# Chapter 4 : Kinetic theory of gases¶

## Example 4.1 page no : 137¶

In [1]:
# Variables
t = 273;			    #temperture of the oxygen molecule in K
m = 32;		        	#molecular mass of the gas in gm
r = 8.32*10**7;			#molar gas consmath.tant in ergs per mole
v2 = 33200.;			#velocity of the gas in cm/sec

# Calculations
v1 = ((3*r*t)/m)**(1./2);			#rms velocity of the molecule in cm/s
T = ((v2*v2*m)/(3*r));			#temperature of the molecule with sound has velocity in K

# Result
print 'the rms velocity of the molecule is %.2e cm/s  \
\nthe temperature of the molecule is %3.0f K'%(v1,T)

the rms velocity of the molecule is 4.61e+04 cm/s
the temperature of the molecule is 141 K


## Example 4.2 page no : 137¶

In [2]:
# Variables
t1 = 308.;			#temperature of the nitrogen molecule in K
m1 = 28.;			#molecular mass of the nitrogen in gm
m2 = 2.;			#molecular mass of the hydrogen molecule in gm

# Calculations
t2 = (t1*m2/m1);			#temperature of the hydrogen molecule in K

# Result
print 'the temperature of the hydrogen molecule is %3.0fK'%(t2)

the temperature of the hydrogen molecule is  22K


## Example 4.3 pageno : 138¶

In [3]:
# Variables
y = 0.00129;			#density of the air in gm/cc
p = 76;			#pressure of the nitrogen molecule in cm
g = 981;			#accelaration due to gravity in cm/sec**2
m = 13.6;			#density of the mercury in gm/cc

# Calculations
v = ((3*p*g*m)/y)**(1./2);			#rms velocity of air at ntp in cm/sec

# Result
print 'the rms velocity of the air is %.2e cm/sec'%(v)

the rms velocity of the air is 4.86e+04 cm/sec


## Example 4.4 pageno : 138¶

In [5]:
# Variables
d = 16*0.000089;			#density of the oxygen molecule in gm/cc
p = 76;			#pressure of the air in cm
g = 981;			#gravitaitonal accelaration in cm/sec**2
m = 13.6;			#density of the mercury in gm/cc

# Calculations
v = ((3*p*g*m)/d)**(1./2);			#velocuty of the oxygen molecule in cm/sec

# Result
print 'velocity of oxygen molecule is %.2e cm/sec'%(v)

velocity of oxygen molecule is 4.62e+04 cm/sec


## Example 4.5 pageno : 138¶

In [7]:
# Variables
t = 273;			#temperature of the hydrogen molecule in K
n = 6.03*10**23;			#1 mole of hydrogen molecules
r = 8.31*10**7;			#universal gas consmath.tant in erg/K/mole

# Calculations
e = (1.5*r*t)/n;			#kinetic energy of the hydrogen molecule in erg

# Result
print "the kinetic energy of the hydrogen molecule is %.2e erg"%e

the kinetic energy of the hydrogen molecule is 5.64e-14 erg


## Example 4.6 page no : 138¶

In [9]:
# Variables
m = 1.;			#mass of the oxygen in gm
r = 8.31*10**7;			#universal gas consmath.tant in erg/K/mole
t = 320.;			#temperature of the oxygen in K
#for 1gm mole k.e is 1.5rt then for 1 gm oxygen (1/32)(k.e)

# Calculations
e = (m/32)*(3*r*t/2);			#kinetic energy of the oxygen in erg

# Result
print 'the kinetic energy of the oxygen is %.2e erg'%(e)

the kinetic energy of the oxygen is 1.25e+09 erg


## Example 4.7 pageno : 138¶

In [11]:
# Variables
t = 273;			#temperature at ntp in K
#rms velocity of oxygen is 3/2 times its rms velocity at ntp then e1 = (3/2)*e

# Calculations
t1 = (9.*t/4.);			#temperature of the oxygen molecule in K

# Result
print 'temperature of the oxygen in %3.1f K'%(t1)

temperature of the oxygen in 614.2 K


## Example 4.8 page no : 139¶

In [12]:
# Variables
p = 10;			#pressure of the gas in atm
v = 5000;			#volume of the gas in ml
l = 76;			#length of the mercury in barometer in cm
g = 981;			#accelaration due to gravity in cm/sec**2
d = 13.6;			#density of the mercury in gm/cc

# Calculations
e = 3*p*v*l*g*d;			#kinetic energy of the molecule in ergs

# Result
print 'the kinetic energy of the molecule is %.2e ergs'%(e)

the kinetic energy of the molecule is 1.52e+11 ergs


## Example 4.9 page no : 139¶

In [13]:
# Variables
t = 323;			#temperature of the hydrogen molecule in K
m1 = 1;			#mass of the hydrogen molecule in gm
m2 = 2;			#molecular weight of the hydrogen in gm
r = 8.3*10**7;			#universal gas consmath.tant in erg/K/mole

# Calculations
e = (m1*r*t*3/(m2*2));			#kinetic enrgy of the hydrogen molecule in ergs

# Result
print 'the kinetic energy of the molecule is %.0e ergs'%(e)

the kinetic energy of the molecule is 2e+10 ergs


## Example 4.10 page no : 139¶

In [14]:
# Variables
t1 = 273;			#temperature of the hydrogen molecule at n.t.p in K

# Calculations
#rms value of hydrogen molecule is double to its rms value at n.t.p, so 3rt/m = 4(3rt/m)
t2 = 4*t1;			#temperature of the hydrogen molecule in K

# Result
print 'the temperature of the hydrogen molecule is %.f K'%(t2)

the temperature of the hydrogen molecule is 1092 K


## Example 4.11 page no : 139¶

In [16]:
# Variables
t1 = 273.;			#temperature of the hydrogen molecule in K
t2 = 373;			#temperature of the hydrogen molecule in K
d = 0.0000896;			#density of the hydrogen molecule in gm/cc
p = 76*13.6*981;			#pressure of the hydrogen molecule in gm/cm/sec**2

# Calculations
v0 = (3*p/d)**(0.5);			#rms velocity at 0deg.C
v100 = v0*(t2/t1)**(0.5);			#rms velocity at 100deg.C

# Result
print 'the rms velocity at 0deg.C is %.2e cm/sec  \
\nthe rms velocity at 100deg.C is %.3e cm/sec'%(v0,v100)

the rms velocity at 0deg.C is 1.84e+05 cm/sec
the rms velocity at 100deg.C is 2.154e+05 cm/sec


## Example 4.12 page no : 140¶

In [17]:
# Variables
cp = 6.84;			#specific heat at consmath.tant pressure in cal/gm mole/deg.C
r = 8.31*10**7;			#universal gas constant in ergs/gm mole/deg.C
v = 130000;			#velocity of sound in cm/sec
j = 4.2*10**7;			#joules constant in ergs/cal

#CALCULATION
cv = cp-(r/j);			#specific heat at constant volume in gm-mole/deg.C
y = (cp/cv);			#index of co-efficient
v1 = (3/y)**(0.5)*v;			#rms velocity in cm/sec

# Result
print 'the rms velocity of gas molecule is %.3e cm/sec'%(v1)

the rms velocity of gas molecule is 1.898e+05 cm/sec


## Example 4.13 page no : 140¶

In [18]:
# Variables
t = 300;			#temperature of the oxygen molecule in K
n = 6.02*10**23;			#avagdrao's number
m = 32/n;			#mass of each molecule in oxygen
k = 1.38*10**(-16);			#boltzmann consmath.tant in erg/deg

# Calculations
v = (8*k*t/(3.14*m))**(0.5);			#average velocity of oxygen molecule in cm/sec
v2 = v*0.022384;			#velocity in miles/hrs

# Results
print 'the avg velocity of  oxygen molecule is %.f miles/hour'%(v2)

the avg velocity of  oxygen molecule is 997 miles/hour


## Example 4.14 page no : 140¶

In [19]:
# Variables
v1 = 2.4;			#velocity of first particle in km/sec
v2 = 2.6;			#velocity of second particle in km/sec
v3 = 3.7;			#velocity of third particle in km/sec

# Calculations
rv = ((v1**2+v2**2+v3**2)/(3))**(0.5);			#rms velocity of the particles in km/sec
mv = (v1+v2+v3)/(3);			                #mean velocity of the particles in km/sec
r = rv/mv;			                            #ratio of the rms to mean velocity

# Results
print 'the ratio of rms to mean velocity is %3.3f'%(r)

the ratio of rms to mean velocity is 1.019


## Example 4.15 pageno : 141¶

In [20]:
# Variables
n = 2.76*10**19;			#no.of molecules per cc
d = 3.36*10**(-8);			#diameter of the helium molecule in cm

# Calculations
mf = 1/((2**(0.5))*3.14*(d**2)*n)

# Result
print 'the mean free path of the hydrogen molecue is %.2e cm'%(mf)

the mean free path of the hydrogen molecue is 7.23e-06 cm


## Example 4.16 page no : 141¶

In [3]:
# Variables
n = 85*10**(-6);			#coefficent of vismath.cosity in dynes/cm**2/velocity gradient
c = 16*10**4;			#velocity in cm/sec
p = 0.000089;			#density in gm/cc
N = 6.06*10**23/22400;			#avagadro number
a = (2)**(0.5)*(22./7);			#constant

# Calculations
mf = (3*n/(p*c));			#mean free path in cm
cr = c/mf;			#collision rate
d = (1/(a*N*mf))**(0.5);			#molecular diameter of hydrogen gas in cm

# Results
print 'the mean free path is %.2e cm  \
\nthe collision rate is %.1e  \
\nthe molecular diameter of hydrogen gas is %.1e cm'%(mf,cr,d)

the mean free path is 1.79e-05 cm
the collision rate is 8.9e+09
the molecular diameter of hydrogen gas is 2.2e-08 cm


## Example 4.17 page no : 141¶

In [22]:
# Variables
d = 2*10**(-8);			#diameter of the molecule in cm
k = 1.38*10**(-6);			#boltzmann constant in ergs/deg
t = 273;			#temperature at ntp in K
p = 76*13.6*981;			#pressure at ntp in gm/cm/sec**2

# Calculations
mf = ((k*t)/(2**(0.5)*3.14*(d**2)*p));			#mean free path in cm

# Result
print 'mean free path at ntp is %.1e cm'%(mf)

mean free path at ntp is 2.1e+05 cm


## Example 4.18 page no : 141¶

In [2]:
# Variables
t = 288;			#temperature in K
k = 1.38*10**(-16);			#boltzmann constant in erg/deg
N = 6.02*10**23;			#avagadro number
m = 32/N;			#mass of each oxygen molecule in gm
v = 196*10**-6;			#viscosity in poise

# Calculations
av = ((8*k*t/(3.14*m))**0.5);			#average velocity in cm/sec
d = (m*av/(3*3.14*2**(0.5)*v))**0.5;			#diameter of the molecule in cm

# Results
print 'diameter of the molecule is %.1e cm'%(d)
print "Note : answer is slightly different because of rounding error"

diameter of the molecule is 3.0e-08 cm
Note : answer is slightly different because of rounding error


## Example 4.19 page no : 142¶

In [27]:
# Variables
mf = 15;			#mean free path in cm
t = 300;			#temperature of oxygen molecule in K
d = 3*10**(-8);			#diameter of the molecule in cm
N = 6.02*10**23;			#avagadro number
r = 8.32*10**7;			#universal gas constant in ergs/mole/deg
a = (2**(0.5))*(22./7);

#CALCULATIONS
p = (r*t)/(N*a*(d**2)*mf);			#pressure of the oxygen molecule in dynes/sq.cm

# Result
print 'the pressure of the oxygen molecule is %3.3f dynes/sq.cm'%(p)

the pressure of the oxygen molecule is 0.691 dynes/sq.cm


## Example 4.20 pageno : 142¶

In [30]:
# Variables
k = 5.64*10**-14;			#kinetic energy of the hydrogen molecule ergs
t = 273;			#temperature of the oxygen molecule in K
r = 8.32*10**7;			#universal gas constant in ergs

# Calculations
N = (3./2)*(r*t/k);			#avagadro number

# Result
print 'the avagadro number is %.2e'%(N)
print "Note : answer is slightly different because of rounding error"

the avagadro number is 6.04e+23
Note : answer is slightly different because of rounding error


## Example 4.21 pageno : 143¶

In [5]:
# Variables
q = 5000;			#total number of molecules
e = 2.7183;			#constant value
t1 = 0.5;			#distance travled to the mean free path
t2 = 1;			#distance travelled to the mean free path

#CALCULATONS
p1 = q*(e**-t1);			#n0.of molecules having no collision in traversing a dismath.tance t1
p2 = q*(e**-t2);			#n0.of molecules having no collision in traversing a dismath.tance t2

#OUPUT
print 'the no. of molecules having no collision in traversing distance equal to 0.5 times the mean free path is %.f  \
\nthe no. of molecules having no collision in traversing a distance equal to 1 time the mean free path is %.f'%(p1,p2)

the no. of molecules having no collision in traversing distance equal to 0.5 times the mean free path is 3033
the no. of molecules having no collision in traversing a distance equal to 1 time the mean free path is 1839


## Example 4.22 page no : 143¶

In [32]:
# Variables
t = 38380;			#temperature of the molecule in K
k = 1.38*10**-16;			#boltzman consmath.tant of one electron in ergs/K
e = 1.6*10**-12;			#charge of one electron volts

#CALCULATIOS
mk = 1.5*k*t/e;			#mean kinetic energy per atom in ev

# Result
print 'the mean kinetic energy of the molecule is %3.2f ev'%(mk)

the mean kinetic energy of the molecule is 4.97 ev


## Example 4.23 pageno : 143¶

In [36]:
# Variables
v = 1.7*10**-4;			#vismath.cosity of the air molecule in cgs
d = 0.00129;			#density of the molecule in gm/ml
p = 76*13.6*981;			#pressure of the molecule in gm/cm/sec**2

# Calculations
r = (3*p/d)**(0.5);			#rms velocity of the molecule in cm/sec
mf = (3*v/(d*r));			#mean free path in cm
cf = r/mf;			#collision frequency

# Result
print 'the mean free path is %.1e cm  \
\nthe collision frequency is %.e'%(mf,cf)
print "Note : answer is slightly different because of rounding error"

the mean free path is 8.1e-06 cm
the collision frequency is 6e+09
Note : answer is slightly different because of rounding error


## Example 4.24 page no : 143¶

In [40]:
# The pressure of the gas

# Variables
t2 = 296.4;			#temperature of the first plate in K
t1 = 304.7;			#temperature of the second plate in K
f = 1.6*10**-2;			#force repelled cold is dynes/sq.cm

# Calculations
p = (4*f*t2/(t1-t2));			#pressure of the gas in dynes/sq.cm

# Result
print 'the pressure of the gas is %3.3f dynes/sq.cm'%(p)
print "Note : mistake in answer in book. Please calculate manually."

the pressure of the gas is 2.285 dynes/sq.cm
Note : mistake in answer in book. Please calculate manually.


## Example 4.25 page no : 144¶

In [42]:
# Variables
mf = 28.5*10**-6;			#mean free path in cm
d = 0.000178;			#density of helium in gm/ml
m = 6*10**-24;			#mass of the helium atom in gm
a = (2**(0.5))*3.14;			#constant

# Calculations
d = (m/(a*d*mf))**(0.5);			#diameter of the size in cm

# Result
print 'the size of the helium atom is %.2e cm'%(d)
print "Note : answer is slightly different because of rounding error"

the size of the helium atom is 1.63e-08 cm
Note : answer is slightly different because of rounding error


## Example 4.26 page no : 144¶

In [43]:
# Variables
a1 = 0*10**-4;			#first horizontal print lacement in cm
a2 = 5.6*10**-4;			#second horizontal print lacement in cm
a3 = -4.7*10**-4;			#third horzontal print lacement in cm
a4 = -10.8*10**-4;			#fourth horizontal print lacement in cm
a5 = 6.6*10**-4;			#fifth horizontal print lacement print lacement in cm
a6 = -9.8*10**-4;			#sixth horizontal print lacement in cm
a7 = -11.2*10**-4;			#7th horizontal print lacement in cm
a8 = -4.0*10**-4;			#8th horizontal print lacement in cm
a9 = 15.0*10**-4;			#9thhorizontal print lacement in cm
a10 = 19.1*10**-4;			#10th horizontal print lacement in cm
a11 = 16.0*10**-4;			#11ht horizontal print lacement in cm
T = 293;			#temperature of the particle in K
v = 0.01;			#viscosity in cgs
r = 1.15*10**-5;			#radius of the particle in cm
R = 8.32*10**7;			#universal gas constant in kj/kg mole
t = 30;			#time for observation of each in sec

# Calculations
x = (a1**2+a2**2+a3**2+a4**2+a5**2+a6**2+a7**2+a8**2+a9**2+a10**2+a11**2)/11
n = R*T*t/(x*3*3.14*v*r);			#no.of molecules in the observation

# Result
print 'the value of n is %.1e'%(n)

the value of n is 5.7e+23


## Example 4.27 page no : 144¶

In [45]:
# Variables
m = 6.*10**-24;			#mass of the helium atom in gm
k =  1.38*10**-16;			#boltzmann consmath.tant in erg
t1 = 100.;			#temperature in K
t2 = 900.;			#temperature in K

# Calculations
r = (t1/t2)**(3./2)*(2.7183**(m*(1./(2*k))*10**8*(1-(1./9))));			#fractional change in the no.of helium atoms

#OUPUT
print 'the fractional change in the no.of helium atoms %.3f'%(r)

the fractional change in the no.of helium atoms 0.256