A rigid tank contains 1 kg of oxygen (O2) at p1 = 35 bar, T1 = 180 K. The gas is cooled until the temperature drops to 150 K. Determine the volume of the tank, in m3, and the final pressure, in bar, using the:

(a) ideal gas equation of state.

(b) Redlich-Kwong equation.

a. Volume = 13.36 x 10^-3 m³ Pressure = 29.17 bar b. Volume = 14.06 x 10^-3 m³ Pressure = 22.5 bar

Explanation:

Mass of O₂ = 1kg, Pressure (P1) = 35bar, T1 = 180K, T2 = 150k Molecular weight of O₂ = 32kg/Kmol

Volume of tank and final pressure using a) Ideal Gas Equation and b) Redlich - Kwong Equation

a. PV=mRT

V = {1 x (8314/32) x 180} / (35 x 10⁵) = 13.36 x 10^-3

Since it is a rigid tank the volume of the tank must remain constant and hnece we can say

T2/T1 = P2/P1, solving for P2

P2 = (150/180) x 35 = 29.17bar

b. P1 = {RT1 / (v1-b) } - {a/v1 (v1+b) (√T1) }

where R, a and b are constants with the values of, R = 0.08314bar. m³/kmol. K, a = 17.22 (m³/kmol) √k, b = 0.02197m³/kmol

solving for v1

35 = { (0.08314 x 180) / (v1 - 0.02197) } - {17.22 / (v1) (v1 + 0.02197) (√180) }

35 = {14.96542 / (v1-0.02197) } - {1.2835/v1 (v1 + 0.02197) }

Using Trial method to find v1

for v1 = 0.5

Right hand side becomes = {14.96542 / (0.5-0.02197) } - {1.2835/0.5 (0.5 + 0.02197) } = 31.30 ≠ Left hand side

for v1 = 0.4

Right hand side becomes = {14.96542 / (0.4-0.02197) } - {1.2835/0.4 (0.4 + 0.02197) } = 39.58 ≠ Left hand side

for v1 = 0.45

Right hand side becomes = {14.96542 / (0.45-0.02197) } - {1.2835/0.45 (0.45 + 0.02197) } = 34.96 ≅ 35

Specific Volume = 35 m³/kmol

V = m x Vspecific/M = (1 x 0.45) / 32 = 14.06 x 10^-3 m³

For Pressure P2, we know that v2 = v1

P2 = {RT2 / (v2-b) } - {a/v2 (v2+b) (√T2) } = { (0.08314 x 150) / (0.45 - 0.02197) } - {17.22 / (0.45) (0.45 + 0.02197) (√150) } = 22.5 bar