Chapter 11 : Thermodynamic Relations, Equilibrium and Stability

Example 11.3 Page No : 418

In [1]:
import math 

# Variables
Tb = 353.;
T = 303.;
R = 8.3143;

# Calculation
P = 101.325*math.exp((88/R)*(1-(Tb/T)));

# Results
print "Vapour pressure of benzene is %.1f kPa"%P

Vapour pressure of benzene is 17.7 kPa

Example 11.4 Page No : 418

In [3]:
import math 

# Variables
T = (3754.-3063)/(23.03-19.49);
P = math.exp(23.03-(3754/195.2));

# Calculation
R = 8.3143;
Lsub = R*3754;
Lvap = 3063*R;
Lfu = Lsub-Lvap;

# Results
print "Temperature of triple point is %.1f K"%T
print "Pressure of triple point is %.2f mm Hg"%P
print "Latent heat of sublimation is %.0f kJ/Kg mol"%round(Lsub,-2)
print "Latent heat of vapourization is %.0f kJ/kg mol"%round(Lvap,-2)
print "Latent heat of fusion is %.0f kJ/kg mol"%round(Lfu,-2)

Temperature of triple point is 195.2 K
Pressure of triple point is 44.63 mm Hg
Latent heat of sublimation is 31200 kJ/Kg mol
Latent heat of vapourization is 25500 kJ/kg mol
Latent heat of fusion is 5700 kJ/kg mol

Example 11.6 Page No : 420

In [1]:
import math 

# Variables
R = 8.314;
N1 = 0.5e-03; 
N2 = 0.75e-03; 			# Mole number of system 1 and 2 in kg/mol
T1 = 200.; 
T2 = 300.; 
V = 0.02;

# Calculation
Tf = ((N1*T1)+(N2*T2))/(N1+N2); 			# Final temperature
Uf1 = (3./2)*R*N1*Tf;
Uf2 = (3./2)*R*N2*Tf;
Pf = (R*Tf*(N1+N2))/V;
Vf1 = (R*N1*Tf)/Pf;
Vf2 = V - Vf1;

# Results
print ("System 1")
print "Volume is",Vf1,"m3"
print "Energy is %.2f kJ"%Uf1
print "\nSystem 2"
print "Volume is",Vf2,"m3"
print "Energy is",round(Uf2*1000,-1),"kJ"
print "Final temperature is",Tf,"K"
print "Final Pressure is",round(Pf),"kN/m**2"

System 1
Volume is 0.008 m3
Energy is 1.62 kJ

System 2
Volume is 0.012 m3
Energy is 2430.0 kJ
Final temperature is 260.0 K
Final Pressure is 135.0 kN/m**2