Chapter7-Thermodynamics and thermochemistry

Ex1-pg205

In [1]:
import math
##Intitalisation of variables

v= 1. ##cc
p= 1.013*10**6 ##dyne cm^-2
r= 4.184*10**7 ##ergs
##CALCULATIONS
W= v*p/r
##RESULTS
print'%s %.2f %s'% ('Work done = ',W,' cal')

Work done =  0.02  cal

Ex2-pg210

In [2]:
import math
##Intitalisation of variables

R= 8.314 ##J/mole K
n= 1. ##mole
v1= 10. ##lit
v2= 20. ##lit
T= 25. ##C
##CALCULATIONS
W= R*10**7*(273.2+T)*math.log(v2/v1)
##RESULTS
print'%s %.2e %s'% ('Maximum work done = ',W,' ergs mole^-1')

Maximum work done =  1.72e+10  ergs mole^-1

Ex3-pg216

In [3]:
import math
##Intitalisation of variables

T= 18. ##C
n1= 7.5 
n2= 3.
n3= 6.
R= 2.*10**-3 ##kcal
dH= -783.4 ##kcal
##CALCULATIONS
dE= dH+R*(273.+T)*(n2+n3-n1)
##RESULTS
print'%s %.2f %s'% ('Heat of the reaction = ',dE,' kcal')

Heat of the reaction =  -782.53  kcal

Ex4-pg219

In [4]:
import math
##Intitalisation of variables

dH= -256.2 ##kcal
hf= -98.3 ##kcal
n= 6.
##CALCULATIONS
x= n*hf-dH
##RESULTS
print'%s %.2f %s'% ('Heat of the formation = ',x,' kcal')

Heat of the formation =  -333.60  kcal

Ex5-pg219

In [5]:
import math
##Intitalisation of variables

dH= -327. ##kcal
n1= 2. ##moles
n2= 3. ##moles
hf= 68.4 ##kcal
hf1= 94. ##kcal
##CALCULATIONS
x= -n1*hf1-n2*hf-dH
##RESULTS
print'%s %.2f %s'% ('Heat of the formation = ',x,' kcal')

Heat of the formation =  -66.20  kcal

Ex6-pg224

In [6]:
import math
##Intitalisation of variables

n= 5. ##moles
h1= 10.55 ##kcal
h2= -18.69 ##kcal
##CALCULATIONS
dH= h2-n*h1
##RESULTS
print'%s %.2f %s'% ('Heat of the hydration = ',dH,' kcal')

Heat of the hydration =  -71.44  kcal

Ex7-pg226

In [7]:
import math
##Intitalisation of variables

cp= 18. ##cal/deg
co2= 6.97 ##cal/deg
ch2= 6.89 ##cal/deg
T1= 25. ##C
T2= 100. ##C
dH1= -68.4 ##kcal
##CALCULATIONS
dCp= (cp-(co2*0.5+ch2))*10**-3
dH2= dH1+(T2-T1)*dCp
##RESULTS
print'%s %.2e %s'% ('dCp = ',dCp,' kcal deg^-1')
print'%s %.2f %s'% ('\n Heat of formation = ',dH2,' kcal') 

dCp =  7.62e-03  kcal deg^-1

 Heat of formation =  -67.83  kcal

Ex8-pg227

In [8]:
import math
##Intitalisation of variables

k1= 9.92 ##kcal/deg
T2= 125. ##C
T1= 25. ##C
k2= 1.15*10**-3 ##kcal deg^-2
k3= 3.4*10**-6 ##kcal deg^-3
dH1= -22.1 ##kcal
##CALCULATIONS
dH= 10**-3*(-k1*(T2-T1)-k2*((273.+T2)**2-(273.+T1)**2)+k3*((273.+T2)**3-(273.+T1)**3))
dH2= dH1+dH
##RESULTS
print'%s %.2f %s'% ('dH2-dH1 = ',dH,' kcal')
print'%s %.2f %s'% ('\n Heat of reaction = ',dH2,' kcal')

dH2-dH1 =  -0.95  kcal

 Heat of reaction =  -23.05  kcal

Ex9-pg228

In [9]:
import math
##Intitalisation of variables

a= -9.92
b= -2.3*10**-3
c= 10.2*10**-6
T= 25. ##C
dH= -22100. ##cal
##CALCULATIONS
dH1= dH-(a*(273.+T)+b*0.5*(273.+T)**2+c*0.33*(273.+T)**3)
##RESULTS
print'%s %.2f %s'% ('Heat of reaction = ',dH1+1,' cal')

Heat of reaction =  -19129.79  cal

Ex10-pg229

In [10]:
import math
##Intitalisation of variables

m= 1.247 ##gm
hc= 2745. ##cal deg^-1
mw= 122.12 ##gm
dT= 2.87 ##C
##CALCULATIONS
mh= dT*hc*mw/(m*1000.)
##RESULTS
print'%s %.2f %s'% ('molar heat of combustion of benzoic acid = ',mh,' kcal mole^-1')
 

molar heat of combustion of benzoic acid =  771.52  kcal mole^-1