# Chapter 6 : Change of state¶

## Example 6.1 page no : 194¶

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)

the change in melting point of water is -0.00748


## Example 6.2 page no : 195¶

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."

the latent heat of vapourisation is 539.0 cal
Note: Answer is slightly different because of rounding error.


## Example 6.3 page no : 195¶

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)

the value of k is 8.4e+07 ergs/mol.deg


## Example 6.4 pageno : 196¶

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)

temperature for the triple point is 0.007188 deg.C


## Example 6.5 pageno : 196¶

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)

the slope of vapourisation curve is 4.45e+02 dyne/sq.cm/deg.C
the slope of sublimation curve is 5.03e+02 dyne/sq.cm/deg.C