V1=14 #initial volume of cylinder in m3
V2=9 #final volume of cylinder in m3
P=2000 #pressure during the operation in N/m2
U=(-6000) #internal energy of the system in J
W=-P*(V2-V1) #work done during the operation in J
Q=U-W #energy tranfered in form of heat in J
print"energy tranfered in form of heat is",Q,"J"
R=8.314 #universal gas constant [J/K/mol]
T=300 #temperture for the process [K]
U=0 #change in internal energy [J]
V1=2.28 #initial volume [m3]
V2=4.56 #final volume[m3]
import math
W=2.303*R*T*math.log10(V2/V1) #work done during the process[J]
Q=W #heat lost or gained by the system[J]
print"The heat gained by the system is",round(Q),"J mol^-1"
H=29.2 #latent heat of vaporisation[KJ/mol]
T=332 #temperature of the system[K]
R=8.314 #universal gas constant [J/K/mol]
Qp=H #at constant pressure [KJ]
W=-R*0.001*T #workdone [KJ]
U=Qp+W #change in internal energy[KJ]
print"Heat absorbed by the bromine vapours is",Qp,"KJ"
print"\nWorkdone during the process is",round(W,2),"KJ"
print"\nChange in internal energy of the system is",round(U,2),"KJ"
print"C7H16(l) + 11O2(g) -> 7CO2(g) + 8H2O(l)"
n=-4 #change in no. of moles when reaction proceeds from reactants to products
T=298 #temperature of the process [K]
R=8.314 #universal gas constant [J/K/mol]
Qv=-4800 #heat energy at constant volume [KJ]
U=Qv #change in internal energy of system [KJ]
H=U+n*R*0.001*T #change in enthalpy of the system[KJ]
print"the change in enthalpy of system is",round(H,2),"kJ"
n=1 #number of moles of an given ideal gas
T=298 #temperature for the process[K]
V1=8.3 #initial volume of the ideal gas[m3]
V2=16.8 #final volume of the ideal gas[m3]
R=8.314 #universal gas constant[J#K#mol]
import math
W=-2.303*R*T*math.log10(V2/V1) #[J]
Q=-W #[J]
print"H=U+PV ,where U is change in internal energy which is zero due to isothermal process"
print"PV where V is change in volume of system ,PV=RT & RT==0 since T i.e change in temp is zero for system"
print"Therefore,the change in enthalpy is 0J"
print"The workdone by system is",round(W,1),"J mol^-1"
print"\nThe heat evolved is",round(Q,1),"J mol^-1"
T1=323 #intial temperature of water[K]
T2=373 #final temperature of water[K]
Cp=75.29 #specific heat of water[J/K/mol]
w=100.0 #weight of water[g]
mol_wt=18.0 #molecular weight of water[g/mol]
n=w/mol_wt #no. of moles of water[moles]
H=(n*Cp*(T2-T1))*0.001 #change in enthalpy of water[J]
print"The change in enthalpy of water is",round(H,2),"kJ"
print"SO2 + 0.5O2 -> SO3"
U=-97030 #heat of reaction[J]
n=1-(1+0.5) #change in no. of moles
R=8.314 #universal gas constant[J/K/mol]
T=298 #temperature during the reaction[K]
H=U+n*R*T #change inenthalpy of reaction[J]
print"The change in enthalpy of reaction is",round(H),"J(approx)"
print"i.C(s) + O2(g) -> CO2(g)"
H1=-393.5 #change in enthalpy [KJ/mol]
T1=298 #temperature [K]
n1=0 #change in no. of moles in reaction moving in forward direction
R=0.008314 #universal gas constant [KJ/K/mol]
U1=H1-n1*R*T1 #change in internal energy [KJ]
print"The change in internal energy is",round(U1,1),"KJ/mol"
print"ii.C(s) + 0.5O2 -> CO(g)"
H2=-110.5 #change in enthalpy[KJ/mol]
T2=298 #temperature[K]
n2=1-0.5 #change in no. of moles in reaction moving in forward direction
R=0.008314 #universal gas constant [KJ/K/mol]
U2=H2-n2*R*T2 #change in internal energy [KJ]
print"The change in internal energy is",round(U2,3),"KJ/mol"
print"The standard heat of combustion of"
print"2C6H6(l)+ 15O2(g)-> 12 CO2(g)+ 6 H2O(l)"
print"H1(standard heat of combustion)=-6536 KJ/mol"
H1=-6536 #standard heat of combustion [KJ/mol]
print"C6H6(l)+ 7.5 O2(g)-> 6 CO2(g)+ 6 H2O(l)"
H2=H1/2 #standard heat of combustion[KJ/mol]
print"H2(standard heat of combustion for 1 mole of C6H6)=",H2,"kJ/mol"
print"N2(g)+3H2(g)-> 2NH3(g)"
H=-92.22 #standard heat of reaction [KJ/mol]
H1=H/2 #standard heat of formation of 1 mole [KJ/mol]
print"H(heat of formation of 1 mole of product)=",H1,"kJ mol^-1"
print"C2H5OH(l)+3O2(g)->2CO2(g)+3H2O(l)"
T=298 #temperature during the reaction[K]
Hw=-285.83 #standard heat of formation of liquid water [KJ/mol]
He=-277.69 #standard heat of formation of liquid ethanol[KJ/mol]
Hco2=-393.51 #standard heat of formation of carbon dioxide[KJ/mol]
Ho2=0 #standard heat of formation of oxygen gas[KJ/mol]
H=2*Hco2+3*Hw-He-3*Ho2 #standard heat of reaction
print"H(standard heat of reaction)=",H,"kJ"
print"CO(g)+NO(g)->0.5N2(g)+CO2(g)"
Hrxn=-374 #standard heat of reaction[KJ/mol]
Hno=90.25 #standard heat of formation of NO[KJ/mol]
Hco2=-393.51 #standard heat of formation of CO2[KJ/mol]
Hn2=0 #standard heat of formation of N2[KJ/mol]
T=298 #temperature of reaction [K]
Hco=0.5*Hn2+Hco2-Hno-Hrxn #standard heat of formation of CO[KJ/mol]
print"Hco(standard heat of formation)=",Hco,"kJ mol^-1"
H1=-29.6 #the standard heat of hydrogenation of gaseous propylene to propane[Kcal]
H2=-530.6 #the heat of combustion of propane[Kcal]
H3=-94.0 #the heat of formation of carbon dioxide[Kcal]
H4=-68.3 #the heat of formation of liquid water[Kcal]
print"C3H6(g)+4.5O2(g)->3CO2(g)+3H2O(l)"
H5=(3*H3+4*H4)-(H1+H2)#[Kcal]
print"\n H(standard heat of combustion)=",H5,"Kcal"
print"3C(s)+3H2(g)->C3H6(g)"
H6=-H5+3*H3+3*H4 #[Kcal]
print"\n H(standard heat of formation)=",H6,"Kcal"
H1=-114.1 #standard heat of reaction:2NO(g)+O2(g)->2NO2(g) [KJ/mol]
H2=-110.2 #standard heat of reaction:4NO2(g)+O2(g)->2N2O5(g) [KJ/mol]
H3=180.5 #standard heat of reaction:N2(g)+O2(g)->2NO(g) [KJ/mol]
#reacton:N2(g)+2.5O2(g)->N2O5(g)
H4=(2*H1+H2+2*H3)/2 #standard heat of formation of N2O5[KJ/mol]
print"H(standard heat of formation of N2O5)=",H4,"kJ/mol"
Hc=-5645 #standard enthalpy of combustion of reaction:C12H22O11(s)+12O2(g)->12CO2(g)+11H2O(l) [KJ/mol]
Hf1=-393.51 #standard heat of formation of CO2: C(s)+O2(g)->CO2(g) [KJ/mol]
Hf2=-285.83 #standard heat of formation of H2O: H2(g)+0.5O2(g)->H2O(l) [KJ/mol]
#reaction:12C(s)+11H2(g)+5.5O2(g)->C12H22O11(s)
Hf=12*Hf1+11*Hf2-Hc #[KJ/mol]
print"Hf(standard heat of formation of solid sucrose)=",Hf,"KJ/mol(approx)"
Hf1=-46.11 #standard heat of formation of NH3 at 298K #reaction:0.5N2(g)+1.5H2(g)->NH3(g) [KJ/mol]
Cp1=29.125 #molar heat capacity at constant pressure for N2(g)[J/K/mol]
Cp2=28.824 #molar heat capacity at constant pressure for H2(g)[J/K/mol]
Cp3=35.06 #molar heat capacity at constant pressure for NH3(g)[J/K/mol]
T1=298 #initial temperature[K]
T2=400 #final temperature[K]
Cp=Cp3-0.5*Cp1-1.5*Cp2 #[J/K/mol]
T=T2-T1 #[K]
Hf2=Hf1+Cp*0.001*T #standard heat of formation for NH3 at 400K[KJ/mol]
print"\n Hf2(standard heat of formation for NH3 at 400K =",round(Hf2,3),"kJ/mol"
from scipy.optimize import fsolve
from scipy import integrate
dH_298=-241.82 #Std Heat of formation at 298 K [kJ mol^-1]
dH_298=dH_298*1000 # in [J mol^-1]
T1=298 #[K]
T2=1273 #[K]
def f(T):
Cp_H2g=(29.07-((0.836*10**-3)*T)+((20.1*10**-7)*T**2))
Cp_O2g=25.72+(12.98*10**-3)*T-(38.6*10**-7)*T**2
Cp_H2Og=30.36+(9.61*10**-3)*T+(11.8*10**-7)*T**2
delta_Cp=(Cp_H2Og-(Cp_H2g+(1.0/2.0)*Cp_O2g))
return(delta_Cp)
dHK=integrate.quad(f,T1,T2)
dH_1273=dH_298+dHK[0]
dH_1273=dH_1273/1000
print"Heat of formation of H2O(g) at 1000 C=",round(dH_1273,1),"kJ mol^-1 (APPROXIMATE)"
print"NOTE:"
print"Slight variation in answer,because integration is not done precisely in the book"
print"In the book,it is written as:-7497.46 instead of -7504.3"
H1=435.0 #bond dissociation energy for: CH4->CH3+H [KJ/mol]
H2=364.0 #bond dissociation energy for:CH3->CH2+H [KJ/mol]
H3=385.0 #bond dissociation energy for:CH2->CH+H [KJ/mol]
H4=335.0 #bond dissociation energy for:CH->C+H [KJ/mol]
H=(H1+H2+H3+H4)/4 #the bond energy for C-H bond in CH4 [KJ/mol]
print"\n H(the C-H bond energy in CH4)=",round(H,1),"kJ/mol"
H1=-84.68 #heat of formation : 2C(s)+3H2(g)->C2H6(g) [KJ/mol]
H2=2*716.68 #heat of formation : 2C(s)->2C(g) [KJ]
H3=3*436 #heat of formation : 3H2(g)->6H(g) [KJ]
H4=412 #taking it as bond energy for one C-H bond[KJ/mol]
H=H2+H3-H1 #heat of reaction : C2H6(g)->2C(g)+6H(g) [KJ/mol]
H5=H-6*H4 #bond energy for one C-C bond in ethane bond [KJ/mol]
print"\n Hc-c(bond energy for one C-C bond in ethane bond)=",H5,"kJ/mol"
#MgBr2(s)-->Mg(s)+Br2(l)-->Mg(g)+Br2(l)-->Mg(g)+Br2(g)-->Mg(g)+2Br(g)-->Mg+2(g) + 2e(g) + 2Br(g)-->Mg+2(g) + 2Br-(g)
H1=-524 #enthalpy of formation of MgBr2(s) from its element [KJ/mol]
H2=148 #enthalpy of sublimation of Mg(s) [KJ/mol]
H3=31 #enthalpy of vaporization of Br2(l) [KJ/mol]
H4=193 #enthalpy of dissociation Br2 to 2Br(g) [KJ/mol]
H5=2187 #enthalpy of ionization of Mg(g) to Mg+2(g) [KJ/mol]
H6=-650 #enthalpy of formation of Br-(g) [KJ/mol]
H=-H1+H2+H3+H4+H5+H6 #lattice enthalpy [KJ/mol]
print"\n H(lattice enthalpy of magnesium bromide)=",H,"kJ/mol"
from scipy.optimize import fsolve
from scipy import integrate
dH1_298=-881.25 #[kJ/mol]
dH2_298=43.60 #[kJ/mol]
dH3_298=2*dH2_298 #[kJ/mol]
dH4_298=dH1_298+dH3_298 #[kJ/mol]
dH_heat=-dH4_298*1000 #[J/mol]
def f(T2):
def g(T):
Cp_CO2g=26.0+((43.5*10**-3)*T)-((148.3*10**-7)*T**2)
Cp_H2Og=30.36+((9.61*10**-3)*T)+((11.8*10**-7)*T**2)
Cp_N2g=27.30-((5.23*10**-3)*T)-((0.04*10**-7)*T**2)
sig_nCpf=Cp_CO2g+2*Cp_H2Og+8*Cp_N2g
return(sig_nCpf)
crt=integrate.quad(g,298,T2)
ct=crt[0]-dH_heat
return(ct)
T2=fsolve(f,2)
print "T2,maximum flame temperature is :",round(T2[0],2),"K"
print"Calculation mistake in book,wrongly written as:2250 K"