Beer is stored at 38 0 F and 14 psig. Assuming P atm is 12.7 psi and that beer i
ID: 1060112 • Letter: B
Question
Beer is stored at 38 0F and 14 psig. Assuming Patm is 12.7 psi and that beer is essentially composed of water, estimate the composition of He, N2, and CO2 (individually) that could be achieved in beer (i.e. water) at these conditions.
The composition of beer (i.e. water) in the gas phase at these conditions is 0.0042.
(Know one of the answers is xHe = 1.35E-05, but not sure how to get that.)
lnPsat = A + B/T + ClnT + DT + ET^2; (T in K) Gas T Range (K) A B C D E Helium 273 - 348 -105.977 4259.62 14.0094 Nitrogen 273 - 348 -181.587 8632.129 24.79808 Carbon Dioxide 273 - 373 -4957.824 105288.4 933.17 -2.854886 1.48E-03Explanation / Answer
For Helium ,
ln Psat = -105.977 + (4529.62 / 276.483) + 14.0094 ln (276.483)
T = 38oF = 276.483 K
Psat = e-10.73106 = 1.97755 * 10-5 psi
From Dalton's law , PHe / Ptotal = XHe
1.97755*10-5 psi / (14-12.7) psi = 1.5211 * 10-5 = XHe
For N2 ,
ln Psat = -181.587 + (8632.129/ 276.483) + 24.79808 ln( 276.483)
Psat = 1.7605514*10-5 psi
XN2 = (1.7605514*10-5) / (14-12.7) psi = 1.3542*10-5 = XN2
For Carbon dioxide,
ln Psat = -4957.824 + (105288.4 / 276.483) + 933.17 ln (276.483) - 2.8548869(276.483) + 1.48*10-3( 276.483)2
ln Psat = -6.781894
Psat = 1.13412*10-3 psi
1.13412*10-3 / (14-12.7) = 8.72402 *10-4 = XCO2
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