In [1]:

```
# Variables
t = 33.18; #critical temperature in K
pc = 12.80*76*981*13.6; #critical pressure in dynes/sq.cm
r = 83.15; #universal gas constant in kj/kg.K
d = 0.08987; #density of hydrogen in gm/lit
v = 2000/0.08987; #gram molecular volune of hydrogen in cc
# Calculations
b = r*10**6*t/(8*pc); #vanderwaal constant in cm**3/mol
to = 2*27*t*(1-(b/v))/8; #inversion temperature of the hydrogen in K
# Result
print 'the inversion temperature of hydrogen is %3.2f K'%(to)
```

In [2]:

```
# Variables
b = 0.00136; #vanderwaal constant in suv/gm
a = 0.011; #vanderwaal constant in atm(suv)**2/gm**2
r = 0.003696; #universal gas constant in atm(suv)/gm.deg
t = 423; #temperature of steam in K
cp = -0.674/0.024205; #specific heat at 423K in atm(cc)gm(deg)
# Calculations
dt = (-b+(2*a/(r*t)))/cp; #change of temperature per atm drop of pressure in deg/atm
# Result
print 'the change of temperature per atmosphere drop of pressure is %3.7f deg/atm'%(dt)
```

In [2]:

```
# Variables
r = 8.3*10**7; #universal gas constant in ergs/deg.C
a = 1.36*10**6*76*13.6*981; #vanderwaal constant in atm.(suv**2)/(gm**2)
b = 32; #vanderwaal constant in cc
cp = 7.03; #specific heat at constant pressure in cal
j = 4.18*10**7; #joules constant in ergs/cal
t = 273; #temperature of the gas in K
# Calculations
dt = ((2*a/(r*t))-b)*10**6/(cp*j); #change of temperature in atmosphere drop of pressure in deg/atm/cm**3
# Result
print 'the change of temperature in atmosphere drop of pressure is %3.2f deg C/atm/cm**2'%(dt)
```

In [6]:

```
# Variables
u = 1.08;
cp = 8.6; #specific heat in kj/kg.K
j = 4.2; #joules constant in j/cal
p1 = 1*1.013*10**6; #pressure at intial in N/sq.m
p2 = 20*1.013*10**6; #pressure at final in N/sq.m
# Calculations
dh = -u*cp*j*(p1-p2); #change in enthalpy in joules
# Result
print 'the change in enthalpy is %3.3e joules'%(dh)
```

In [7]:

```
# Variables
tc = 5.26; #critical temperature of the helium in K
# Calculations
ti = 27*tc/4; #inversion temperature of the helium in K
# Result
print 'the inversion temperature of the helium is %3.2f K'%(ti)
```

In [9]:

```
# Variables
a = 0.245*10**6*10**6; #vanderwaal constant in cm**4.dyne/mole**2
b = 2.67*10; #vanderwaal constant in cc/mole
r = 2*4.2*10**7; #universal gas constant in ergs/mole.K
# Calculations
ti = 2*a/(b*r); #inversion temperature in K
# Result
print 'inversion temperature of hydrogen is %.f K'%(round(ti,-1))
```

In [11]:

```
# Variables
dp = 50*10**6; #change in pressure in dynes/sq.cm
cp = 7*4.2*10**7; #specific heat constant pressure in ergs/mole.K
a = 1.32*10**12; #vanderwaal constant in cm**4.dyne/mole**2
b = 31.2; #vanderwaal constant in cm**2/mole
t = 300; #inital temperature in K
r = 2*4.2*10**7; #ergs/mole.K
# Calculations
dt = ((2*a/(r*t))-b)*dp/cp; #change in temperature in K
# Result
print 'the change in temperature is %3.1f K'%(dt)
```

In [16]:

```
# Variables
p1 = 1.; #inital pressure in atm
p2 = 51.; #final pressure in atm
t1 = 300.; #inital temperature in K
y = 1.4; #coefficient of expansion
# Calculations
t2 = t1*(p2/p1)**((1-y)/y); #final temperature in K
dt = t1-t2; #drop in temperature in K
# Results
print 'the drop in temperature is %3.2f K'%(dt)
print "Note : answer is slightly different because of rounding error"
```