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Thermodynamics 3-Day Inservice Course

Learn the fundamentals of thermodynamics and master key formulas in this 3-day inservice course. Gain a solid understanding of concepts such as heat transfer, work, internal energy, enthalpy, entropy, and more. Numerical problem-solving examples included.

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Thermodynamics 3-Day Inservice Course

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  1. THERMODYNAMICS 3day inservvice course

  2. FORMULA THERMODYNAMIC U=q+w W=Pex.(-V)= - Pex V= Pex(Vf-Vi) W=-∫Pin dV Wrev=-∫ Pex dV =-∫ (PmC dp) dV Wrev =-∫ Pm dV

  3. Wrev =-∫ nRT dV/V=-nRT In Vf/Vi =-2.303nRT log Vf/Vi • qp=(U2+pV2)-(U1+pV1) • H=U+pV • H= U + pV • PV= ngRT

  4. H= U + ngRT • q=c x m x ∆T=CT • H= U + RT • Cp-Cv=R • rH=(sum of enthalpies of product)-( sum of enthalpies of reactant) =∑ai H products-∑bi H reactants i i • rHº==∑ai jHºproducts-∑bi jHº reactants

  5. rHº==∑ai jHºproducts-∑bi jHº reactants rH= rH1+ rH2+ rH3 • rHv=∑ bond enthalpies reactants-∑ bond enthalpies products • S= qrev/T • Stotal= Ssystem+ Ssuit>0

  6. G=H-TS • G= H- TS •  G= H- TS<0 • rGº=-2.303RTlogK • rGº= rHº- T rSº=-RTInK

  7. Numericals on thermodynamics • In a process,701J of heat is absorbed by a system . what is the change in internal energy for the process? Sol. Heat absorbed by the system (q)=+701J Work done by the system (w)=-304J Change in internal energy (∆U)=q+w=701-394= 307J

  8. The reaction of cyanamide, NH2CN with oxygen was affected in a bomb calorimeter and ∆U was found to be -742.7KJ /mol of cyanamide at 298K. Calculate the enthalpy change for the reaction at 288K. NH2CN(s)+3/2O2(g) N2(g)+CO2(g)+H2O Sol. ∆U=-742.7KJ/mol;∆n(g)=2-3/2=+0.5 R=8.314*10-3kJ/K/mol;T=298.5K ∆H=-742.7+0.5*8.314*10-3 *298.15 =-742.7+1.239kJ=-741.15kJ

  9. Calculate the number of kJ necessary to rise the temperature of 60ginium from 35 to 55 C. molar heat capacity of Al is 24J/mol/K Sol. Moles of Al(n)=60/27=2.22mol Molar heat capacity (Cm)=24J/mol/K ∆T=55-35=20K q=Cm*n*∆T =24*2.22*20= 1065.6J

  10. Calculate the enthalpy change on freezing of 1.0 mol of water at -10 C to ice at-10C .∆fus=6.03kJ /mol at 0 C. Cp[H2o(l)]=75.3J /mol/k :Cp[H2o(s)]=36.8J /mol/K . Sol. The freezing process is represented as H2O(l) H2O(s) T1=-10 C =263.15K; T2=0 C=273.15K;∆T=T2-T1=10K Now according to Kirchoff’s equation , (∆H2-∆H1)/T2-T1=Cp (ice)-Cp (water) (-6030-∆H1)/10=36.8-75.3 -6030-∆H1=10*38.5 ∆H1= 5625J /mol

  11. Enthalpy of combustion of carbon to carbon dioxide is -3393.5kJ /mol . Calculate the heat released upon formation of 35.2g of Co2 from carbon to oxygen gas . Sol. The combustion eq is C(s) + O2 (g) Co2(g);∆cH=-393.5kJ/mol Heat released in the formation of 44g of Co2=393.5kJ Heat released in the formation of 35.2g of Co2=(393.5*35.2)/44=314.8kJ

  12. The enthapy of formation of Co(g),N2O(g) are -110,393,81 and 9.7kJ /mol respectively. Find the value of ∆fH for the reaction,N2O4(g)+3CO(g) N2O(g)+3CO2(g). Sol. ∆fH for the given reaction is given as ∆fH= ∑∆fH(products)-∑∆fH(reactants) =∆fH (CO2)+ ∆fH (N2O) -3 ∆fH (CO)- ∆fH (N2O4) =3(-393)+81-3(-110)-9.7 =-1179+810+330-9.7= -777.7kJ /mol

  13. The eqilibrium constant fpr the vreaction is 10.caculate the value of ∆G;given R=8J/K/mol;T=300K. Sol. ∆G=-2.303RT log K Here R =8J/K/Mol; T=300K ;K=10 ∆G=-2.303*8*300*log 10 = -5527J/mol • Calculate the entropy change in surroundings when 1 mol of H2o(l) formed under standard conditions. Given ∆fH=-286kJ/mol . Sol. ∆fH of H2O(l)=-286kJ/mol which refers to heat lost to surrounds q(surr)=- ∆fH of H2O(l)=-(-286)=+286kJ/mol ∆Ssurr=q(surr)/t=(+286*10³)/298.15= 959.24J/K/mol

  14. Fir the reaction;2A(g)+B(g) 2D(g) ∆U298=-10.5kJ and ∆S=-44.1J Calculate ∆G298 for the reaction and predict whether the reaction is spontaneous or not. Sol. ∆H= ∆U+ ∆n(g)RT ∆ n(g)=2- ∆ 3=-1mol;T=298K; U=10.5kJ R=8.314kJ/K/mol ∆H=-10.5+[-1*8.314*10-³*298] =12.298kJ ∆G= ∆H-T∆S =-12.978-298*0.044 =+0.134kJ/mol Since ∆G is positive, the reaction is spontaneous ∆

  15. For a reaction at 298K; 2A+B C H=400KJ/mol and ∆S=2kJ/mol. At what temperaturewill be the reaction become spontaneous considering ∆Hand ∆S to be constant over the temperature range. Sol. ∆G= ∆H-T∆S At equilibrium, ∆G=0 , Therefore , ∆H=T∆ S T=400/2=200K Thus reaction will be in astate of equilibrium at 200K and will become spontaneous as the temperature becomes>200K.

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