Question 19.1: Employing the Lennard–Jones 6–12 potential, determine the se...
Employing the Lennard–Jones 6–12 potential, determine the second virial coefficient, in cm³/mol, for CH_{4} at 222 K.
From Appendix O, the force constants for CH_{4} are ε/k = 148 K and σ = 3.81 Å. Hence, from Eq. (19.42), the reduced temperature is
T^{*} \equiv \frac{k T}{\varepsilon}, (19.42)
T^{*}=\frac{k T}{\varepsilon}=\frac{222}{148}=1.5.
Consequently, from Appendix Q, the reduced second virial coefficient, B^{*}\left(T^{*}\right), is −1.20. On this basis, the second virial coefficient, in cm³/mole, can be determined from Eqs. (19.37) and (19.44), i.e.,
b_{\circ}=\frac{2 \pi}{3} N_{A} \sigma^{3} (19.37)
B^{*}\left(T^{*}\right) \equiv \frac{B(T)}{b_{\circ}}, (19.44)
B(T)=\frac{2 \pi}{3} N_{A} \sigma^{3} B^{*}\left(T^{*}\right).
Evaluating this expression, we have
B(T)=\frac{2 \pi}{3}\left(6.02 \times 10^{23} mol ^{-1}\right)\left(3.81 \times 10^{-8} cm \right)^{3}(-1.20)and thus
B(T) = −1.20(69.7 cm³/mol) = −83.7 cm³/mol .
The negative value of B(T) in this case underscores the importance of attractive forces at low temperatures, as confirmed by Fig. 19.3.
