# variables
V = 0.6 #m**3 vessel
T = 473. #K temperature
N = 1. * 10 ** 3 #mol
R = 8.314 #Pa * m**3/molK
# Calculation
P = N * R * T / (V * 10**5)
# Result
print "(a)Pressure calculated using ideal gas equation = %.2f bar"%P
a = 0.4233 #N * m**4 / mol**2
b = 3.73 * 10**-5 #m**3/mol
P1 = (R*T/(V/N - b)-a/(V/N)**2)/10**5
print "(a)Pressure calculated using van der waals equation = %.2f bar"%P1
# variables
P = 10.**7 #Pa
T = 500. #K molar volume
R = 8.314 #Pa * L / mol K
N = 1000.
# Calculation
V = N * R * T / ( P * 1000)
# Result
print "(a)Volume of CO2 calculated using ideal gas equation = %.4e m**3"%V
# variables
V = 0.6 * 10**-3 #m**3
T = 473. #K vellel of volume
Tc = 405.5 #K temperature
Pc = 112.8 * 10 ** 5 #Pa pressure
R = 8.314
# Calculation
a = 0.4278 * (R**2) * (Tc ** 2.5)/Pc
b = 0.0867 * R * Tc / Pc
P1 = (R*T/(V - b) - a/((T**0.5)*V*(V + b)))/10**5
# Result
print "Pressure developed by gas = %.2f bar"%P1
# variables
P = 10.**6 #Pa
T = 373. #K molar volume
Tc = 405.5 #K temperature
Pc = 112.8 * 10 ** 5 #Pa pressure
R = 8.314
# Calculation
a = 0.4278 * (R**2) * (Tc ** 2.5)/Pc
b = 0.0867 * R * Tc / Pc
#P1 = (R*T/(V - b) - a/((T**0.5)*V*(V + b)))/10**5
#10**6=((8.314*373)/(V-2.59*10**-5))-8.68/((373**0.5)*V*(V+2.59*10**-5)
#solving this we get,
V = 3.0 * 10**-3 #m**3/mol
# Result
print "molar volume of gas = %.e m**3/mol"%V
# variables
B = -2.19 * 10**-4 #m**3/mol virial coefficients
C = -1.73 * 10**-8 #m**6/mol**2
P = 10. #bar molar volume
T = 500. #K
# Calculation
#virial equation is given as, Z = PV/RT = 1 + B/V + C/V**2
#V = (RT/P)*(1 + B/V + C/V**2)
# now by assuming different values for V on RHS and checking for corresponding V on LHS, we have to assume such value of V on RHS by which we get the same value for LHS V
#by trial and error we get,
V = 3.92 * 10**-3 #m**3
# Result
print "Molar volume of methanol = %.3e m**3"%V
# variables
T = 510. #K molar volume
P = 26.6 #bar
Tc = 425.2 #K temperature
Pc = 38. #bar pressure
Zc = 0.274 # compressibility factor
R = 8.314
# Calculation
Pr = P / Pc
Tr = T / Tc
# Result
print "Pr = ",Pr
print "Tr = %.2f"%Tr
#From fig. 5.4 and 5.5 from the text book
Z = 0.865
D = 0.15
Z1 = Z + D * ( Zc - 0.27)
V = R * T * Z1 / (P * 10**5)
print "Molar volume of n-butane = %.4e m**3/mol"%V
# variables
T = 510. #K molar volumes
P = 26.6 #bar
Tc = 425.2 #K temperature
Pc = 38. #bar pressure
w = 0.193 # acentric factor
R = 8.314
# Calculation
Pr = P / Pc
Tr = T / Tc
# Result
print "Pr = ",Pr
print "Tr = %.2f"%Tr
#From fig. 5.6 and 5.7 from the text book
Z0 = 0.855
Z1 = 0.042
Z = Z0 + w*Z1
print "Z = %.4f"%Z
V = R * T * Z / (P * 10**5)
print "Molar volume of n-butane = %.4e m**3/mol"%V
# note : answer may vary because of rounding error.
from numpy import *
# variables
P = 6000. #kPa ethane
T = 325. #K ethane
xn2 = 0.4 # gas mixture
xethane = 0.6
an2 = 0.1365 #N m**4 / mol**2 nitrogen
# Calculation
bn2 = 3.86 * 10**-5 #m**3/mol
aethane = 0.557 #N m**4 / mol**2
bethane = 6.51 * 10**-5 #m**3/mol
Pcn2 = 3394. #kPa
Tcn2 = 126.2 #K
Pcethane = 4880. #kPa
Tcethane = 305.4 #K
R = 8.314
# Result
V = R * T / (P*1000)
print "(a)Molar volume by ideal gas equation = %.3e m**3/mol"%V
a = (xn2 * (an2**0.5) + xethane * (aethane**0.5))**2
b = (xn2*bn2 + xethane*bethane)
#substituting the above values in van der waals equation, and solving, we get
#V^3 - 5.0484*10^-4V^2+5.9117*10^-8V-3.2214*10^-12=0
#Using co-efficients from the above equation,
coeff = [1, -5.0484e-4, 5.9117e-8, -3.2217e-12]
x = roots(coeff)
y = x[0]
print "(b)Molar volume by van der waals equation =",round((y.real)/1e-4,3),"*10^-4 m**3/mol"
Prin2 = P/Pcn2
Trin2 = T/Tcn2
Priethane = P/Pcethane
Triethane = T/Tcethane
# using compressibilty chart,
Zn2 = 1
Zethane = 0.42
Z = xn2 * Zn2 + xethane * Zethane
V2 = Z * R * T / P * (10**-3)
print "(c)Molar volume based on compressibilty factor = %.2e m**3/mol"%V2
Pri1n2 = xn2*P/Pcn2
Tri1n2 = T/Tcn2
Pri1ethane = xethane*P/Pcethane
Tri1ethane = T/Tcethane
# using compressibilty chart,
Zn21 = 1.
Zethane1 = 0.76
Z1 = xn2 * Zn21 + xethane * Zethane1
V3 = (Z1 * R * T / P ) * (10**-3)
print "(c)Molar volume based on daltons law = %.3e m**3/mol"%V3
Tc = xn2 * Tcn2 + xethane * Tcethane
Pc = xn2 * Pcn2 + xethane * Pcethane
Zc = 0.83
V4 = Zc * R *T / P * (10**-3)
print "(d)Molar volume by kays method = %.3e m**3/mol"%V4
# variables
P1 = 40. #% ( nitrogen )
P2 = 60. #% ( ethane )
T = 325. #K ethane
# Calculation
V = 4.5 * 10**-4 #m**3/mol
a1 = 0.1365 #N*m**4/mol**2
b1 = 3.86 * 10 ** -5 #m**3/mol
a2 = 0.557 #N*m**4/mol**2
b2 = 6.51 * 10 ** -5 #m**3/mol
Pc1 = 3394. #kPa
Tc1 = 126.1 #K
Pc2 = 4880. #kPa
Tc2 = 305.4 #K
R = 8.314
Pideal = R * T / (V * 1000) #kPa
# Result
print "(a)Pressure of Gas by the ideal gas equation = %.f kPa"%Pideal
y1 = P1/100.
y2 = P2/100.
a = (y1 * (a1**(1./2)) + y2 * (a2**(1./2)))**2
b = y1 * b1 + y2 * b2
Pv = ((R * T / (V - b)) - a / (V**2))/1000
print "(b)Pressure of Gas by Van der waals equation = %.f kPa"%Pv
Tc = y1*Tc1 + y2*Tc2
Pc = y1*Pc1 + y2*Pc2
Vc = R * Tc / Pc #Pseudo critical ideal volume
Vr = V / Vc #Pseudo reduced ideal volume
Tr = T / Tc #Pseudo reduced temperature
#From fig 5.3, we get Pr = 1.2
Pr = 1.2
Pk = Pr * Pc
print "(c)Pressure of Gas by the Kays method = %.f kPa"%Pk