Question 9.1: Determine the contribution to the Helmholtz free energy (kJ/...
Determine the contribution to the Helmholtz free energy (kJ/mole) from the translational energy mode at 500 K and 2 bars for monatomic helium.
From Eq. (4.38), the translational contribution to the Helmholtz free energy for an ideal gas is
\left(\frac{a}{RT}\right) = -\left[\ln\left(\frac{Z}{N}\right)+1\right] (4.38)
\left(\frac{a}{RT}\right)_{tr} = -\left[\ln\left(\frac{Z_{tr}}{N}\right)+1\right].
Employing Eqs. (9.4) and (9.10), we find that
Z_{tr} = \left(\frac{2\pi_\ mkT}{h^{2}}\right)^{{3}/{2}}V. (9.4)
PV = NkT, (9.10)
\frac{Z_{tr}}{N} = \left(\frac{2\pi_\ mkT}{h^{2}}\right)^{{3}/{2}}\left(\frac{kT}{P}\right).
Substituting for the given conditions and for appropriate fundamental constants, we obtain
\frac{Z_{tr}}{N} = \left(2\pi_\ m\right)^{{3}/{2}}\frac{\left(kT\right)^{{5}/{2}}}{h^{3}P}
= \left[2\pi(4.0026)(1.6605\times10^{-27})\right]^{{3}/{2}}\left\{\frac{\left[(1.3807\times10^{-23})(500)\right]^{{5}/{2}}}{(6.6261\times 10^{-34})^{3}(2\times10^{5})}\right\},
from which we calculate Z_{tr}/N = 5.808\times10^{5}. On the other hand, because this rather cumbersome expression has already been simplified via Problem 4.1, we could also find the same answer more quickly from
\frac{Z_{tr}}{N} = 0.02595\frac{T^{{5}/{2}}M^{{3}/{2}}}{P},
where M is the molecular weight and P is the pressure in bars. In either case, we now have
\left(\frac{a}{RT}\right)_{tr} = -\left[\ln\left(\frac{Z_{tr}}{N}\right)+1\right] = – (13.272 + 1.000) = – 14.272.
Evaluating the Hemholtz free energy, we thus obtain, for the translational mode,
a_{tr} = −14.272(8.3145 J/K · mol)(500 K) = −58.333 kJ/mol.