In [2]:

```
# Variables
vl = 1; #volume of water in cc
vs = 1.0908; #volume of ice in cc
t = 273; #temperature in k
p = 76*13.6*981; #pressure in dynes/sq.cm
l = 80; #latent heat of fusion in cal
j = 4.2*10**7; #joules consmath.tant in erg/cal
# Calculations
v = vl-vs; #change in volume
T = (v*t*p)/(j*l); #change in melting point of water
# Result
print 'the change in melting point of water is %.5f'%(T)
print "there is wrong answer printed in book. Please calculate manually."
```

In [6]:

```
# Variables
vv = 1674.; #volume of vapour in cc
vl = 1.; #volume of liquid in cc
p = 760.; #pressure of steam and water in mm
t = 373.; #temperature in K
p1 = 27.12; #superincumbent pressure in mm
# Calculations
v = vv-vl; #change in volume
l = (v*p1*t*0.024203/(p)); #latent heat of vapourisation in cal
# Result
print 'the latent heat of vapourisation is %3.1f cal'%(l)
print "Note: Answer is slightly different because of rounding error."
```

In [7]:

```
# Variables
m = 1./(342*100); #molar concentration of water
t = 289.; #temperature in K
p = 53.5*13.6*981; #pressure in dynes/sq.cm
# Calculations
k = p/(t*m); #the value of k in ergs/mol.deg
# Result
print 'the value of k is %.1e ergs/mol.deg'%(k)
```

In [8]:

```
# Variables
p1 = 4.60; #presure at 0deg.C in mm per deg.C
p2 = 4.94; #pressure at 1deg.C in mm per deg.C
t = 0.0072; #lowering the melting point in deg.C
t1 = 7.1563979*10**(-3); #rise in melting point in deg.C
p = 760; #atmospheric pressure in mm hg
# Calculations
dp = p2-p1; #rate of increase of pressure in mm per deg.C
p3 = (t1*p)/t; #pressure in mm
dt = (755.4-p3)/dp; #tmperature for the triple point in deg.C
# Result
print 'temperature for the triple point is %3.6f deg.C'%(dt)
```

In [10]:

```
# Variables
v = 21*10**4; #change in volume from vapour to liquid in cc
Ls = 687; #latent heat of sublimation in cal
lv = 607; #latent heat of vapourisation in cal
t = 273; #temperature of water in deg.C
j = 4.2*10**7; #joules constant in ergs/cal
# Calculations
sv = lv*j/(t*(v)); #slope of vapourisation curve at 0 deg.C in dyne/sq.cm/deg.C
ss = Ls*j/(t*(v)); #slope of sublimation curve at 0 deg.C in dyne/sq.cm/deg.C
# Result
print 'the slope of vapourisation curve is %.2e dyne/sq.cm/deg.C \
\nthe slope of sublimation curve is %.2e dyne/sq.cm/deg.C'%(sv,ss)
```