Answer:
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
