A storage tank with a volume of 25ft3 contains 52 lbm of CO2

A storage tank with a volume of 25ft^3 contains 52 lb_m of CO_2 at a temperature of 105 degree F. Determine the pressure in the storage tank using: the Ideal Gas equation van der Waals equation the Redlieh-K wrong equation the generalized compressibility chart

Solution

Given data

volume,v=25feet^3

1 ft^3=0 .02831 m^3

25 ft^3=0.0709m^3

mass,m=52lbm of CO2

1 lbm =0.45359 kg

52 lbm = 0.45359 * 52=23.588 kg.

temperature,T = 105\'F =313.56 k

a) pressure according to ideal gas law

PV=mRT

P=mRT/p

R=R₀/m =8314/44

              = 188.95 j/kgk

p=(23.586*188.95*313.56)/0.7079

=1.974*10^6 Pa =1.974 MPa.

b)pressure according to vanderwaal\'s equation

P ={R₀T/(V-b) - a/V^2}       - eqn 1

Van der waal\'s constant a =0.3658 pa(m^3/mol)

Vander waal\'s constant b=4.286*10^-5 m^3/mol

volume occupied by 1 kg of CO₂,V

=(0.7079/23.5868)*44 { molar mass of CO₂ = 44}

=1.320 m^3

substituting above values in eqn 1

P = [((8314*313.56)/(1.320-4.28 *10^-5)) - 0.3658/1.320^2]

=1.975*10^6 Pa

=1.975 MPa

C) Pressure according to Redlich - kwong equation

P = {(RT/V-b)-(a/(T^.5*V[V+b]))}    eqn 2

substituting vanderwaal\'s values of a,b in above eqn. we get,

((8.314*313.56)/(1.320-4.288*10^-5) - 0.368/((313.56^.5)* 1.320[1.320(1.320+4.28*10^-5)]))

=1.93*10^6

=1.93 MPa