Question 6.4: Values of the second and third virial coefficients of nitrog...

(a) Here, we employ Eqs. (6.55) and (6.56) for H_R  and  S_R, respectively. These require the temperature derivative of the second virial coefficient. For this, we use the values at 290 K and 310 K:

\frac{H^R}{R T}=\frac{P}{R}\left(\frac{B}{T}-\frac{d B}{d T}\right)        (6.55)

\frac{S^R}{R}=-\frac{P}{R} \frac{d B}{d T}     (6.56)

\frac{d B}{d T} \approx \frac{-80.4+94.3}{20}=0.695  cm ^3 \cdot mol ^{-1} \cdot K ^{-1}

Using this value in Eq. (6.55) gives:

\frac{H^R}{R T}=\frac{P}{R}\left(\frac{B}{T}-\frac{d B}{d T}\right)=\frac{100}{83.14}\left(\frac{-87.1}{300}-0.695\right)=-1.19

from which H^R = −2960  J·mol^−1.

Similarly,

\frac{S^R}{R}=-\frac{P}{R} \frac{d B}{d T}=-\frac{100}{83.14}(0.695)=-0.836

from which S^R=-6.95  J \cdot mol ^{-1} \cdot K ^{-1}

(b) In this case we require the temperature derivatives of both B and C, along with the value of the density of NF_3 at 300 K and 100 bar. Solving the three-term virial coefficient iteratively as described in Ex. (3.8) gives V = 178.0  cm^3·mol^−1,  or  ρ = 0.00562  mol·cm^3. This corresponds to a compressibility factor of Z = 0.714. Applying the same approach as in part (a) for the temperature derivative of C:

\frac{d C}{d T} \approx \frac{6740-6090}{20}=32.5  cm ^6 \cdot mol ^{-2} \cdot K ^{-1}

With these values, Eqs. (6.62) and (6.63) give:

\frac{H^R}{R T}=T\left[\left(\frac{B}{T}-\frac{d B}{d T}\right) \rho+\left(\frac{C}{T}-\frac{1}{2} \frac{d C}{d T}\right) \rho^2\right]        (6.62)

\frac{S^R}{R}=\ln Z-T\left[\left(\frac{B}{T}+\frac{d B}{d T}\right) \rho+\frac{1}{2}\left(\frac{C}{T}+\frac{d C}{d T}\right) \rho^2\right]         (6.63)

\frac{H^R}{R T}=300\left[\left(\frac{-87.1}{300}-0.695\right) 0.00562+\left(\frac{6430}{300}-\frac{32.5}{2}\right) 0.00562^2\right]=-1.61

\frac{S^R}{R}=\ln (0.714)-300\left[\left(\frac{-87.1}{300}+0.695\right) 0.00562+\frac{1}{2}\left(\frac{6430}{300}+32.5\right) 0.00562^2\right]=-1.27

from which H^R=-4021  J \cdot mol ^{-1} \text { and } S^R=-10.60  J \cdot mol ^{-1} \cdot K ^{-1}

For comparison, values obtained from data in the NIST Chemistry WebBook are H^R=-3540  J \cdot mol ^{-1} \text { and } S^R=-8.83  J \cdot mol ^{-1} \cdot K ^{-1} . The expression based on the two-term virial equation underestimates these, while the expression based on the three-term virial equation overestimates them. Note that the conditions selected correspond to T_r = 1.28  and  P_r = 2.24. These are outside the range where we would expect the two-term virial expansion to work well, and it  ppears that adding a third term produces an overcorrection.