Answer:
[tex]PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...][/tex]
B = b -a/RT
C = b^2
a = 1.263 atm*L^2/mol^2
b = 0.03464 L/mol
Explanation:
In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:
Using the van deer Waals equation of state:
[tex]P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }[/tex]
With further simplification, we have:
[tex]P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }][/tex]
Then, we have:
[tex]P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }][/tex]
Therefore,
[tex]PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }][/tex]
Using the expansion:
[tex](1-x)^{-1} = 1 + x + x^{2} + ....[/tex]
Therefore,
[tex]PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }][/tex]
Thus:
[tex]PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...][/tex] equation (1)
Using the virial equation of state:
[tex]P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...][/tex]
Thus:
[tex]PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...][/tex] equation (2)
Comparing equations (1) and (2), we have:
B = b -a/RT
C = b^2
Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.
[tex]b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol[/tex][/tex] = 0.03464 L/mol
a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2
