Chapter 10 : Thermodynamic relations

Example 10.1 pageno: 329

In [1]:
# Variables
t = 289.6;			#temperature in K
dt = 0.0244;			#raise in temperature in deg.C
v1 = 0.00095;			#volume occupied in liquid state in litres
v2 = 0.00079;			#volume occupied in solid state in litres

# Calculations
l = t*(v1-v2)/dt;			#latent heat of fusion in lit.atm

# Result
print 'the latent heat of fusion is %3.2f lit.atm'%(l)
the latent heat of fusion is 1.90 lit.atm

Example 10.2 pageno : 329

In [3]:
# Variables
t = 295.    			#temperature of water in K
dp = 10**6.	    		#cahnge in pressure in dyne/sq.cm
j = 4.2*10**7;			#joules constant in ergs/cal

# Calculations
dc = -t*10**-5*dp/j;			#change in specific heat

# Result
print 'the change in specific heat is %.e cal/degree'%(dc)
the change in specific heat is -7e-05 cal/degree

Example 10.3 pageno: 329

In [4]:
# Variables
cp = 0.0909	    		#specific heat at consmath.tant pressure in cal/degree
t = 273;		    	#temperature in K
v = 0.112;			    #specific volume in lit/deg.C
a = 5.01*10**(-6);		#coefficient of linear expansion
k = 8*10**-7;			#compressibility of copper in per atoms

# Calculations
cv = cp+(9*a**2*v*t*0.024142*10**3/k);			#specific heat at constant volume in cal/deg.C

# Result
print 'specific heat at constant volume is %3.2f cal/deg.C'%(cv)
specific heat at constant volume is 0.30 cal/deg.C

Example 10.5 pageno : 331

In [5]:
# Variables
t = 289.6;			#temperature in K
dt = 0.0244;			#raise in temperature in deg.C
v1 = 0.00095;			#volume occupied in liquid state in litres
v2 = 0.00079;			#volume occupied in solid state in litres

# Calculations
l = t*(v1-v2)/dt;			#latent heat of fusion in lit.atm

# Result
print 'the latent heat of fusion is %3.3f lit.atm'%(l)
the latent heat of fusion is 1.899 lit.atm

Example 10.6 pageno : 331

In [6]:
# Variables
l = 539;			#latent heat of water at 100deg.C in cal
j = 4.2*10**7;			#joules constant in ergs/cal
t = 373;			#temperature of water in K
v2 = 1670;			#volume of steam formed in cc
v1 = 1;			#intial volume in cc
g = 981;			#acceleration due to gravity in cm/sec**2
d = 13.6;			#specific gravity of hg

# Calculations
dp = l*j/(t*(v2-v1)*g*d);			#rate of change of saturation pressure in cm of mercury

# Result
print 'the rate of change of saturation pressure is %3.1f cm of hg'%(dp)
the rate of change of saturation pressure is 2.7 cm of hg

Example 10.7 pageno : 331

In [8]:
# Variables
p1 = 77.371;			#pressure at 100.5deg.C in cm of hg
p2 = 74.650;			#pressure at 99.5deg.C in cm of hg
g = 981;			#universal gas constant in cm/sec**2
d = 13.6;			#specific gravity
l = 537;			#latent heat of vapourisation in cal/gm
t = 373;			#temperature of water in K
j = 4.2*10**7;			#joules constant in ergs/cal
v1 = 1;			#intial volume in cc

# Calculations
v2 = v1+(l*j/(t*(p1-p2)*g*d));			#volume of gram of steam at 100deg.C in cc

# Result
print 'volume of gram of steam at 100deg.C is %.f cc'%(v2)
volume of gram of steam at 100deg.C is 1667 cc

Example 10.8 pageno : 332

In [9]:
# Variables
t = 350;			#boiling point temperature in K
l = 46;			#latent heat of vapourisation in cal/gm
v1 = 1/1.6;			#intial volume in cc
dp = 2.3;			#change in pressure with temperature in cm of hg/deg.C
d = 13.6;			#specific gravity of mercury
g = 981;			#acceleration due to gravity in cm/sec**2
j = 4.2*10**7;			#joukes constant in ergs/cal

#CALCULTIONS
v2 = v1+(l*j)/(t*dp*d*g);			#specific volume in cc

# Result
print 'specific volume of vapour of carbon is %3.3f cc'%(v2)
print "Note : Answer is slightly different because of rounding error"
specific volume of vapour of carbon is 180.513 cc
Note : Answer is slightly different because of rounding error

Example 10.9 pageno : 332

In [10]:
# Variables
l = 536.;			#latent heat of vapourisation in cal/gm
v1 = 1.;			#volume of 1 gm of water in cc
v2 = 1600.;			#volume of steam in cc
t = 373.;			#boiling point of water in K
p = 1.;			#pressure in cm of hg
d = 13.6;			#specific gravity of mercury
g = 981.;			#gravitational constant in cm/sec**2s/cal
j = 4.2*10**7;			#joules constant in erg/cal

# Calculations
dt = (t*(v2-v1)*d*g)/(l*j);			#change in temperature in deg.C

# Result
print 'change in temperature is %3.2f deg.C'%(dt)
change in temperature is 0.35 deg.C

Example 10.10 page no : 332

In [12]:
# Variables
t = 353;			#temperature in K
p = 76*13.6*981;			#pressure in dynes/sq.cm
v = 0.146;			#specific volume in cc/kg
l = 35.6;			#latent heat of fusion in cal/gm
j = 4.18*10**7;			#joules constant in ergs/cal

# Calculations
dt = t*p*v/(l*j);			#change in melting point per atmosphere

# Result
print 'the rate of change in melting point is %.3f per atmosphere'%(dt)
the rate of change in melting point is 0.035 per atmosphere

Example 10.11 pageno : 333

In [1]:
# Variables
l = 79.6*4.18*10**7;			#latent heat of water in ergs/gm
t = 273.16;			#temperature of water in K
v1 = 1.0001;			#specific volume of water at 0deg.C in cc
v2 = 1.0908;			#specific volume of ice at 0deg.C in cc
p = 1.013*10**6;			#pressure of atmosphere in dyne/sq.cm

# Calculations
dt = t*(v1-v2)*p/l;			#change in freezing point of water in deg.C

# Result
print 'change in freezing point of water is %3.4f deg.C'%(dt)
change in freezing point of water is -0.0075 deg.C