Question 3.11: What pressure is generated when 500 mol of methane is stored...
What pressure is generated when 500 mol of methane is stored in a volume of 0.06 m^{3 }at 50°C? Base calculations on each of the following:
(a) The ideal-gas state.
(b) The Redlich/Kwong equation.
(c) A generalized correlation.
The molar volume of the methane is V = 0.06 / 500 = 0.0012 m^{3}· mol^{−1} .
(a) For the ideal-gas state, with R = 8.314 \times 10^{−5} bar·m^{3}·mol^{−1}·K^{−1} :
P =\frac{RT}{V} = \frac{( 8.314 \times 10^{−5} ) ( 323.15)}{0.00012}= 223.9 bar
(b) The pressure as given by the Redlich/Kwong equation is:
P=\frac{RT}{V-b} – \frac{a(T)}{V(V+b)} (3.40)
Values of b and a(T) come from Eqs. (3.44) and (3.45), with Ω, Ψ, and \alpha (T_{r} )=x^{-1/2}_{r} from Table 3.1. With values of T_{c} and P_{c} from Table B.1, we have:
Table 3.1: Parameter Assignments for Equations of State
| Eqn. of State | α(T_{r}) | σ | ε | Ω | Ψ | Z_{c} |
| vdW (1873) | 1 | 0 | 0 | 1/8 | 27/64 | 3/8 |
| RK (1949) | T_{ r}^{-1/2} | 1 | 0 | 0.08664 | 0.42748 | 1/3 |
| SRK (1972) | α_{SRK} ( T_{ r} ; ω )^{†} | 1 | 0 | 0.08664 | 0.42748 | 1/3 |
| PR (1976) | α_{PR} ( T_{ r} ; ω )^{†} | 1 +\sqrt{2} | 1 – \sqrt{2} | 0.07780 | 0.45724 | 0.30740 |
T_{r}=\frac{323.15}{190.6}= 1.695
b = 0.08664 \frac{(8.314 \times 10^{−5} ) ( 190.6 )}{45.99} = 2.985 \times 10^{−5} m^{3}·mol^{−1}
a = 0.42748 \frac{( 1.695 )^{−0.5} (8.314 \times 10^{−5} )^{2} ( 190.6 )^{2}}{45.99}= 1.793 × 10^{-6} bar· m^{6} mol^{-2}
Substitution of numerical values into the Redlich/Kwong equation now yields:
P = \frac{(8.314 \times 10^{−5} ) ( 323.15 )}{0.00012 − 2.985 \times 10^{−6}} – \frac{1.793\times 10^{−5} }{0.00012(0.00012 + 2.985 \times 10^{−5})} = 198.3 bar(c) Because the pressure here is high, the full Lee/Kesler generalized compressibilityfactor correlation is the proper choice. In the absence of a known value for P_{r}, an iterative procedure is based on the following equation:
P = \frac{ZRA}{V} = \frac{ Z(8.314 \times 10^{-5} ) ( 323.15 )}{0.00012}= 223.9 ZBecause P = P_{c} P_{r} = 45.99 P_{r} , this equation becomes:
\frac{}{} z = \frac{ 45.99 P_{r} }{223.9}= 0.2054P_{r} or P_{r}= \frac{Z}{0.2054}One now assumes a starting value for Z, say Z = 1. This gives P_{r} = 4.68, and allows a new value of Z to be calculated by Eq. (3.53) from values interpolated in Tables D.3 and D.4 at the reduced temperature of T_{r} = 1.695. With this new value of Z, a new value of P_{r} is calculated, and the procedure continues until no significant change occurs from one step to the next. The final value of Z so found is 0.894 at P_{r} = 4.35. This is confirmed by substitution into Eq. (3.53) of values for Z^{0 } and Z^{1} from Tables D.3 and D.4 interpolated at P_{r} = 4.35 and T_{r} = 1.695. With ω = 0.012 ,
Z = Z^{0} + \omega Z^{1} = 0.891 + (0.012)(0.268)= 0.894
P = \frac{ZRT}{V}=\frac{ ( 0.894 ) (8.314 \times 10^{−5} ) ( 323.15 )}{0.00012}= 200.2 bar
Because the acentric factor is small, the two- and three-parameter compressibilityfactor correlations are little different. The Redlich/Kwong equation and the generalized compressibility-factor correlation give answers within 2% of the experimental value of 196.5 bar.