diff options
Diffstat (limited to 'Thermodynamics:_A_Core_Course/CH5.ipynb')
| -rw-r--r-- | Thermodynamics:_A_Core_Course/CH5.ipynb | 210 |
1 files changed, 0 insertions, 210 deletions
diff --git a/Thermodynamics:_A_Core_Course/CH5.ipynb b/Thermodynamics:_A_Core_Course/CH5.ipynb index 865af391..2b811195 100644 --- a/Thermodynamics:_A_Core_Course/CH5.ipynb +++ b/Thermodynamics:_A_Core_Course/CH5.ipynb @@ -27,19 +27,14 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.1\n",
"\n",
- "#to find the workdone , heat rejected , and efficiency of the engine\n",
- "#Variable declaration\n",
"T1=373.0 \t\t\t\t#initial temperature [K]\n",
"T2=573.0 \t\t\t\t#final temperature [K]\n",
"Q2=750.0 \t\t\t\t#Heat absorbed by carnot engine[J]\n",
"\n",
- "#Calculation\n",
"e=(T2-T1)/T2 \t\t\t#efficiency of the engine\n",
"W=e*Q2 \t\t\t\t#Workdone by the engine[J]\n",
"Q1=T1*Q2/T2 \t\t\t#Heat rejected by the engine[J]\n",
- "#Result\n",
"print\"Efficiency of the engine =\",round(e,3) \n",
"print\"\\n Workdone by the engine =\",round(W),\"J\"\n",
"print\"\\n Heat rejected by the engine =\",round(Q1),\"J\"\n"
@@ -73,15 +68,10 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.2\n",
- "#to analyse the efficiency of the engine \n",
"\n",
- "#Variable declaration\n",
"T1=250.0 #temperature of heat rejection[K]\n",
"T2=1000.0 #temperature of heat absorption[K]\n",
- "#Calculation\n",
"e=1-(T1/T2) \n",
- "#Result\n",
"print\"Efficiency of the corresponding carnot engine =\",e,\"or\",e*100,\"%\"\n",
"print\" Therefore , the inventors claim of 80% efficiency is absurd.The patent application should be rejected\" \n"
],
@@ -111,16 +101,11 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.3\n",
- "#To find the minimum heat required from heat source to yield the above work\n",
- "#Variable declaration\n",
"T1=323.0 \t\t\t\t#temperature [K]\n",
"T2=423.0 \t\t\t\t#temperature [K]\n",
"W=1.3 \t\t\t\t#work [KJ]\n",
- "#Calculation\n",
"e=(T2-T1)/T2 \t\t\t#efficiency\n",
"Q2=W/e \t\t\t\t#minimum heat withdrawal from heat source[KJ]\n",
- "#Result\n",
"print\"Minimum heat withdrawal from heat source=\",round(Q2,2),\"kJ\""
],
"language": "python",
@@ -148,18 +133,13 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.5\n",
- "#to find the molar entropy change \n",
- "#Variable declaration\n",
"T=298 \t\t\t#Temperature [K]\n",
"n=1 \t\t\t#no. of moles\n",
"V1=500 \t\t\t#initial volume [cm3]\n",
"V2=1000 \t\t#final volume [cm3]\n",
"R=8.314 \t\t#Universal gas constant [J/mol/K]\n",
"import math\n",
- "#Calculation\n",
"S=R*math.log(V2/V1)\t\t#molar entropy change at constant temperature[J/K]\n",
- "#Result\n",
"print\"Molar entropy change of argon =\",round(S,1),\"J/K\"\n"
],
"language": "python",
@@ -187,16 +167,11 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.6 \n",
"\n",
- "#to find the change in molar entropy when the gas expands isothermally and reversibly\n",
- "#Variable declaration\n",
"W=1728.0 \t\t\t#Isothermal and reversible work done[J/mol]\n",
"T=298.0 \t\t\t#Isothermal temperature[K]\n",
- "#Calculation\n",
"\n",
"S=W/T \t\t\t#change in molar entropy for isothermal and reversible process\n",
- "#result\n",
"print\"The change in molar entropy =\",round(S,1),\"JK^-1mol^-1\"\n"
],
"language": "python",
@@ -224,17 +199,12 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.7\n",
"\n",
- "#To find the change in entropy of the surroundings at 298K\n",
- "#Variable declaration\n",
"H=-92.22 \t\t\t#Standard reaction enthalpy[KJ]\n",
"T=298 \t\t\t\t#Temperature [K]\n",
"\n",
- "#Calculation\t\n",
"\t#standard reaction enthalpy is H.Therefore, heat gained by the surroundings at 298K is -H\n",
"S=-H*1000/T \t\t\t#Change in entropy[J/K]\n",
- "#Result\n",
"print\"Change in entropy of the surroundings at 298k =\",round(S,1),\"J/K\"\n"
],
"language": "python",
@@ -262,19 +232,14 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.8\n",
"\n",
- "#To find the change in entropy for argon gas\n",
- "#Variable declaration\n",
"T1=298.0 \t\t\t\t#Initial Temperature[K]\n",
"T2=573.0 \t\t\t\t#Final Temperature[K]\n",
"Cv=29.1 \t\t\t#Specific Heat capacity of argon gas [J/K/mol]\n",
"n=1 \t\t\t\t#no. of moles\n",
"\n",
"import math\n",
- "#calculation\n",
"S=n*Cv*math.log(T2/T1) \t\t#Change in entropy [J/K]\n",
- "#Result\n",
"print\"The change in entropy of the argon gas is\",round(S,2),\"J/K\"\n",
"\n"
],
@@ -303,10 +268,7 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.9\n",
"\n",
- "#To find the total entropy change of solid\n",
- "#Variable declaration\n",
"T1=276.0 \t\t\t\t#Initial temperature[K]\n",
"Tf=278.7 \t\t\t#Freezing point temperature[K]\n",
"Tb=353.3 \t\t\t#Boiling point temperature[K]\n",
@@ -318,7 +280,6 @@ "mass=200.0\t\t\t#weight of solid benzene[g]\n",
"print\"Cp doesnot change within this temp limit\" \n",
"import math\n",
- "#calculation\n",
"n=mass/mol_wt \t\t\t#no. of moles\n",
"\n",
"S1=n*Cp*math.log(Tf/T1) \t#entropy change in heating [J/K]\n",
@@ -327,7 +288,6 @@ "S4=n*Hv/Tb \t\t\t#entropy change in vaporization[J/K]\n",
"S5=n*Cp*math.log(T2/Tb) \t#entropy change in heating[J/K]\n",
"S=S1+S2+S3+S4+S5 \t\t#total entropy change in heating from 276 to 373K\n",
- "#Result\n",
"print\"Total entropy change in heating 200g benzene from 3 to 100`C is\",round(S,1),\"J/K or\",round(S/1000,3),\"KJ/K\"\n",
"print\"\\nNOTE:In textbook the value of 'n' is wrongly calculated as 25.64 instead of 2.564,SO there is a error in answer shown in book\" "
],
@@ -359,22 +319,16 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.10\n",
- "#To find the change in entropy of the gas\n",
- "#Isothermal process\n",
- "#Variable declaration\n",
"mass=32 \t\t\t#weight of methane gas[gm]\n",
"P1=6*10**5 \t\t\t#Initial temperature[N/m2]\n",
"P2=3*10**5 \t\t\t#Final pressure[N/m2]\n",
"mol_wt=16 \t\t\t#molecular weight of methane gas[g/mol]\n",
"T=298 \t\t\t\t#Temperature[K]\n",
- "#calculation\t\n",
"R=8.314 \t\t\t#Universal gas constant[J/K/mol]\n",
"import math\t\n",
"n=mass/mol_wt \t\t\t#no. of moles\n",
"S=n*R*math.log(P1/P2) \t\t#change in entropy of gas[J/K]\n",
"\n",
- "#Result\n",
"print\"The change in entropy of the gas is\",round(S,2),\"J/K\""
],
"language": "python",
@@ -402,37 +356,25 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#Calculate:\n",
- "#1.Total no. og possible configuration\n",
- "#2.Probability of getting a configuration\n",
- "#3.Total energy and average energy of the system\n",
- "#4.Change in energy of the system\n",
"\n",
"import math\n",
- "#Variable declaration\n",
"black=2 #No. of black balls\n",
"white=1 #No. of white ball\n",
"\n",
- "#CALCULATION\n",
"\n",
- "#Total no. of config\n",
"W=math.factorial(black+white)/(math.factorial(black)*math.factorial(white))\n",
- "#Probability of getting a config\n",
"P=1.0/W\n",
- "#Total and Average energy of system\n",
"E1=0+1+2\n",
"E2=E1\n",
"E3=E2\n",
"E=E1+E2+E3\n",
"E_av=E/3\n",
- "#Change in total energy of system\n",
"E1_dash=1+2+3\n",
"E2_dash=E1_dash\n",
"E3_dash=E2_dash\n",
"E_dash=E1_dash+E2_dash+E3_dash\n",
"change=E_dash-E\n",
"\n",
- "#RESULT\n",
"print\"1.Total No. of possible configuration:\",W\n",
"print\"2.Probability of getting a configuration=\",P,\"or 1/3\"\n",
"print\"3.Total energy of system=\",E\n",
@@ -470,20 +412,14 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.12\n",
"\n",
- "#To find the relative number of distinguishable quantum states in 1 mole of water and ice at 273K \n",
- "#Variable declaration\n",
"n=1.0 \t\t\t#no. of moles\n",
"T=273.0 \t\t\t#temperature [K]\n",
"Hf=6000.0 \t\t#enthalpy of fusion at 273K [J/mol]\n",
"k=1.38*(10**-23) \t#boltzmann constant[J/K]\n",
"\n",
- "#calculation\n",
"p=Hf/(k*T)/2.303 \n",
"print\"\\nTHE RESULT IS 10^24,which is too large to be displayed by ipython \"\n",
- "#w=10**(p) \t\t#w is the relative no. of distinguishable quantum states\n",
- "#Result\n",
"print\"This value of w is very large to calculate for python,because it's in the range of 10^24\"\n",
"print\"The relative no. of distinguishable quantum states in 1 mole of water and ice at 273K is 10^24\" \n",
"print\"\\nTHE RESULT IS 10^24,which is too large to be displayed by ipython \"\n",
@@ -519,19 +455,14 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.13\n",
"\n",
- "#To find the total entropy change of solid\n",
- "#Variable declaration\n",
"T=300 \t\t\t#temperature[K]\n",
"n=4 \t\t\t#no. of moles of an ideal gas\n",
"P1=2.02*10**5 \t\t#initial pressure[N/m2]\n",
"P2=4.04*10**5 \t\t#final pressure[N/m2]\n",
"R=8.314 \t\t#Universal gas constant[J/K/mol]\n",
"import math\n",
- "#calculation\n",
"G=n*R*T*2.303*math.log10(P2/P1) \t#[J]\n",
- "#Result\n",
"print\" The change in Gibbs free energy is\",round(G,1),\"J\"\n"
],
"language": "python",
@@ -559,20 +490,15 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.14\n",
"\n",
- "#To find the work function or Helmholts free energy\n",
- "##Variable declaration\n",
"n=1 \t\t\t#no. of moles\n",
"T=300 \t\t\t#temperature[K]\n",
"V1=2 \t\t\t#initial volume[m3]\n",
"V2=20 \t\t\t#final volume[m3]\n",
"R=8.314 \t\t#Universal gas constant[J/K/mol]\n",
"import math\n",
- "#calculation\n",
"\t\n",
"A=-n*R*T*2.303*math.log10(V2/V1) \t#Change in work function[J/mol]\n",
- "#Result\n",
"print\"The change in Helmholts free energy is\",round(A),\"J/mol\""
],
"language": "python",
@@ -600,23 +526,18 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.15\n",
- "#To find the energy change that can be extracted as heat and work \n",
"print\"C6H12O6(s) + 6O2(g) --> 6CO2(g) + 6H2O(l)\"\n",
- "#Variable declaration\n",
"T=298 \t\t\t\t#Temperature[k]\n",
"R=8.314 \t\t\t#Universal gas constant[J/K/mol]\n",
"S=182.45 \t\t\t#standard entropy change at 298K [J/K]\n",
"U=-2808 \t\t\t#change in internal energy at 298K[KJ/mol]\n",
"\t#reaction is taking place in bomb calorimeter so no volume change \n",
"\t#therefore U=Q at constant volume\n",
- "#calculation\n",
"\t\n",
"A=U-T*S*0.001 \t\t\t#Energy extracted as heat[KJ/mol]\n",
"Wmax=A \t\t\t\t#work done [KJ/mol]\n",
"dn=6-6 \t\t\t\t#change in no. of moles\n",
"H=U+dn*R*T \t\t\t#Change in enthalpy of the bomb calorimeter[KJ]\n",
- "#Result\n",
"print\"The energy change that can be extracted as heat is\",round(A),\"KJ/mol\"\n",
"print\"\\nThe energy change that can be extracted as work is\",round(-A),\"KJ/mol\"\n",
"print\"\\nThe change in enthalpy of bomb calorimeter is\",round(H),\"KJ/mol\"\n"
@@ -651,10 +572,7 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.16\n",
"\n",
- "#To find the helmholts free energy and Gibbs free energy\n",
- "#Variable declaration\n",
"print\"C8H18(g)+12.5O2(g)-->8CO2(g)+9H2O(l)\" \n",
"\n",
"T=298.0 \t\t\t\t#temperature[K]\n",
@@ -662,12 +580,10 @@ "H=-5109000.0 \t\t\t#Heat of reaction[J]\n",
"R=8.314 \t\t\t#Universal gas constant[J/K/mol]\n",
"dn=8-(1+12.5) \t\t\t#change in no. of moles\n",
- "#calculation\n",
"\n",
"U=H \t\t\t\t#[J]\n",
"A=U-T*S \t\t\t#Change in helmholts free energy[J]\n",
"G=A+dn*R*T \t\t\t#Change in Gibbs free energy[J]\n",
- "#Result\n",
"print\"The change in Helmholts free energy is\",round(A),\"J\"\n",
"print\"\\nThe change in Gibbs free energy is\",round(G),\"J\"\n",
"print\"The calculation is not precise in book,that's why a slight change in answer\""
@@ -701,10 +617,7 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.17\n",
- "#To find the Helmholts free energy and Gibbs free energy\n",
"print\"C3H6(g)+4.5O2(g)-->3CO2(g)+3H2O(l)\" \n",
- "##Variable declaration\n",
"S=-339.23 \t\t\t#standard change in entropy [J/K]\n",
"T=298 \t\t\t\t#temperature[K]\n",
"Hf1=20.42 \t\t\t#enthalpy of formation of C3H6(g)[J]\n",
@@ -712,13 +625,11 @@ "Hf3=-285.83 \t\t\t#enthalpy of formation of H2O(l)[J]\n",
"dn=3-4.5-1 \t\t\t#change in no. of moles\n",
"R=8.314 \t\t\t#Universal gas constant[J/K/mol]\n",
- "#calculation\n",
"\n",
"H=3*Hf2+3*Hf3-Hf1 \t\t#Enthalpy of the reaction[J]\n",
"U=H-dn*R*0.001*T \t\t#Change in internal energy of the reaction[J]\n",
"A=U-T*S*0.001 \t\t\t#Helmholts free energy change[J]\n",
"G=A+dn*R*0.001*T \t\t#Gibbs free energy change[J]\n",
- "#Result\n",
"print\"The change in Helmholts free energy is\",round(A,2),\"kJ\"\n",
"print\"\\nThe change in Gibbs free energy is\",round(G,2),\"kJ\""
],
@@ -750,22 +661,17 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.19\n",
- "#To find the total entropy change\n",
"print\"CH4(g)+2O2(g)-->CO2(g)+2H2O(l)\"\n",
"\n",
- "#Variable declaration\n",
"S1=-242.98 \t\t\t\t#standard entropy change for the combustion reaction[J/K]\n",
"Hf1=-74.81 \t\t\t\t#Enthalpy of formation of CH4(g)[KJ/mol]\n",
"Hf2=-393.51 \t\t\t\t#Enthalpy of formation of CO2(g)[KJ/mol]\n",
"Hf3=-285.83 \t\t\t\t#Enthalpy of formation of H2O(l)[KJ/mol]\n",
"T=298 \t\t\t\t\t#temperature[K]\n",
- "#calculation\n",
"\t \n",
"H=Hf2+2*Hf3-Hf1 \t\t\t#Change in enthalpy of reaction[KJ]\n",
"S2=-H*1000/T \t\t\t\t#Change in entropy of the surrounding[J/K]\n",
"Stotal=(S1+S2)*0.001 \t\t\t#Total entropy change \n",
- "#Result\n",
"print\"The total change in entropy is\",round(Stotal,2),\"KJ/K\"\n"
],
"language": "python",
@@ -794,19 +700,14 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.20\n",
- "#To find the spontanity of the reaction\n",
"print\"2H2(g)+O2(g)-->2H2O(l)\" \n",
- "##Variable declaration\n",
"Hf1=-285.83 \t\t\t\t#standard enthalpy of formation of H2O(l)[KJ/mol]\n",
"S=-327 \t\t\t\t\t#Standard entropy change for the same reaction[J/K]\n",
"T=298 \t\t\t\t\t#temperature[K]\n",
"\n",
- "#calculation\n",
"\t\n",
"H=2*Hf1-0-0 \t\t\t\t#Enthalpy of the reaction[KJ/mol]\n",
"G=H-T*S*0.001 \t\t\t\t#Change in Gibbs free energy[KJ]\n",
- "#Result\n",
"print\"The change in Gibbs free energy is\",round(G,2),\"KJ\\n \"\n",
"print\"As change in Gibbs free energy is negative.Therefore,the reaction is spontaneous\"\n"
],
@@ -838,22 +739,17 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.21\n",
"\n",
"\n",
- "#To find the standard enthalpy of reaction \n",
"print\"CH4(g)+2O2(g)-->CO2(g)+2H2O(l)\" \n",
- "#Variable declaration\n",
"S=-242.98 \t\t\t#standard entropy change for reaction [J/K]\n",
"T=298 \t\t\t\t#temperature[K]\n",
"Gf1=-50.72 \t\t\t#standard Gibbs free energy of formation for CH4(g)[KJ/mol]\n",
"Gf2=-394.36 \t\t\t#standard Gibbs free energy of formation for CO2(g)[KJ/mol]\n",
"Gf3=-237.13 \t\t\t#standard Gibbs free energy of formation for H2O(l)[KJ/mol]\n",
- "#calculation\n",
"\n",
"G=Gf2+2*Gf3-Gf1 \t\t#Standard Gibbs free energy for reaction[KJ/mol]\n",
"H=G+T*S*0.001 \t\t\t#Standard enthalpy of reaction [KJ]\n",
- "#Result\n",
"print\"The standard enthalpy of reaction is\",round(H,2),\"kJ\"\n"
],
"language": "python",
@@ -882,22 +778,17 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.22\n",
- "#To find the maximum energy that can be extracted as non-expansion work is equal to the change in free energy of the system\n",
"print\"C6H12O6(s)+6O2(g)-->6CO2(g)+6H2O(l)\"\n",
- "#Variable declaration\n",
"mass=25.0 \t\t\t#mass of glucose for combustion under standard condition[gm]\n",
"T=298 \t\t\t\t#temperature[K]\n",
"Gf1=-910 \t\t\t#Standard Gibbs free energy of formation for C6H12O6[KJ/mol]\n",
"Gf2=-394.4 \t\t\t#Standard Gibbs free energy of formation for CO2(g)[KJ/mol]\n",
"Gf3=-237.13 \t\t\t#Standard Gibbs free energy of formation for H2O(l)[KJ/mol]\n",
"mol_wt=180.0 \t\t\t#molecular weight of glucose[gm/mol]\n",
- "#calculation\n",
"\t\n",
"G=6*Gf2+6*Gf3-Gf1\n",
"n=mass/mol_wt \t\t\t#no. of moles\n",
"Gactual=G*n \t\t\t#Gibbs free energy for the combustion of 0.139mol of glucose \n",
- "#Result\n",
"print\"The energy that can be extracted as non-expansion work is\",round(-Gactual),\"KJ\" \n"
],
"language": "python",
@@ -926,16 +817,11 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.23\n",
- "#To find the value of the inversion temperature for the gas\n",
- "#Variable declaration\n",
"a=1.39*10**-2 \t\t#constant for a vanderwaal's gas[lit2.atm/mol2]\n",
"b=3.92*10**-2 \t\t#constant for a vanderwaal's gas[lit2.atm/mol2]\n",
"R=0.082 \t\t#Universal gas constant[lit.atm/deg/mol]\n",
- "#calculation\n",
"\t\n",
"Ti=(2*a)/(R*b) \t\t#inversion temperature [K]\n",
- "#Result\n",
"print\"The inversion temperature for the gas is\",round(Ti,3),\" K\"\n"
],
"language": "python",
@@ -963,19 +849,14 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.26\n",
- "#To find the Enthalpy of vaporization of ethylene\n",
- "#Variable declaration\n",
"T=169.25 \t\t\t#Boiling point[K]\n",
"R=8.314 \t\t\t#Universal gas constant[J/K/mol]\n",
"print\"dlnP/dT=He/R*T**2\" \n",
"print\"dlnP/dT=(2.303*834.13/T**2)+(1.75/T)-(2.30*8.375*10**-3)\" \n",
"print\"Therefore using these two equations we calculate the He(enthalpy) of ethylene\" \n",
- "#calculation\n",
"\n",
"x=(2.303*834.13/T**2)+(1.75/T)-(2.30*8.375*10**-3) #it is dlnP/dT\n",
"He=R*0.001*T**2*x #Enthalpy of vaporization[J/mol]\n",
- "#Result\n",
"print\"\\n\\nThe Enthalpy of vaporization of ethylene at its boiling point is\",round(He,3),\"KJ/mol\"\n",
"\n"
],
@@ -1009,21 +890,16 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.27\n",
- "#To find the boiling point of b/enzene at 60KPa\n",
"\n",
- "#Variable declaration\n",
"P1=101.3 \t\t\t#Initial Pressure[KPa]\n",
"P2=60 \t\t\t\t#Final Pressure[KPa]\n",
"He=31.8 \t\t\t#Enthalpy of vaporization[KJ/mol]\n",
"R=8.314 \t\t\t#Universal gas constant[J/K/mol]\n",
"T1=353.2 \t\t\t#boiling point of benzene at 101.3KPa[K]\n",
"import math\n",
- "#calculation\n",
"\n",
"x=(T1**-1)-(R*0.001*math.log(P2/P1)/He) \n",
"T2=x**-1 \t\t\t#Boiling point of benzene at 60KPa\n",
- "#Result\n",
"print\"The boiling point of benzene at 60KPa is\",round(T2,1),\"K\""
],
"language": "python",
@@ -1051,24 +927,19 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.28\n",
"\n",
- "#To find the molar enthalpy of vapourization\n",
- "##Variable declaration\n",
"P1=0.016 \t\t\t#Vapour pressure of pure ethanol at 273K[bar]\n",
"P2=0.470 \t\t\t#Vapour pressure of pure ethanol at 333K[bar]\n",
"T1=273 \t\t\t\t#initial temperature [K]\n",
"T2=333 \t\t\t\t#final temperature[K]\n",
"R=8.314 \t\t\t#Universal gas constant[J/K/mol]\n",
"P=1.01 \t\t\t\t#vapour pressure at normal boiling point[bar]\n",
- "#calculation\n",
"import math\n",
"\t\n",
"x=(T2**-1)-(T1**-1) \n",
"He=-R*0.001*math.log(P2/P1)/x \t#molar enthalpy of vaporization[J/mol]\n",
"t=(T2**-1)-(R*0.001*math.log(P/P2)/He) \n",
"T=(t**-1)-273 \t\t\t#normal boiling point [C]\n",
- "#Result\n",
"print\"\\n\\nThe normal boiling point for pure ethanol is \",round(T,1),\"C\"\n",
"print\"The molar enthalpy of vapourization is\",round(He,2),\"J/mol\""
],
@@ -1100,10 +971,7 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.29\n",
"\n",
- "#To find the vapour pressure of benzene at 298K\n",
- "#Variable declaration\n",
"T2=353.2 \t\t\t#normal boiling point of benzene at 1.01325bar[K]\n",
"T1=298\t \t\t\t#temperature [K]\n",
"R=8.314 \t\t\t#Universal gas constant[J/K/mol]\n",
@@ -1112,13 +980,11 @@ "\t#benzene obey's Trouton's rule\n",
"print\" from Troutons rule , \" \n",
"print\" He/Tb=85J/K/mol\" \n",
- "#calculation\n",
"\n",
"He=85*T2 \t\t\t#molar enthalpy of vapourization[J/K/mol]\n",
"x=(T2**-1)-(T1**-1) \n",
"t=-He*x/R \n",
"P1=P2/math.exp(t) \n",
- "#Result\n",
"print\"\\nThe vapour pressure of benzene at 298K is\",round(P1,3),\" bar\"\n"
],
"language": "python",
@@ -1149,15 +1015,10 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.30\n",
- "#To find the degrees of freedom \n",
- "#Variable declaration\n",
"c=1 \t\t\t#no. of components(only CO2)\n",
"p=2 \t\t\t#no. of phases(liquid + gas)\n",
- "#calculation\n",
"\n",
"F=c-p+2 \t\t#degree of freedom\n",
- "#Result\n",
"print\"Degrees of freedom is\",F \n",
"print\"Degrees of freedom 1 means that either pressure or temperature can be varied independently,i.e.when temperature is fixed,pressure is automatically fixed\" "
],
@@ -1187,16 +1048,11 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.31\n",
"\n",
- "#To find the values of degrees of freedom\n",
- "#Variable declaration\n",
"c=1 \t\t\t#no. of components\n",
"p=1 \t\t\t#no. of phases\n",
- "#calculation\n",
"\n",
"F=c-p+2 \t\t#Degrees of freedom\n",
- "#Result\n",
"print\"Degrees of freedom,F is\",F \n",
"print\"Degrees of freedom 2 means both the pressure and temperature can be varied independently\" \n",
"\n",
@@ -1228,16 +1084,11 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.32\n",
"\n",
- "#To find the mole fraction of sucrose,so that the vapour pressure of water will be lowered by dP\n",
- "#Variable declaration\n",
"P=1.75*10**-5 \t\t\t#Vapour pressure of pure water at 293K[torr]\n",
"dP=1.1*10**-7 \t\t\t#Lowering in vapour pressure of water\n",
- "#calculation\n",
"\n",
"x=dP/P \t\t\t\t#mole fraction of sucrose\n",
- "#Result\n",
"print\"The mole fraction of sucrose is\",round(x,6) \n",
"\n",
"\n"
@@ -1267,18 +1118,13 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.33\n",
"\n",
- "#To find the partial vapour pressure of benzene over the solution\n",
- "#Variable declaration\n",
"P=94.6 \t\t\t\t#The vapour pressure of pure benzene at 298K[torr]\n",
"n1=20.0 \t\t\t\t#no. of moles of pure benzene\n",
"n2=5.0 \t\t\t\t#no. of moles of pure naphthalene\n",
- "#calculation\n",
"\n",
"x=n1/(n1+n2) \t\t\t#(mole fraction of benzene)\n",
"p=x*P \t\t\t\t#the partial vapour pressure of benzene[torr]\n",
- "#Result\n",
"print\"The partial vapour pressure of benzene is\",p,\"torr\""
],
"language": "python",
@@ -1306,17 +1152,12 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.34\n",
- "#To find the reduction in chemical potential\n",
- "#Variable declaration\n",
"x=0.28 \t\t\t\t#mole fraction of solute\n",
"R=8.314 \t\t\t#Universal gas constant[J/K/mol]\n",
"T=298 \t\t\t\t#temperature[K]\n",
"import math\n",
- "#calculation\n",
"\n",
"du=R*T*math.log(1-x) \t\t#reduction in chemical potential[J/mol]\n",
- "#Result\n",
"print\"The reduction in chemical potential is\",round(-du,1),\"J/mol\"\n"
],
"language": "python",
@@ -1344,17 +1185,12 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.35\n",
- "#To find the boiling point of the solution which is made by dissolving 155g of glucose in 1000g of water\n",
- "#Variable declaration\n",
"Kb=0.51 \t\t\t#ebullioscopic constant of water [K*Kg/mol]\n",
"n=155.0/180.0 \t\t\t#no. of moles of glucose\n",
"m=n/1 \t\t\t\t#[mol/Kg]\n",
"Ti=373.0 \t\t\t\t#Boiling point temperature of water[K]\n",
- "#calculation\n",
"\n",
"Tf=(Ti+Kb*m)-273 \t\t#boiling point temperature of the solution[C]\n",
- "#Result\n",
"print\"The boiling point of the solution is\",round(Tf,2),\"degree C\""
],
"language": "python",
@@ -1382,20 +1218,15 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.36\n",
"\n",
- "#To find the molar mass of solute(M2)\n",
- "#Variable declaration\n",
"Ti=5.44 #freezing point of pure benzene[K]\n",
"Tf=4.63 #freezing point of solution[K]\n",
"m1=2.12 #mass of the solute[gm]\n",
"m2=125.0 #mass of the benzene[gm]\n",
"Kf=5.12 #cryoscopic constant of pure benzene[K*Kg/mol]\n",
- "#calculation\n",
"\n",
"dTf=Ti-Tf \t#depression in freezing point[K]\n",
"M2=(m1*1000*Kf)/(m2*dTf) #molar mass of solute\n",
- "#Result\n",
"print\"The molar mass of solute is\",round(M2),\"(approx)\"\n"
],
"language": "python",
@@ -1423,20 +1254,15 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.38\n",
- "#To find the Kp value of the above reaction\n",
"print\"N2(g)+3H2(g)<=>2NH3(g)\"\n",
- "#Variable declaration\n",
"T=298 \t\t\t#Temperature[K]\n",
"Gf1=-16450 \t\t#Gibb's free energy of formation for NH3(g)[J/mol]\n",
"R=8.314 \t\t#Universal gas constant[J/K/mol]\n",
"import math\t\n",
- "#calculation\n",
"\t\n",
"Gf=2*Gf1\t\t\t#Gibb's free energy for the reaction[KJ]\n",
"x=Gf/R/T\n",
"Kp=math.exp(-x) \n",
- "#Result\n",
"print\"The Kp for above reaction is\",round(Kp),\"or 5.85*10^5,in scientific notation(APPROX)\" "
],
"language": "python",
@@ -1465,10 +1291,7 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.39\n",
- "#To find the reaction Gibb's energy\n",
"print\"0.5N2(g)+1.5H2(g)<=>NH3(g)\" \n",
- "#Variable declaration\n",
"T=298 #Temperature[K]\n",
"Kp=900 #Equilibrium constant for above reaction\n",
"P1=0.32 #partial pressure of N2(g)[bar]\n",
@@ -1476,13 +1299,11 @@ "P3=0.98 #partial pressure of NH3(g)[bar]\n",
"R=8.314 #Universal gas constant[J/K/mol]\n",
"import math\n",
- "#calculation\n",
"\n",
"G=-R*T*math.log(Kp) \n",
"x=(P1**0.5)*(P2**1.5) \n",
"p=P3/x \n",
"Gr=(G+R*T*math.log(p))*0.001 \n",
- "#Result\n",
"print\"The reaction Gibbs free energy is\",round(Gr*1000),\"J/mol \""
],
"language": "python",
@@ -1511,24 +1332,19 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.40\n",
- "#To find the Kp at 423K temperature\n",
"print\"N2(g)+3H2(g)<=>2NH3(g)\"\n",
"\n",
- "#Variable declaration\n",
"Kp1=5.85*10**5 #equilibrium constant at 298K\n",
"H1=-46.11 #standard enthalpy of formation of NH3(g)[KJ/mol]\n",
"T1=298 #Initial temperature[K]\n",
"T2=423 #Final temperature[K]\n",
"R=8.314 #Universal gas constant[J/K/mol]\n",
- "#calculation\n",
"import math\n",
"\n",
"H=2*H1 #enthalpy for reaction [KJ]\n",
"t=(T1**-1)-(T2**-1) \n",
"x=-H*t/(R*0.001) \n",
"Kp2=Kp1*math.exp(x) \n",
- "#Result\n",
"print\"The Equilibrium constant for reaction at 423K is\",round(Kp2) \n"
],
"language": "python",
@@ -1557,11 +1373,7 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.41\n",
- "#To find the Standard Free energy and equilibrium constant \n",
"print\"Zn(s)|ZnCl2(aq)||CdSO4(aq)|Cd(s)\"\n",
- "#For Zn(s)|ZnCl2(aq)||CdSO4(aq)|Cd(s)\n",
- "#Variable declaration\n",
"T=298.0 \t\t\t#Temperature[K]\n",
"R=8.314 \t\t#Universal gas constant[J/K/mol]\n",
"E1=-0.7618 \t\t#Standard electrode potential for Zn2+/Zn [volts]\n",
@@ -1570,25 +1382,18 @@ "n=2.0 \t\t\t#no. of electrons balancing\n",
"\n",
"Ei=E2-E1 \t\t#Standard potential for the reaction[volts]\n",
- "#calculation\n",
"import math\n",
"Gi=-n*F*Ei \t\t#Standard Gibb's Free Energy [KJ/mol] \n",
"Ki=math.exp(-Gi/R/T) \t#Equilibrium constant\n",
- "#Result\n",
"print\"The Free energy for the rection is\",Gi*0.001,\"KJ/mol\"\n",
"print\"The value of equilibrium constant is\",Ki \n",
"\n",
- "#To find the standard free energy and equilibrium constant\n",
- "#Variable declaration\n",
"print\"Cd(s)|CdSO4(aq),Hg2SO4(s)|Hg(l)\" \n",
- "#For Cd(s)|CdSO4(aq),Hg2SO4(s)|Hg(l)\n",
"E3=0.6141 \t\t#Standard electrode potential for Hg2SO4(s),SO4^2-/Hg(l) [volts]\n",
- "#calculation\n",
"\n",
"Eii=E3-E2 \t\t#Standard potantial for the reaction[volts]\n",
"Gii=-n*F*Eii \t\t#Standard Gibb's free energy[KJ/mol]\n",
"Kii=math.exp(-Gii/R/T) \t#Equilibrium constant\n",
- "#Result\n",
"print\"The Free energy for the rection is\",round(Gii*0.001,1),\"KJ/mol\"\n",
"print\"The value of equilibrium constant is\",Kii\n",
"print\"PLEASE REDO the last line calculation,It is showing wrong result in my PC\"\n"
@@ -1624,9 +1429,6 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.42\n",
- "#To find the overall e.m.f of the above cell\n",
- "#Variable declaration\n",
"print\"Zn(s)|ZnCl2(soln)||AgCl(s)|Ag-Ag|AgCl(s)|ZnCl2(soln)|Zn(s)\" \n",
"\n",
"m1=0.02 \t\t\t#concentration[M]\n",
@@ -1637,10 +1439,8 @@ "T=298 \t\t\t\t#Temperature [K]\n",
"F=96500 \t\t\t#Faraday's constant[coulomb/mol]\n",
"import math\n",
- "#calculation\n",
"\t\n",
"E=R*T*(math.log(m2*Y2/m1/Y1))*3/2/F \t#[volts]\n",
- "#Result\n",
"print\"The overall e.m.f of the cell is\",round(E,4),\"volt\""
],
"language": "python",
@@ -1669,9 +1469,6 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.43\n",
- "#To find the e.m.f of the cell\n",
- "#Variable declaration\n",
"print\"H2(g,1atm)|HCl(aq)|HCl(aq)|H2(g,1atm)\" \n",
"m1=0.02 \t\t\t#concentration[M]\n",
"Y1=0.88 \t\t\t#mean ionic activity coefficient\n",
@@ -1682,10 +1479,8 @@ "F=96487 \t\t\t#Faraday's constant[coulombs/mol]\n",
"t=0.178 \t\t\t#Tranference number of Cl-1\n",
"import math\n",
- "#calculation\n",
"\n",
"E=-2*t*R*T*(math.log(m1*Y1/m2/Y2))/F \t#e.m.f of the cell[volts]\n",
- "#Result\n",
"print\"The e.m.f of the cell is\",round(E,3),\" volts\" \n",
"print\"\\nWrongly calculated in book as 0.351 volt\""
],
@@ -1717,22 +1512,17 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#example 5.44\n",
"\n",
- "#To find the values of dG,dS and dH\n",
"print\"The values for reaction that goes on within the cadmium cell\" \n",
- "#Variable declaration\n",
"n=2 \t\t\t#no. of moles\n",
"E=1.01463 \t\t#standard cadmium cell potential[volts]\n",
"d=-5*10**-5 \t\t#i.e d=dE/dT[V/K]\n",
"F=96500 \t\t#[coulomb/mol]\n",
"T=298 \t\t\t#Temperature [K]\n",
- "#calculation\n",
"\n",
"dG=-n*E*F \t\t#Change in Gibb's free energy[J]\n",
"dS=n*F*d \t\t#Change in entropy [J/K]\n",
"dH=dG+T*dS \t\t#change in enthalpy[J]\n",
- "#Result\n",
"print\" dG=\",dG,\"J\\nWrongly calculated in book as -195815 J\"\n",
"print\"\\n dS=\",dS,\"J/K\"\n",
"print\"\\n dH=\",dH,\"J\""
|