Chapter5 Second and Third Law of Thermodynamics

Example 5.1, Page no.20

In [4]:
import math
#initialisation of variables
q2=1000      #in cal
T2=100       #in C
T1= 20       #in C
#CALCULATIONS
wmax= q2*(T2-T1)/(273.1+T2)
#RESULTS
wmax=round(wmax,2)
print 'maximum work=',wmax,'cal'

maximum work= 214.42 cal

Example 5.2, Page no.20

In [5]:
import math
#initialisation of variables
dH= 6896   #in cal moleˆ−1
T= 68.7    #in C
#CALCULATIONS
dS= dH/(273.1+T)
#RESULTS
dS=round(dS,2)
print 'entropy change per mole=',dS,'cal degˆ−1 moleˆ−1'

entropy change per mole= 20.18 cal degˆ−1 moleˆ−1

Example 5.3, Page no.21

In [6]:
import math
#initialisation of variables
Cp=6.09    #in cal degˆ−1 moleˆ−1
T1=30      #in C
T2=0       #in C
#CALCULATIONS
k=0.0452799815  #log10((273+T1)/(273+T2)))
dS=2.303*Cp*k
#RESULTS
dS=round(dS,2)
print 'increase in entropy=',dS,'cal degˆ−1 mole ˆ−1'

increase in entropy= 0.64 cal degˆ−1 mole ˆ−1

Example 5.4, Page no.21

In [7]:
import math
#initialisation of variables
T1=25    #in C
T2=600   #in C
k1= 6.0954
k2= 3.2533*10**-3      #in K
k3= -10.71*10**-7      #in Kˆ−1
#CALCULATIONS
dS=k1*2.303*math.log10((273+T2)/(273+T1))+k2*(T2-T1)+(k3 /2)*((273+T2)**2-(273+T1)**2)
#RESULTS
dS=round(dS,2)
print 'increase in entropy=',dS,'cal degˆ−1 mole^−1'

increase in entropy= 5.74 cal degˆ−1 mole^−1

Example 5.5, Page no.22

In [8]:
import math
#initialisation of variables
n=2      #in mole
R=1.987  #in cal Kˆ−1 moleˆ−1
X1=0.5   #in atm
X2=0.5   #in atm
#CALCULATIONS
S=-2.303*n*R*(X1*math.log10(X1)+X2*math.log10(X2))
#RESULTS
S=round(S,2)
print 'change in entropy=',S,'cal degˆ−1 moleˆ−1'

change in entropy= 2.76 cal degˆ−1 moleˆ−1

Example 5.6, Page no.22

In [9]:
import math
#initialisation of variables
SH2O= 45.106   #in cal degˆ−1 moleˆ−1
SH2= 31.211    #in cal degˆ−1 moleˆ−1
SO2= 49.003    #in cal degˆ−1 moleˆ−1
#CALCULATIONS
dS= SH2O-SH2 -0.5*SO2
#RESULTS
dS=round(dS,2)
print 'change in entropy=',dS,'cal degˆ−1 moleˆ−1'

change in entropy= -10.61 cal degˆ−1 moleˆ−1

Example 5.7, Page no.23

In [10]:
import math
#initialisation of variables
n=2           #in moles
p=1           #in atm
p1=0.1        #in atm
T=25          #in C
R= 1.987      #in cal moleˆ−1 Kˆ−1
#CALCULATIONS 
dG= n*R*2.303*math.log10(p1/p)*(273+T)
#RESULTS
dG=round(dG,2)
print 'change in Gibbs free energy=',dG,'cal'

change in Gibbs free energy= -2727.33 cal

Example 5.8, Page no.23

In [11]:
import math
#initialisation of variables
R=1.987   #in cal moleˆ−1 Kˆ−1
T=-10     #in C
P1=2.149  #in mm
P2=1.950  #in mm
#CALCULATIONS
dG=R*2.303*(273+T)*math.log10(P2/P1)
#RESULTS
dG=round(dG,2)
print 'change in Gibbs free energy=',dG,'cal mole ˆ−1'

change in Gibbs free energy= -50.79 cal mole ˆ−1

Example 5.9, Page no.23

In [12]:
import math
#initialisation of variables
T=100       #in C
R=1.987     #in cal moleˆ−1 Kˆ−1
H=539.7     #in cal gˆ−1
M=18        #in g moleˆ−1
#CALCULATIONS
w=-R*(273+T)
qp=-H*M
dE=qp-w
dA=-w
dS=qp/(273+T)
dG=qp -(273+T)*dS
#RESULTS
dS=round(dS,2)
print 'W=',w,'cal moleˆ−1'
print 'qp=',qp,'cal moleˆ−1'
print 'dE=',dE,'cal moleˆ−1'
print 'dA=',dA,'cal moleˆ−1'
print 'dS=',dS,'cal degˆ−1 moleˆ−1'
print 'dG=',dG,'cal moleˆ−1'

W= -741.151 cal moleˆ−1
qp= -9714.6 cal moleˆ−1
dE= -8973.449 cal moleˆ−1
dA= 741.151 cal moleˆ−1
dS= -26.04 cal degˆ−1 moleˆ−1
dG= 0.0 cal moleˆ−1

Example 5.10, Page no.24

In [13]:
import math
#initialisation of variables
R=1.987     #in cal degˆ−1 moleˆ−1
T=27        #in C
V1=24.62    #in lit
V2=2.462    #in lit
#CALCULATIONS
wmax=2.303*R*(273.1+T)*math.log10(V1/V2)
dA=-wmax
dE=0
q=dE+wmax
dH=0
dG=-R*(273.1+T)*2.303
dS=dG/(273.1+T)
dS1=(dH-dG)/(273.1+T)
#RESULTS
wmax=round(wmax,2)
q=round(q,2)
dA=round(dA,2)
dS1=round(dS1,2)
dG=round(dG,2)
print 'W=',wmax,'cal moleˆ−1'
print 'q=',q,'cal moleˆ−1'
print 'dE=',dE,'cal moleˆ−1'
print 'dA=',dA,'cal moleˆ−1'
print 'dS=',dS1,'cal degˆ−1 moleˆ−1'
print 'dG=',dG,'cal moleˆ−1'

W= 1373.28 cal moleˆ−1
q= 1373.28 cal moleˆ−1
dE= 0 cal moleˆ−1
dA= -1373.28 cal moleˆ−1
dS= 4.58 cal degˆ−1 moleˆ−1
dG= -1373.28 cal moleˆ−1