24.7 24.8
FIGURE 2.10 Quadratic real gas behavior.
Problem 2.1
The van der Waals constants for n-octane (a component of gasoline) are a =37.81 and b=0.2368. Find V for 1.00 mol of n-octane confined at 2.0 bar pressure and 450 K.
Problem 2.3
Find all three roots in the previous problem.
Problem 2.4
What are the units of van der Waals constants a and b?
Problem 2.5
Using commercial graphing software, produce a 3-D plot of p–V–T for the van der Waals gas N2, where a=1.39 and b=0.039. What are the units of a and b? What happens to p as V becomes very small?
PROBLEMS AND EXERCISES 33 Problem 2.6
Find the van der Waals constant b in terms of the critical constants pc,Vc, and Tc.
Problem 2.7
Show that the van der Waals parameter a is a=98RTcVcwhere the subscript c denotes the critical point.
Problem 2.8
Find the molar volume of ethane at 50 bar pressure and 400 K, from the van der Waals parameters a=5.562 and b=0.0638.
Problem 2.9
Draw the pV isotherms for an ideal gas at 300, 400, 600, and 800 K. The results should resemble the corresponding isotherms in the text.
Problem 2.10
The second virial coefficient B [T ] of toluene (methylbenzene) is –1641 cm3mol−1at 350 K. First, convert this unit to the more modern unit of dm3mol−1, and then find the compressibility factor Z at this temperature and 1.00 bar.
Problem 2.11
Continuing with the data above, what is the molar volume of toluene vapor at 1.00 bar at 350 K?
Problem 2.12
A real gas follows the equation pV =24.79−0.0103 p+6.52×10−5p2 for one mol. Plot the curve of pV vs. p from 0 to 100 bar and locate the minimum pV product.
Problem 2.14
Suppose the data set in the previous problem were as shown in Table 2.2. The last four entries in Table 2.2 differ from those of the previous problem in the second and third digits beyond the decimal point. The graph in Fig. 2.11 turns up slightly above
p=70 bar. Fit this data set with a cubic equation.
p, bar
100 80
60 40
20 0
pV, dm3 bar
24.3 24.4 24.5 24.6 24.7 24.8
FIGURE 2.11 Cubic real gas behavior.
Problem 2.15
(a) What is the volume of one mole of CO2at 366 K and p=111 bar according to the ideal gas law?
(b) The critical temperature of CO2is 305.1 K. The critical pressure and volume of CO2are 73.8 bar and 0.0956 dm3(Laidler and Meiser, 1999). If a compress- ibility factor of 0.68 is read from a chart of isotherms, what is the reduced volume of one mole of CO2at 366 K and p=111 bar according to the chart?
(c) The van der Waals parameters for CO2 are a=3.66 and b=0.0429. What is the volume of one mole of CO2at 366 K and p=111 bar according to the van der Waals equation?
(d) What is the volume of one mole of CO2at 366 K and p=111 bar according to the corresponding states equation? What is the volume?
(e) The critical constants for helium are Tc=5.3 K, pc=2.29 bar, and Vc= 0.0577 dm3mol−1. What is the volume of one mole of He at 7.95 K and p=2.75 bar according to the corresponding states equation? What is the volume?