# Chapter 8 : Ideal and Real Gases¶

### Example 8.1 Page no : 392¶

In [1]:
import math

# Variables
R = 287.; 			#J/kg K
V1 = 40.; 			#m**3
V2 = 40.; 			#m**3
p1 = 1.*10**5; 			#Pa
p2 = 0.4*10**5; 			#Pa
T1 = 298.; 			#K
T2 = 278.; 			#K

# Calculations
m1 = p1*V1/R/T1;
m2 = p2*V2/R/T2;
#Let mass of air removed be m
m = m1-m2;

# Results
print ("Mass of air removed  = %.3f")% (m),("kg")

V = m*R*T1/p1;
print ("Volume of gas removed  = %.3f")% (V), ("m**3")

Mass of air removed  = 26.716 kg
Volume of gas removed  = 22.849 m**3


### Example 8.2 Page no : 393¶

In [2]:
# Variables
V = 0.04; 			#m**3
p = 120.*10**5; 			#Pa
T = 293.; 			#K
R0 = 8314.;

# Calculations and Results
print ("(i) kg of nitrogen the flask can hold")
M = 28; 			#molecular weight of Nitrogen
R = R0/M;

m = p*V/R/T;
print ("kg of nitrogen = %.3f")% (m), ("kg")

print ("(ii) Temperature at which fusible plug should melt")
p = 150.*10**5; 			#Pa
T = p*V/R/m; 			#K
t = T-273; 			#0C
print ("Temperature  = %.3f")% (t),("°C")

(i) kg of nitrogen the flask can hold
kg of nitrogen = 5.517 kg
(ii) Temperature at which fusible plug should melt
Temperature  = 93.250 °C


### Example 8.3 Page no : 393¶

In [3]:
import math

# Variables
p1 = 1.*10**5; 			#Pa
T1 = 293.; 			#K
d = 6.; 			#m; diameter of the spherical balloon
p2 = 0.94*p1;
T2 = T1;
cv = 10400.; 			#J/kg K
R = 8314/2.;
r = 3.; 			#m

# Calculations and Results
print ("(i) Mass of original gas escaped")

mass_escaped = (p1-p2)/p1*100;
print ("%mass_escaped  = "), (mass_escaped), ("%")

print ("(ii)Amount of heat to be removed ")
T2 = 0.94*T1;
m = p1*4/3*math.pi*r**3/R/T1;

Q = m*cv*(T1-T2)/10**6;
print ("Q  = %.3f")% (Q),("MJ")

(i) Mass of original gas escaped
%mass_escaped  =  6.0 %
(ii)Amount of heat to be removed
Q  = 1.698 MJ


### Example 8.4 Page no : 394¶

In [2]:
from numpy import *
import math

# Variables
m = 28.; 			#kg
V1 = 3.; 			#m**3
T1 = 363.; 			#K
R0 = 8314.;
M = 28.; 			#Molecular mass of N2
R = R0/m;
V2 = V1;
T2 = 293.; 			#K

# Calculations and Results
print ("(i) Pressure (p1) and specific volume (v1) of the gas")

p1 = m*R*T1/V1/10**5; 			#bar
print ("Pressure  = %.3f")% (p1), ("bar")

v1 = V1/m;
print ("specific volume = %.3f")% (v1), ("m**3/kg")

#cp-cv = R/1000;
#cp-1.4cv = 0;
#solving the above two eqns
A = [[1,-1],[1,-1.4]];
B = [R/1000,0];
X = linalg.inv(A)*B;
cp = X[0,0]
print ("cp = %.3f")% (cp), ("kJ/kg K")

cv = X[1][0];
print ("cv = %.3f")% (cv),("kJ/kg K")

print ("(iii) Final pressure of the gas after cooling to 20°C")
p2 = p1*T2/T1;
print ("p2 = %.3f")% (p2), ("bar")

du = cv*(T2-T1);
print ("Increase in specific internal energy = %.3f")% (du), ("kJ/kg")

dh = cp*(T2-T1);
print ("Increase in specific Enthalpy  = %.3f")%(dh), ("kJ/kg")

v2 = v1;
ds = cv*math.log(T2/T1) + R*math.log(v2/v1);
print ("Increase in specific entropy  = %.3f")%(ds),("kJ/kg K")

W = 0; 			#constant volume process
Q = m*du+W;
print ("Heat transfer  = %.3f")%(Q), ("kJ")

(i) Pressure (p1) and specific volume (v1) of the gas
Pressure  = 10.060 bar
specific volume = 0.107 m**3/kg
cp = 1.039 kJ/kg K
cv = 0.742 kJ/kg K
(iii) Final pressure of the gas after cooling to 20°C
p2 = 8.120 bar
Increase in specific internal energy = -51.963 kJ/kg
Increase in specific Enthalpy  = -72.748 kJ/kg
Increase in specific entropy  = -0.159 kJ/kg K
Heat transfer  = -1454.950 kJ


### Example 8.5 Page no : 396¶

In [3]:
import math

print ("Part (a)")
# Variables
R = 0.287; 			#kJ/kg K
y = 1.4;
m1 = 1.; 			#kg
p1 = 8.*10**5; 			#Pa
T1 = 373.; 			#K
p2 = 1.8*10**5; 			#Pa
cv = 0.717; 			#kJ/kg K
n = 1.2;

# Calculations and Results
#pv**1.2  =  consmath.tant
print ("(i) The final specific volume, temperature and increase in entropy")

v1 = R*10**3*T1/p1;
v2 = v1*(p1/p2)**(1./n);
print ("v2 = %.3f")%(v2), ("m**3/kg")

T2 = p2*v2/R/10**3; 			#K
t2 = T2-273; 			#0C
print ("Final temperature  = %.3f")% (t2), ("0C")

ds = cv*math.log(T2/T1) + R*math.log(v2/v1);
print ("ds = %.3f")%(ds), ("kJ/kg K")

W = R*(T1-T2)/(n-1);
print ("Work done = %.3f")% (W), ("kJ/kg")

Q = cv*(T2-T1) + W;
print ("Heat transfer = %.3f")%(Q),("kJ/kg")

print ("Part (b)")

print ("(i) Though the process is assumed now to be irreversible and adiabatic, the end states are given to be the same as in (a). Therefore, all the properties at the end of the process are the same as in (a).")

Q = 0;
print ("Heat transfer = %.3f")%(Q), ("kJ/kg")

W = -cv*(T2-T1);
print ("Work done = %.3f")%(W),("kJ/kg")

Part (a)
(i) The final specific volume, temperature and increase in entropy
v2 = 0.464 m**3/kg
Final temperature  = 17.897 0C
ds = 0.179 kJ/kg K
Work done = 117.818 kJ/kg
Heat transfer = 58.950 kJ/kg
Part (b)
(i) Though the process is assumed now to be irreversible and adiabatic, the end states are given to be the same as in (a). Therefore, all the properties at the end of the process are the same as in (a).
Heat transfer = 0.000 kJ/kg
Work done = 58.868 kJ/kg


### Example 8.6 Page no : 397¶

In [6]:
# Variables
d = 2.5; 			#m; diameter
V1 = 4./3*math.pi*(d/2)**3; 			#volume of each sphere
T1 = 298.; 			#K
T2 = 298.; 			#K
m1 = 16.; 			#kg
m2 = 8.; 			#kg
V = 2.*V1; 			#total volume
m = m1+m2;
R = 287.; 			#kJ/kg K

# Calculations
p = m*R*T1/V/10**5; 			#bar

# Results
print ("pressure in the spheres when the system attains equilibrium = %.3f")%(p),("bar")

pressure in the spheres when the system attains equilibrium = 1.254 bar


### Example 8.7 Page no : 398¶

In [3]:
# Variables
m = 6.5/60; 			#kg/s
import math
cv = 0.837; 			#kJ/kg K
p1 = 10*10**5; 			#Pa
p2 = 1.05*10**5; 			#Pa
T1 = 453; 			#K
R0 = 8.314;
M = 44.; 			#Molecular mass of CO2

# Calculations and Results
R = R0/M;
cp = cv+R;
y = cp/cv;

T2 = T1*(p2/p1)**((y-1)/y);
print T2
t2 = T2-273;
print ("Final temperature = %.3f")%(t2),("0C")

v2 = R*10**3*T2/p2; 			#m**3/kg
print ("final specific volume  = %.3f")%(v2), ("m**3/kg")

ds = 0; 			#Reversible and adiabatic process
print ("Increase in entropy = "), (ds)

print ("Heat transfer rate from turbine = "), (Q)

W = m*cp*(T1-T2);
print ("Power delivered by the turbine = %.3f")% (W), ("kW")

# Note : answers wont match with the book because of the rounding error.

299.106840283
Final temperature = 26.107 0C
final specific volume  = 0.538 m**3/kg
Increase in entropy =  0
Heat transfer rate from turbine =  0
Power delivered by the turbine = 17.104 kW


### Example 8.8 Page no : 400¶

In [4]:
import math

# Variables
p1 = 8.*10**5; 			#Pa
V1 = 0.035; 			#m**3
T1 = 553.; 			#K
p2 = 8.*10**5; 			#Pa
V2 = 0.1; 			#m**3
n = 1.4;
R = 287.; 			#J/kg K
T3 = 553.; 			#K
cv = 0.71; 			#kJ/kg K

# Calculations and Results
m = p1*V1/R/T1;
T2 = p2*V2/m/R;
p3 = p2/((T2/T3)**(n/(n-1)));
V3 = m*R*T3/p3;

print ("(i) The heat received in the cycle")

#constant pressure process 1-2
W_12 = p1*(V2-V1)/10**3; 			#kJ
Q_12 = m*cv*(T2-T1) + W_12; 			#kJ

#polytropic process 2-3
W_23 = m*R/10**3*(T2-T3)/(n-1);
Q_23 = m*cv*(T3-T2) + W_23;

print ("(ii) The heat rejected in the cycle")

#Isothermal process 3-1
W_31 = p3*V3*math.log(V1/V3)/10**3; 			#kJ
Q_31 = m*cv*(T3-T1) + W_31;
print ("Heat rejected in the cycle  = %.3f")% (-Q_31), ("kJ")

print ("Efficiency of the cycle  = %.3f")% (n), ("%")

(i) The heat received in the cycle
Total heat received in the cycle =  182.0 kJ
(ii) The heat rejected in the cycle
Heat rejected in the cycle  = 102.883 kJ
Efficiency of the cycle  = 43.471 %


### Example 8.9 Page no : 424¶

In [10]:
# Variables
v = 44.; 			#m**3/kg-mol
T = 373.; 			#K

# Calculations and Results
print ("(i) Using Van der Waals’ equation")
a = 362850.; 			#N*m**4/(kg-mol)**2
b = 0.0423; 			#M**3/kg-mol
R0 = 8314.; 			#J/kg K

p = ((R0*T/(v-b)) - a/v**2);
print ("Pressure umath.sing Van der Waals equation = %.3f")%(p), ("N/m**2")

print ("(ii) Using perfect gas equation")

p = R0*T/v;
print ("Pressure using perfect gas equation = %.3f")% (p), ("N/m**2")

(i) Using Van der Waals’ equation
Pressure umath.sing Van der Waals equation = 70360.445 N/m**2
(ii) Using perfect gas equation
Pressure using perfect gas equation = 70480.045 N/m**2


### Example 8.10 Page no : 425¶

In [5]:
# Variables
V = 3.; 			#m**3
m = 10.; 			#kg
T = 300.; 			#K

# Calculations and Results
R0 = 8314.;
M = 44.;
R = R0/M;
p = m*R*T/V;
print ("Pressure Using perfect gas equation  = %.3f")% (p),("N/m**2")

a = 362850; 			#Nm**4/(kg-mol)**2
b = 0.0423; 			#m**3/(kg-mol)
v = 13.2; 			    #m**3/kg-mol

p = R0*T/(v-b) - a/v**2;
print ("Pressure Using Van der Waals’ equation = %.3f")%(p), ("N/m**2")

A0 = 507.2836;
a = 0.07132;
B0 = 0.10476;
b = 0.07235;
C = 66*10**4;
A = A0*(1-a/v);
B = B0*(1-b/v);
e = C/v/T**3;

p = R0*T*(1-e)/v**2*(v+B) - A/v**2;
print ("Pressure Using Beattie Bridgeman equation  =  %.3f")%(p), ("N/m**2")

Pressure Using perfect gas equation  = 188954.545 N/m**2
Pressure Using Van der Waals’ equation = 187479.533 N/m**2
Pressure Using Beattie Bridgeman equation  =  190090.365 N/m**2


### Example 8.11 Page no : 404¶

In [12]:
import math

# Variables
a = 139250; 			#Nm**4/(kg-mol)**2
b = 0.0314; 			#m**3/kg-mol
R0 = 8314; 			#Nm/kg-mol K
v1 = 0.2*32; 			#m**3/kg-mol
v2 = 0.08*32; 			#m**3/kg-mol
T = 333; 			#K
print ("(i) Work done during the process")

# Calculations
def f21( v):
return R0*T/(v-b) - a/v**2

# Results
print ("W = %.3f")% (W),("Nm/kg-mol")

print ("(ii) The final pressure")
p2 = R0*T/(v2-b) - a/v2**2;
print ("p2 = %.3f")%(p2), ("N/m**2")

(i) Work done during the process
W = -2524722.415 Nm/kg-mol
(ii) The final pressure
p2 = 1073651.290 N/m**2


### Example 8.12 Page no : 404¶

In [13]:
# Variables
pr = 20;
Z = 1.25;
Tr = 8.0;
Tc = 282.4; 			#K

# Calculations
T = Tc*Tr;

# Results
print ("Temperature  = %.3f")%(T),("K")

Temperature  = 2259.200 K


### Example 8.13 Page no : 405¶

In [14]:
# Variables
p = 260.*10**5; 			#Pa
T = 288.; 			#K
pc = 33.94*10**5; 			#Pa
Tc = 126.2; 			#K
R = 8314./28;

# Calculations
pr = p/pc;
Tr = T/Tc;
Z = 1.08;
rho = p/Z/R/T;

# Results
print ("Density of N2 = %.3f")% (rho), ("kg/m**3")

Density of N2 = 281.517 kg/m**3


### Example 8.14 Page no : 405¶

In [15]:
# Variables
p = 200.*10**5; 			#Pa
pc = 73.86*10**5; 			#Pa
Tc = 304.2; 			#K
pr = p/pc;
Z = 1;
Tr = 2.48;

# Calculations
T = Tr*Tc;

# Results
print ("Temperature  = "), (T), ("K")

Temperature  =  754.416 K


### Example 8.15 Page no : 405¶

In [16]:
import math

# Variables
d = 12.; 			#m; diameter of spherical balloon
V = 4./3*math.pi*(d/2)**3;
T = 303.; 			#K
p = 1.21*10**5; 			#Pa
pc = 12.97*10**5; 			#Pa
Tc = 33.3; 			#K
R = 8314./2;

# Calculations
pr = p/pc;
Tr = T/Tc;
Z = 1;
m = p*V/Z/R/T;

# Results
print ("Mass of H2 in the balloon  = %.3f")% (m), ("kg")

Mass of H2 in the balloon  = 86.917 kg


### Example 8.16 Page no : 406¶

In [17]:
# Calculations
Z_cp = 3./2-9./8;

# Results
print ("Z_cp = "),(Z_cp)

Z_cp =  0.375