Chapter 2 : Gas Laws Ideal and Real Gases

Example 2.1 Page No : 41

In [2]:
# Variables
p1= 2.;			# in bar
v1= 30.;			# in litre
T1= 27.+273;			# in K
T2= -3.+273;			# in K
v2= v1;			# in litre

# Calculations
# Gas law  p1*v1/T1= p2*v2/T2
p2= p1*v1*T2/(T1*v2);			# in bar

# Results
print "The air pressure in the tyre in bar is :",p2

The air pressure in the tyre in bar is : 1.8

Example 2.2 Page No : 42

In [1]:
# Variables
p= 12.;			# in bar
p=p*10**5;			# in N/m**2
v= 25.;			# in m**3
T= 30.+273;			# in K
# Part (a) Mass of each gas
#Formula p*v=m*R*T
R_U= 8314.;			# in J/kg-mole K
M_N2= 28.016;			# in mole
M_O2= 32.;			# in mole
M_CO2= 44.;			# in mole

# Calculations and Results
R_N2= R_U/M_N2;			# in J/kg K
R_O2= R_U/M_O2;			# in J/kg K
R_CO2= R_U/M_CO2;			# in J/kg K
m_of_N2= p*v/(R_N2*T);			# in kg
m_of_O2= p*v/(R_O2*T);			# in kg
m_of_CO2= p*v/(R_CO2*T);			# in kg
print "The mass of Nitrogen gas stored in the vessel in kg is : %.2f"%(m_of_N2)
print "The mass of Oxygen gas stored in the vessel in kg is : %.2f"%m_of_O2
print "The mass of Carbon dioxide gas stored in the vessel in kg is : %.2f"%round(m_of_CO2)

# Part (b) Molar Volume
# Formula v_molar= M*R*T/p= R_U*T/p
v_molar= R_U*T/p;			# in m**3
print "Molar volume of the gas mixture in m**3 is : %.2f"%v_molar

# Part (c) Average density
# rho_avg= total mass/total volume
rho_avg= (m_of_N2+m_of_O2+m_of_CO2)/v;			# in kg/m**3
print  "density of the gas mixture in kg/m**3 is : %.2f"%rho_avg

The mass of Nitrogen gas stored in the vessel in kg is : 333.64
The mass of Oxygen gas stored in the vessel in kg is : 381.08
The mass of Carbon dioxide gas stored in the vessel in kg is : 524.00
Molar volume of the gas mixture in m**3 is : 2.10
density of the gas mixture in kg/m**3 is : 49.55

Example 2.3 Page No : 47

In [2]:
# Variables
Qp = 1230.;			# kJ/kg
Qv = 795.; 			# kJ/kg
t1 = 16.;			# in °C
t2 = 96.;			# in °C
R_U = 8.314;

# Calculations and Results
delta_T= t2-t1;			# in °C
Cp= Qp/delta_T;			# in kJ/kg °C
print "The value of Cp in kJ/kg°C",Cp

Cv= Qv/delta_T;			# in kJ/kg °C
print "The value of Cv in kJ/kg°C",Cv

R= Cp-Cv;			# in kJ/kg °C
print "The value of R in kJ/kg°C",R

molecular_weight= R_U/R;
print "Molecular weight of the gas is : %.3f"%molecular_weight

The value of Cp in kJ/kg°C 15.375
The value of Cv in kJ/kg°C 9.9375
The value of R in kJ/kg°C 5.4375
Molecular weight of the gas is : 1.529

Example 2.4 Page No : 48

In [2]:
import math 
from scipy.integrate import quad 

# Variables
a= 0.85;
b= 0.00004;
c= 5*10**-5;
T1= 300;			# in K
T2= 2300;			# in K
gama= 1.5;			# the ratio of specific heats
m=1;			# in kg

# Calculations and Results
def f1(T): 
    return a+b*T+c*T**2

delta_H= m* quad(f1,T1,T2)[0]

print "Change in enthalpy in MJ is : %.3f"%(delta_H*10**-3)

# Formula delta_U= integration of m*Cv = integration of m*Cp/gama= delta_H/gama
delta_U= delta_H/gama;			# in kJ
print  "change in internal energy in MJ is : %.3f"%(delta_U*10**-3)

Change in enthalpy in MJ is : 204.137
change in internal energy in MJ is : 136.092

Example 2.5 Page No : 52

In [4]:
# Variables
v= 0.9/3;			# in m**3/kg
v= 2*v;			# in m**3/kg mole (as M_hydrogen = 2)
T=120+273;			# in K
R=8314;			# in J/kg mole K
a=2.51*10**4;			# in Nm**4/(kg mole)**2
b= 0.0262;

# Calculations and Results
# Part (a)
p= R*T/v;			# in N/m**2
p= p*10**-5;			# in bar
print "Using perfect gas law the pressure for unit mass of hydrogen in bar is : %.3f"%p

# Part (b)
p= R*T/(v-b)-a/v**2;			# N/m**2
p= p*10**-5;			# in bar
print "Using Van der waals equation, the pressure in bar is : %.3f"%p

Using perfect gas law the pressure for unit mass of hydrogen in bar is : 54.457
Using Van der waals equation, the pressure in bar is : 56.246

Example 2.6 Page No : 55

In [1]:
# Variables
p1= 0.98;			# in bar
p2= 0.6;			# in bar
v1= 0.45;			# in m**3/kg

# Calculations
# Applying Boyle's law
v2= p1*v1/p2;			# in m**3/kg
rho2= 1/v2; 			# in kg/m**3

# Results
print  "density of the gas under the changed condition in kg/m**3 is : ",round(rho2,2)

density of the gas under the changed condition in kg/m**3 is :  1.36

Example 2.7 Page No : 55

In [5]:
# Exa 2.7
import math 

# Variables
r=5;			    # in cm
R_U= 8314
T= 27+273;			# in K

# Calculations
V= 4./3*math.pi*r**3;			# volume of balloon in cm**3
# atmPressure= 75 cm off mercury = 75/76*1.01325 
atmPressure= round(75./76*1.01325) ;			# in bar
p= atmPressure;			# pressure of hydrogen in balloon in bar
p=p*10**5;			# in N/m**2
R= R_U/2;			# in J/kg K
m1= p*V/(R*T);			# in kg
# The volume of air print laced = the volume of balloon, so
R=287;
T=20+273;			# in K
m2= p*V/(R*T);			# in kg
payload= m2-m1;			# in kg

# Results
print "Payload that can be lifted along with the balloon in kg is : %.2f"%payload

Payload that can be lifted along with the balloon in kg is : 580.67

Example 2.12 Page No : 57

In [3]:
# Variables
AvogadroNo= 6.023*10**23;
n= 5/AvogadroNo;			# number of moles
v=10**-6;			# in m**3
T= -270+273;			# in K
R= 0.287;

# Calculations
p= n*R*T/v;			# in kPa
p= p*10**18;			# in aPa

# Results
print "The pressure in the space is %.3f aPa"%p;

The pressure in the space is 7.148 aPa

Example 2.13 Page No : 57

In [13]:
# Exa 2.13
import math 
from scipy.integrate import quad 

# Variables
T1 = 300;			# in K
T2 = 900;			# in K
m = 2;    			# in kg

# Calculations
def f3(T): 
    return 40-600/math.sqrt(T)+7000/T

delta_H=m*  quad(f3,T1,T2)[0]

delta_H= delta_H/17.03;			# in kJ/kg

# Results
print "Change in enthalpy in kJ/kg is : %.1f"%delta_H

Change in enthalpy in kJ/kg is : 1934.8

Example 2.14 Page No : 58

In [14]:
# Variables
m = 12.;			# in kg mol
v = 723.7;			# in m**3
T = 140.;			# in °C
T = T+273;			# in K
rho = 0.644;			# in kg/m**3
Ro = 8314;			# in J/kg-mole K

# Calculations and Results
# rho= m/v, where m in Kg , so rho= m*M/v
M = rho*v/m;
m = m*M;			# in kg
print "Molecular weight is : %.2f"%M

# Part (b)
R = Ro/M;			# in J/kg K
print "Gas constant in kJ/kg K is : %.3f"%(R*10**-3)

# Part(c)
p = m*R*T/v;			# in N/m**2
p = p*10**-5;			# in bar
print "The pressure of the gas in bar is : %.3f"%p

Molecular weight is : 38.84
Gas constant in kJ/kg K is : 0.214
The pressure of the gas in bar is : 0.569

Example 2.15 Page No : 58

In [15]:
# Variables
p = 0.98;   			# in bar
p = p*10**5;			# in N/m**2
v = 1000;	    		# in m**3
T = 27+273;		    	# in K
g = 9.8;
M = 2;
Ro = 8314;			    # in J/kg-mole K

# Calculations
R = Ro/M;		    	# in kg K
m = p*v/(R*T);			# in kg
W = m*g;			    # in N

# Results
print "The load that can be lifted with the air of aerostat in N is : %.2f"%W

The load that can be lifted with the air of aerostat in N is : 770.11

Example 2.16 Page No : 59

In [16]:
import math 
from scipy.integrate import quad 

# Variables
T1= 500;			# in K
T2= 2000;			# in K
m=1;			# in kg

# Calculations
def f2(T): 
    return 11.515-172/math.sqrt(T)-1530/T

delta_H=m*  quad(f2,T1,T2)[0]

# Results
print "Change in enthalpy in kcal/kg mole is : %.2f"%delta_H

Change in enthalpy in kcal/kg mole is : 7459.40

Example 2.17 Page No : 59

In [1]:
from scipy.misc import derivative

# Variables			
duBydt= 0.718;

# Calculations

def f1(t):
    return 196. + 0.178 * t

Cv = round(derivative(f1,1.0, dx=1e-6),3)

def f2(t):
    return  273.351 + 1.005*t

Cp = round(derivative(f2,1.0, dx=1e-6),3)
# Results
print "The value of Cv in kJ/kg-K is : ",Cv
print "The value of Cp in kJ/kg-K is : ",Cp

The value of Cv in kJ/kg-K is :  0.178
The value of Cp in kJ/kg-K is :  1.005