Question 3.12: A mass of 500 g of gaseous ammonia is contained in a vessel ...
A mass of 500 g of gaseous ammonia is contained in a vessel of 30,000 cm^{3} volume and immersed in a constant-temperature bath at 65°C. Calculate the pressure of the gas by:
(a) The ideal-gas state;
(b) A generalized correlation.
The molar volume of ammonia in the vessel is:
V = \frac{V^{t}}{n} = \frac{V^{T} }{ml ℳ} =\frac{30,000}{500/17.02} =1021.2 cm^{3}·mol^{−1}(a) For the ideal-gas state,
P = \frac{RT}{V}=\frac{ ( 83.14 ) ( 65 + 273.15 )}{1021.2} = 27.53 bar(b) Because the reduced pressure is low (Pr ≈ 27.53/112.8 = 0.244), the generalized virial-coefficient correlation should suffice. Values of B0 and B1 are given by Eqs. (3.61) and (3.62). With T_{r} = 338.15/405.7 = 0.834 ,
B^{0} = 0.083 – \frac{0.422}{T^{1.6}_{r} }(3.61)
B^{1} = 0.139 – \frac{0.172}{T^{4.2}_{r} }(3.62)
B^{0} =−0.482 B^{1} = −0.232
Substitution into Eq. (3.59) with ω = 0.253 yields:
\widehat{B} = B^{0} +\omega B^{1} (3.59)
\widehat{B} = − 0.482 + ( 0.253 )( −0.232 ) = −0.541
B = \frac{\widehat{B} R T_{c} }{P_{c} } = \frac{ − ( 0.541 ) ( 83.14 ) ( 405.7 )}{112.8}= −161.8 cm^{3}·mol^{−1}
By the second equality of Eq. (3.36):
Z = \frac{PV}{RT} = 1 + \frac{BP}{RT} (3.36)
p = \frac{RT}{V – B} = \frac{( 83.14 ) ( 338.15 )}{1021.2 + 161.8} = 23.76 barAn iterative solution is not necessary because B is independent of pressure. The calculated P corresponds to a reduced pressure of P_{r} = 23.76/112.8 = 0.211. Reference to Fig. 3.13 confirms the suitability of the generalized virial-coefficient correlation.
Experimental data indicate that the pressure is 23.82 bar at the given conditions. Thus the ideal-gas state yields an answer high by about 15%, whereas the virial-coefficient correlation gives an answer in substantial agreement with experiment, even though ammonia is a polar molecule.
