# Chapter 18 : Partial Molar Properties¶

## Example 18.1 Page No : 429¶

In [1]:
# Variables
import math
k1 = 16.4 			#ml mole**-1
k2 = 2.5 			#ml mole**-2
k3 = -1.2 			#ml mole**-3
m = 1 			#molal

# Calculations
Ov = k1+k2*m+k3*m**2

# Results
print  'Apparent molar volume  = %.1f ml mole**-1'%(Ov)

Apparent molar volume  = 17.7 ml mole**-1


## Example 18.2 Page No : 443¶

In [2]:
# Variables
n = 1    			#mole
n1 = 400 			#mole
T = 25 	    		#C
H1 = 5410 			#cal
H2 = -5020 			#cal

# Calculations
dH = -(H1+H2)

# Results
print  'Heat required to remove the water  = %.f cal'%(dH)

Heat required to remove the water  = -390 cal


## Example 18.3 Page No : 443¶

In [3]:
# Variables
n = 1    			#mole
n1 = 400 			#mole
T = 25 	    		#C
H1 = 23540 			#cal
H2 = -5410 			#cal

# Calculations
dH = -(H1+H2)

# Results
print  'Heat required to remove the water  = %.f cal'%(dH)

Heat required to remove the water  = -18130 cal


## Example 18.4 Page No : 446¶

In [4]:
# Variables
n1 = 1   			#mole
n2 = 400 			#mole
H1 = 5638 			#cal
H2 = 23540 			#cal
L = -1.54 			#cal/mole
l1 = -2.16 			#cal/mole
l2 = 5842 			#cal/mole

# Calculations
Q1 = n2*L+H1+H2
Q2 = n2*l1+2*l2
Q = Q2-Q1

# Results
print  'Heat change  = %.f cal'%(Q)

Heat change  = -17742 cal


## Example 18.5 Page No : 447¶

In [5]:
# Variables
L2 = 6000. 	    		#cal
v = 3.
T = 25.        			#C
T1 = 0. 	    		#C

# Calculations
R = ((L2/(v*4.576))*(T-T1)/((273+T1)*(273+T)))
r = 10**((L2/(v*4.576))*(T-T1)/((273+T1)*(273+T)))

# Results
print  'Ratio  = %.3f '%(R)
print  ' Relative change in mean ionic coefficient  = %.2f '%(r)

Ratio  = 0.134
Relative change in mean ionic coefficient  = 1.36


## Example 18.6 Page No : 450¶

In [6]:
# Variables
L2 = 4120. 			#cal
l = -108. 			#cal mole**-1
L21 = -306. 			#cal mole**-1
n1 = 55.5 			#moles
n2 = 1. 			#mole

# Calculations
Q = L21+L2

# Results
print  'differential heat of solution  = %.f cal mole**-1'%(Q)

differential heat of solution  = 3814 cal mole**-1


## Example 18.7 Page No : 456¶

In [7]:
# Variables
n1 = 2.  			#moles
n2 = 100. 			#moles
Cp1 = 17.9 			#cal deg**-1 mole**-1
Cp2 = 21.78 			#cal deg**-1 mole**-1
T1 = 30 			#C
T2 = 25 			#C
L1 = 5780. 			#cal
L2 = 5410. 			#cal
h = 5620. 			#cal mole**-1
n3 = 3. 			#moles
Cp3 = 16.55 			#cal deg**-1 mole**-1

# Calculations
Cp = n2*Cp1+n1*Cp2
Q = (T2-T1)*Cp
Q1 = (n1*L1+L2)
Q2 = n3*h
dQ = Q2-Q1
dH = Q+dQ
HC = 300*Cp1+n3*Cp3
t = -dH/HC
Tf = T2+t

# Results
print  'Increase in temperature  = %.2f deg'%(t)
print  ' Final temperature  = %.1f deg'%(Tf)

Increase in temperature  = 1.71 deg
Final temperature  = 26.7 deg

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