Question

Beer is stored at 38 ⁰F and 14 psig. Assuming P_atm 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 x_He = 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-03

Answer 1

For Helium ,
ln P^sat = -105.977 + (4529.62 / 276.483) + 14.0094 ln (276.483)
T = 38^oF = 276.483 K
P^sat = e^-10.73106 = 1.97755 * 10^-5 psi
From Dalton's law , P_He / P_total = X_He
1.97755*10^-5 psi / (14-12.7) psi = 1.5211 * 10^-5 = X_He
For N₂ ,
ln P^sat = -181.587 + (8632.129/ 276.483) + 24.79808 ln( 276.483)
P^sat = 1.7605514*10^-5 psi
X_N2 = (1.7605514*10^-5) / (14-12.7) psi = 1.3542*10^-5 = X_N2
For Carbon dioxide,
ln P^sat = -4957.824 + (105288.4 /  276.483) + 933.17 ln (276.483) - 2.8548869(276.483) + 1.48*10^-3( 276.483)²   
ln P^sat = -6.781894
P^sat = 1.13412*10^-3 psi
1.13412*10^-3 / (14-12.7) = 8.72402 *10^-4 = X_CO2