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)
```

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)
```

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)
```

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)
```

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
```

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)
```

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)
```

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)
```

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)
```

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)
```

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)
```

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)
```

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)
```

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)
```

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)
```

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)
```

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)
```

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"
```

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)
```

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"
```

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)
```

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)
```

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"
```

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."
```

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"
```

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)
```

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)
```