Question

Using the van der Waals equation, determine the pressure, in atm, that exert 100 lb of CO2, at a temperature of 212 ºF if the specific volume is 19.3 ft3. Compare the result obtained if the ideal gas equation and the general compressibility chart are used.

Answer 1

Van der walls

Given that

Given that mass of CO2 = 100 lb = 100 x 453.59 g = 45359 g [ 1 lb = 453.59 g]

molar mass of CO2 = 44 g/mol

moles of CO2, n = mass/ molar mass = 45359 g / 44 g/mol = 1030.88 moles

volume V = 19.3 ft3 = 19.3 x 28.3 L = 546.19 L [ 1ft³ = 28.3 L ]

temperature T = 212 ^oF = 100 oC = 100 + 273 K = 373 K

Van der walls constants for CO2, a = 3.658 L². atm/mol² , b = 0.0428 L/mol

R = 0.0821 L.atm/K/mol

Pressure of CO2:

Van der walls equation is

(P + an²/V²) (V-nb) = nRT

P = [ nRT / (V -nb) ] - (an²/V²)

= [(1030.88)(0.0821)(373) / (546.19 - 1030.88 x 0.0428)] - ( 3.658 x 1030.88² / 546.19²)

= 49.84 atm

Pressure of CO2 = 49.84 atm

Ideal gas:

PV = n RT

P = nRT/V

= 1030.88 mol  x 0.0821 L.atm/K/mol x 373 K / 546.19 L

= 57.79 atm

Pressure of ideal gas = 57.79 atm

------------------------------------------------------------------------------------

Therefore,

Pressure of Ideal gas is greater than Pressure of CO2.