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diff --git a/Thermodynamics:_A_Core_Course/CH3.ipynb b/Thermodynamics:_A_Core_Course/CH3.ipynb
index b72e6afa..fa64788f 100644
--- a/Thermodynamics:_A_Core_Course/CH3.ipynb
+++ b/Thermodynamics:_A_Core_Course/CH3.ipynb
@@ -27,17 +27,12 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.1\n",
- "#to find energy transfered in form of heat:\n",
- "#Variable declaration\n",
"V1=14 \t\t\t\t#initial volume of cylinder in m3\n",
"V2=9 \t\t\t\t#final volume of cylinder in m3\n",
"P=2000 \t\t\t\t#pressure during the operation in N/m2\n",
"U=(-6000) \t\t\t#internal energy of the system in J\n",
- "#Calculation\t\n",
"W=-P*(V2-V1) \t\t\t#work done during the operation in J\n",
"Q=U-W \t\t\t\t#energy tranfered in form of heat in J\n",
- "#Result\n",
"print\"energy tranfered in form of heat is\",Q,\"J\"\n",
" "
],
@@ -66,19 +61,14 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.2\n",
- "#to find the heat lost or gained by the system:\n",
- "#Variable declaration\n",
"R=8.314 \t\t\t\t#universal gas constant [J/K/mol]\n",
"T=300\t\t\t\t\t#temperture for the process [K]\n",
"U=0 \t\t\t\t\t#change in internal energy [J]\n",
"V1=2.28 \t\t\t\t#initial volume [m3]\n",
"V2=4.56 \t\t\t\t#final volume[m3]\n",
- "#Calculation\n",
"import math\n",
"W=2.303*R*T*math.log10(V2/V1) \t\t#work done during the process[J]\n",
"Q=W \t\t\t\t\t#heat lost or gained by the system[J]\n",
- "#Result\n",
"print\"The heat gained by the system is\",round(Q),\"J mol^-1\""
],
"language": "python",
@@ -106,19 +96,13 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.3\n",
"\n",
- "#to find Q,W,U for 1 mole of bromine vaporizes\n",
- "#Variable declaration\n",
- "#given:\n",
"H=29.2 \t\t\t\t\t#latent heat of vaporisation[KJ/mol]\n",
"T=332 \t\t\t\t\t#temperature of the system[K]\n",
"R=8.314 \t\t\t\t#universal gas constant [J/K/mol]\n",
- "#Calculation\n",
"Qp=H \t\t\t\t\t#at constant pressure [KJ]\n",
"W=-R*0.001*T \t\t\t\t#workdone [KJ]\n",
"U=Qp+W \t\t\t\t\t#change in internal energy[KJ]\n",
- "#Result\n",
"print\"Heat absorbed by the bromine vapours is\",Qp,\"KJ\"\n",
"print\"\\nWorkdone during the process is\",round(W,2),\"KJ\"\n",
"print\"\\nChange in internal energy of the system is\",round(U,2),\"KJ\"\n"
@@ -152,19 +136,14 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.4\n",
- "#to find change in enthalpy of the process:\n",
"\n",
- "#variable declaration\n",
"print\"C7H16(l) + 11O2(g) -> 7CO2(g) + 8H2O(l)\" \n",
"n=-4 \t\t\t\t#change in no. of moles when reaction proceeds from reactants to \t\t\t\tproducts\n",
"T=298 \t\t\t\t#temperature of the process [K]\n",
"R=8.314 \t\t\t#universal gas constant [J/K/mol]\n",
"Qv=-4800 \t\t\t#heat energy at constant volume [KJ]\n",
- "#Calculation\t\n",
"U=Qv \t\t\t\t#change in internal energy of system [KJ]\n",
"H=U+n*R*0.001*T \t\t#change in enthalpy of the system[KJ]\n",
- "#Result\n",
"print\"the change in enthalpy of system is\",round(H,2),\"kJ\"\n"
],
"language": "python",
@@ -193,19 +172,14 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.5\n",
- "#to find the Q,W,H\n",
- "#Variable declaration\n",
"n=1 \t\t\t\t#number of moles of an given ideal gas\n",
"T=298 \t\t\t\t#temperature for the process[K]\n",
"V1=8.3 \t\t\t\t#initial volume of the ideal gas[m3]\n",
"V2=16.8 \t\t\t#final volume of the ideal gas[m3]\n",
"R=8.314 \t\t\t#universal gas constant[J#K#mol]\n",
"import math\n",
- "#Calculation\n",
"W=-2.303*R*T*math.log10(V2/V1) #[J]\n",
"Q=-W \t\t\t\t#[J]\n",
- "#Result\n",
"print\"H=U+PV ,where U is change in internal energy which is zero due to isothermal process\" \n",
"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\" \n",
"print\"Therefore,the change in enthalpy is 0J\" \n",
@@ -242,19 +216,14 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.6\n",
"\n",
- "##to find the change in enthalpy(H) of water \n",
- "#Variable declaration\n",
"T1=323 \t\t\t#intial temperature of water[K]\n",
"T2=373 \t\t\t#final temperature of water[K]\n",
"Cp=75.29 \t\t#specific heat of water[J/K/mol]\n",
"w=100.0 \t\t\t#weight of water[g]\n",
"mol_wt=18.0 \t\t#molecular weight of water[g/mol]\n",
- "#Calculation\t\n",
"n=w/mol_wt \t\t#no. of moles of water[moles]\n",
"H=(n*Cp*(T2-T1))*0.001 \t#change in enthalpy of water[J]\n",
- "#Result\n",
"print\"The change in enthalpy of water is\",round(H,2),\"kJ\"\n"
],
"language": "python",
@@ -282,17 +251,12 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.7\n",
- "#to find the change in enthalpy of \treaction(H)\n",
"print\"SO2 + 0.5O2 -> SO3\"\n",
- "#Variable declaration\n",
"U=-97030 \t\t\t#heat of reaction[J]\n",
"n=1-(1+0.5) \t\t\t#change in no. of moles \n",
"R=8.314 \t\t\t#universal gas constant[J/K/mol]\n",
"T=298 \t\t\t\t#temperature during the reaction[K]\n",
- "#Calculation\t\n",
"H=U+n*R*T \t\t\t#change inenthalpy of reaction[J]\n",
- "#result\n",
"print\"The change in enthalpy of reaction is\",round(H),\"J(approx)\"\n"
],
"language": "python",
@@ -321,30 +285,22 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.8\n",
- "#to find the change in internal energy (U) of given reaction\n",
"print\"i.C(s) + O2(g) -> CO2(g)\"\n",
- "#Variable declaration\n",
"H1=-393.5 \t\t#change in enthalpy [KJ/mol]\n",
"T1=298 \t\t\t#temperature [K]\n",
"n1=0 \t\t\t#change in no. of moles in reaction moving in forward direction\n",
"R=0.008314 \t\t#universal gas constant [KJ/K/mol]\n",
"\n",
- "#Calculation\t\n",
"U1=H1-n1*R*T1 \t\t#change in internal energy [KJ]\n",
"print\"The change in internal energy is\",round(U1,1),\"KJ/mol\"\n",
"\n",
- "#to find the change in internal energy (U) of given reaction\n",
"print\"ii.C(s) + 0.5O2 -> CO(g)\" \n",
- "#Variable declaration\n",
"H2=-110.5 \t\t#change in enthalpy[KJ/mol]\n",
"T2=298 \t\t\t#temperature[K]\n",
"n2=1-0.5 \t\t#change in no. of moles in reaction moving in forward direction\n",
"R=0.008314 \t\t#universal gas constant [KJ/K/mol]\n",
"\n",
- "#Calculation\n",
"U2=H2-n2*R*T2 \t\t#change in internal energy [KJ]\n",
- "#result\n",
"print\"The change in internal energy is\",round(U2,3),\"KJ/mol\""
],
"language": "python",
@@ -375,17 +331,12 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.9\n",
- "#to find the standard heat of rxn for combustion of 1 mole of C6H6\n",
"print\"The standard heat of combustion of\"\n",
"print\"2C6H6(l)+ 15O2(g)-> 12 CO2(g)+ 6 H2O(l)\" \n",
"print\"H1(standard heat of combustion)=-6536 KJ/mol\" \n",
- "#Variable declaration\n",
"H1=-6536 \t\t\t#standard heat of combustion [KJ/mol]\n",
- "#Calculation\t\n",
"print\"C6H6(l)+ 7.5 O2(g)-> 6 CO2(g)+ 6 H2O(l)\" \n",
"H2=H1/2 \t\t\t#standard heat of combustion[KJ/mol]\n",
- "#Result\n",
"print\"H2(standard heat of combustion for 1 mole of C6H6)=\",H2,\"kJ/mol\"\n"
],
"language": "python",
@@ -417,14 +368,9 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.10\n",
- "#to find the standard heat of formation of one mole of product\n",
"print\"N2(g)+3H2(g)-> 2NH3(g)\" \n",
- "#Variable declaration\n",
"H=-92.22 \t\t\t#standard heat of reaction [KJ/mol]\n",
- "#Calculation\n",
"H1=H/2 \t\t\t\t#standard heat of formation of 1 mole [KJ/mol]\n",
- "#Result\n",
"print\"H(heat of formation of 1 mole of product)=\",H1,\"kJ mol^-1\"\n"
],
"language": "python",
@@ -453,19 +399,14 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.11\n",
"\n",
- "#to find the standard heat of reaction\n",
"print\"C2H5OH(l)+3O2(g)->2CO2(g)+3H2O(l)\" \n",
- "#Variable declaration\n",
"T=298 \t\t\t\t#temperature during the reaction[K]\n",
"Hw=-285.83 \t\t\t#standard heat of formation of liquid water [KJ/mol]\n",
"He=-277.69 \t\t\t#standard heat of formation of liquid ethanol[KJ/mol]\n",
"Hco2=-393.51 \t\t\t#standard heat of formation of carbon dioxide[KJ/mol]\n",
"Ho2=0 \t\t\t\t#standard heat of formation of oxygen gas[KJ/mol]\n",
- "#Calculation\t\n",
"H=2*Hco2+3*Hw-He-3*Ho2 \t\t#standard heat of reaction\n",
- "#Result\n",
"print\"H(standard heat of reaction)=\",H,\"kJ\"\n"
],
"language": "python",
@@ -494,19 +435,14 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.12\n",
"\n",
- "#to find the value of standard heat of formation of CO\n",
"print\"CO(g)+NO(g)->0.5N2(g)+CO2(g)\" \n",
- "#Variable declaration\n",
"Hrxn=-374 \t\t#standard heat of reaction[KJ/mol]\n",
"Hno=90.25 \t\t#standard heat of formation of NO[KJ/mol]\n",
"Hco2=-393.51 \t\t#standard heat of formation of CO2[KJ/mol]\n",
"Hn2=0 \t\t\t#standard heat of formation of N2[KJ/mol]\n",
"T=298 \t\t\t#temperature of reaction [K]\n",
- "#Calculation\n",
"Hco=0.5*Hn2+Hco2-Hno-Hrxn \t#standard heat of formation of CO[KJ/mol]\n",
- "#result\n",
"print\"Hco(standard heat of formation)=\",Hco,\"kJ mol^-1\""
],
"language": "python",
@@ -535,24 +471,15 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.13\n",
- "#to find the heat of combustion and formation of propylene\n",
"\n",
- "#Variable declaration\n",
- "#(\"C3H6(g)+H2(g)->C3H8(g) H1=-29.6 Kcal\") \n",
- "#(\"C3H8(g)+5O2(g)->3CO2(g)+4H2O(l) H2=-530.6 Kcal\") \n",
- "#(\"C(s)+O2(g)->CO2(g) H3=-94.0 Kcal\") \n",
- "#(\"H2(g)+0.5O2(g)->H2O(l) H4=-68.3 Kcal. \") \n",
"H1=-29.6 \t\t#the standard heat of hydrogenation of gaseous propylene to propane[Kcal]\n",
"H2=-530.6 \t\t#the heat of combustion of propane[Kcal] \n",
"H3=-94.0 \t\t#the heat of formation of carbon dioxide[Kcal]\n",
"H4=-68.3 \t\t#the heat of formation of liquid water[Kcal]\n",
"\n",
- "#Calculation\n",
"\n",
"print\"C3H6(g)+4.5O2(g)->3CO2(g)+3H2O(l)\" \n",
"H5=(3*H3+4*H4)-(H1+H2)#[Kcal]\n",
- "#Result\n",
"print\"\\n H(standard heat of combustion)=\",H5,\"Kcal\"\n",
"print\"3C(s)+3H2(g)->C3H6(g)\" \n",
"H6=-H5+3*H3+3*H4 #[Kcal]\n",
@@ -588,18 +515,13 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.14\n",
"\n",
- "#to find the heat of formation of N2O5\n",
- "#Variable declaration\n",
"H1=-114.1 \t\t\t#standard heat of reaction:2NO(g)+O2(g)->2NO2(g) [KJ/mol]\n",
"H2=-110.2 \t\t\t#standard heat of reaction:4NO2(g)+O2(g)->2N2O5(g) [KJ/mol]\n",
"H3=180.5 \t\t\t#standard heat of reaction:N2(g)+O2(g)->2NO(g) [KJ/mol]\n",
"\n",
"\t#reacton:N2(g)+2.5O2(g)->N2O5(g)\n",
- "#Calculation\n",
"H4=(2*H1+H2+2*H3)/2 \t\t#standard heat of formation of N2O5[KJ/mol]\n",
- "#Result\n",
"print\"H(standard heat of formation of N2O5)=\",H4,\"kJ/mol\" "
],
"language": "python",
@@ -627,17 +549,12 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.15\n",
- "#to find the standard heat of formaton of solid sucrose\n",
- "#Variable declaration\n",
"Hc=-5645 \t\t#standard enthalpy of combustion of \t\t\treaction:C12H22O11(s)+12O2(g)->12CO2(g)+11H2O(l) [KJ/mol]\n",
"Hf1=-393.51 \t\t#standard heat of formation of CO2: C(s)+O2(g)->CO2(g) [KJ/mol]\n",
"Hf2=-285.83 \t\t#standard heat of formation of H2O: H2(g)+0.5O2(g)->H2O(l) [KJ/mol]\n",
"\n",
- "#Calculation\n",
"\t#reaction:12C(s)+11H2(g)+5.5O2(g)->C12H22O11(s)\n",
"Hf=12*Hf1+11*Hf2-Hc \t#[KJ/mol]\n",
- "#Result\n",
"print\"Hf(standard heat of formation of solid sucrose)=\",Hf,\"KJ/mol(approx)\""
],
"language": "python",
@@ -665,10 +582,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.16\n",
"\n",
- "#to find the standard heat of formation of NH3 at 400K for same reaction\n",
- "#Variable declaration\n",
"Hf1=-46.11 \t\t\t#standard heat of formation of NH3 at 298K \t\t\t\t#reaction:0.5N2(g)+1.5H2(g)->NH3(g) [KJ/mol]\n",
"Cp1=29.125 \t\t\t#molar heat capacity at constant pressure for N2(g)[J/K/mol]\n",
"Cp2=28.824 \t\t\t#molar heat capacity at constant pressure for H2(g)[J/K/mol]\n",
@@ -676,12 +590,10 @@
"T1=298 \t\t\t\t#initial temperature[K]\n",
"T2=400 \t\t\t\t#final temperature[K]\n",
"\n",
- "#Calculation\n",
"\t\n",
"Cp=Cp3-0.5*Cp1-1.5*Cp2 \t\t#[J/K/mol]\n",
"T=T2-T1 \t\t\t#[K]\n",
"Hf2=Hf1+Cp*0.001*T \t\t#standard heat of formation for NH3 at 400K[KJ/mol]\n",
- "#result\n",
"print\"\\n Hf2(standard heat of formation for NH3 at 400K =\",round(Hf2,3),\"kJ/mol\""
],
"language": "python",
@@ -710,11 +622,9 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Calculate heat of formation of H2O\n",
"\n",
"from scipy.optimize import fsolve\n",
"from scipy import integrate\n",
- "#Variable declaration\n",
"dH_298=-241.82 #Std Heat of formation at 298 K [kJ mol^-1]\n",
"dH_298=dH_298*1000 # in [J mol^-1]\n",
"T1=298 #[K]\n",
@@ -732,7 +642,6 @@
"dH_1273=dH_1273/1000\n",
"print\"Heat of formation of H2O(g) at 1000 C=\",round(dH_1273,1),\"kJ mol^-1 (APPROXIMATE)\"\n",
"\n",
- "#NOTE\n",
"print\"NOTE:\"\n",
"print\"Slight variation in answer,because integration is not done precisely in the book\"\n",
"print\"In the book,it is written as:-7497.46 instead of -7504.3\""
@@ -765,16 +674,11 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.18\n",
- "#to find the C-H bond energy of CH4 from the above bond energies\n",
- "#Variable declaration\n",
"H1=435.0 \t\t\t#bond dissociation energy for: CH4->CH3+H [KJ/mol]\n",
"H2=364.0 \t\t\t#bond dissociation energy for:CH3->CH2+H [KJ/mol]\n",
"H3=385.0 \t\t\t#bond dissociation energy for:CH2->CH+H [KJ/mol] \n",
"H4=335.0 \t\t\t#bond dissociation energy for:CH->C+H [KJ/mol]\n",
- "#Calculation\t\n",
"H=(H1+H2+H3+H4)/4 \t#the bond energy for C-H bond in CH4 [KJ/mol]\n",
- "#Result\n",
"print\"\\n H(the C-H bond energy in CH4)=\",round(H,1),\"kJ/mol\"\n"
],
"language": "python",
@@ -803,20 +707,15 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.19\n",
"\n",
- "#to find the C-C bond energy in ethane bond\n",
- "#Variable declaration\n",
"H1=-84.68 \t\t\t#heat of formation : 2C(s)+3H2(g)->C2H6(g) [KJ/mol]\n",
"H2=2*716.68 \t\t\t#heat of formation : 2C(s)->2C(g) [KJ]\n",
"H3=3*436 \t\t\t#heat of formation : 3H2(g)->6H(g) [KJ]\n",
"H4=412 \t\t\t\t#taking it as bond energy for one C-H bond[KJ/mol]\n",
"\n",
- "#Calculation\n",
"\n",
"H=H2+H3-H1 \t\t\t#heat of reaction : C2H6(g)->2C(g)+6H(g) [KJ/mol]\n",
"H5=H-6*H4 \t\t\t#bond energy for one C-C bond in ethane bond [KJ/mol]\n",
- "#Result\n",
"print\"\\n Hc-c(bond energy for one C-C bond in ethane bond)=\",H5,\"kJ/mol\"\n"
],
"language": "python",
@@ -845,9 +744,6 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#example 3.20\n",
- "#to find the lattice enthalpy of magnesium bromide\n",
- "#Variable declaration\n",
"\t#MgBr2(s)-->Mg(s)+Br2(l)-->Mg(g)+Br2(l)-->Mg(g)+Br2(g)-->Mg(g)+2Br(g)-->Mg+2(g) + 2e(g) + \t2Br(g)-->Mg+2(g) + 2Br-(g)\n",
"H1=-524 \t\t#enthalpy of formation of MgBr2(s) from its element [KJ/mol]\n",
"H2=148 \t\t\t#enthalpy of sublimation of Mg(s) [KJ/mol]\n",
@@ -856,9 +752,7 @@
"H5=2187 \t\t#enthalpy of ionization of Mg(g) to Mg+2(g) [KJ/mol]\n",
"H6=-650 \t\t#enthalpy of formation of Br-(g) [KJ/mol]\n",
"\n",
- "#calculation\n",
"H=-H1+H2+H3+H4+H5+H6 \t#lattice enthalpy [KJ/mol]\n",
- "#Result\n",
"print\"\\n H(lattice enthalpy of magnesium bromide)=\",H,\"kJ/mol\"\n"
],
"language": "python",
@@ -887,24 +781,17 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Calculate maximum flame temperature\n",
"\n",
"from scipy.optimize import fsolve\n",
"from scipy import integrate\n",
"\n",
- "#Variable declaration\n",
"\n",
- "#CH4(g)+2*O2(g)--->CO2(g)+2*H2O(l)\n",
"dH1_298=-881.25 #[kJ/mol]\n",
- "#H2O(l)-->H2O(g)\n",
"dH2_298=43.60 #[kJ/mol]\n",
- "#2H2O(l)-->2*H2O(g)\n",
"dH3_298=2*dH2_298 #[kJ/mol]\n",
- "#Adding (1) and (3)\n",
"dH4_298=dH1_298+dH3_298 #[kJ/mol]\n",
"dH_heat=-dH4_298*1000 #[J/mol]\n",
"\n",
- "#CALCULATION\n",
"\n",
"def f(T2):\n",
" def g(T):\n",
generated by cgit (git 2.34.1) at 2025-04-04 09:20:35 +0000