Thermodynamics and an Introduction to Thermostatistics problems, Exercises of Chemistry

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1 1 1 1 1 Problem 1.1 LEVEL 1 PROBLEMS o o o o kJ kg T T o kJ kgK T C T C U U T U T W Q W V Q U . T C . T C Q U . T T dS Q T ∂W Q dU . dT dS . dT T S . dT T . T T Problem Set 1 Solutions 3.20 MIT Professor Gerbrand Ceder Fall 2001 = 40 = 315  = ( ) ( ) = + = 0  = 0 =  = (0 17 + ) (0 17 + ) =  = 0 17( ) = = = 0 = = 0 17 = 0 17  = 0 17 = 0 17 ln = Gas is heating in a rigid container from to (First Law) (a) Since only PV work is possible & since the container is rigid. 46.75 (b) Entropy change: (for reversible heating) since or, 0.107 1 p p p tot       ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ → ′ → → → Problem 1.2 → → →



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1 1 1 1 1 1 1 2 2 1 1 1 1 2 1 1 1 2 1 1  = 0 = ( ) = = = = =  = ln = ln = 36 53 1 2 = = 7 2 = = = 7 2  = 7 2 ln = 15 59  =  + = 20 94 U U U T Q W Q pdV nRT V dV dS Q T nR V dV S nR V V nR P P . J mole K Q c dT C n R dS Q T c dT T nR dT T S nR T T . J mole K S S S . J mole K Calculate entropy change by going from 1 to 2 according to: ( ) A reversible isothermal compression ( ) Followed by an isobaric heating Entropy change ( ) or Entropy change (isobaric heating) Now for the total process 2 ′ ′ ′ tot Problem 1.5 →








∫ ∫  ∫ ∫ ∫   ∫ ∫   ∫ ∫   ∫ ∫   2 2 1 2 2 1 2 2 2 1 1 2 3 3 1 2 2 2 2 1 0 2 2 2 2 1 0 3 3 1 0 3 4 1 0 7 4 2 1 0 2 2 4 2 4 1 0 6 4 5 7 1 0 12 7 2 2 2 1 0 2 2 1 0 4 3 1 0 2 1 0 2 2 2 2 2 1 0 4 5 1 0 w pdV p dV w p V V V . P V P V V m w Pa . m w kJ w kJ df y x y dx x x y dy y x dy dx x x x dx x x x dx x x dx x dx x y x dy xdx x x x dx x x x xdx x x dx x x df y x y dx x x y dy y x dy dx x x y dx x x x dx x x dx x x y x dy xdx x x x dx x x x xdx x dx x 1 2 = = = ( ) = = 15 = 101300 15 6 919 = 819 = 2030 + 819 = 2849 = (3 + ) + ( + 2 ) = = ) (3 + ) + ( + 2 ) (4 + 3 ) = 7 = 7 4 = = = 2 ) (3 + ) + ( + 2 )2 (5 + 5 ) = + 5 7 = = (3 + ) + ( + 2 ) = = ) (3 + ) + ( + 2 ) (4 3 + 3 2) = 4 4 + 3 3 = = = 2 ) (3 + ) + ( + 2 )2 10 = 10 5 = This process is at constant pressure, so the work is given by We just have to fine That can be done easily by using the equation of state of an ideal gas along an isotherm So, The total work done is then - Function 1: (1) Integrate along ( (2) Integrate along ( Function 2: (1) Integrate along ( 2 (2) Integrate along ( 2 Since function 2 is path independent it is an exact differential. 5 J K ∫        


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Problem 1.6 1 2 2 1 2 1 1 1 1 1 2 2 2 2 1 2 1 2 1 2 1 1 2 2 1 2 2 1 2 1 = + = 0 = = = = ln = ln = ln(15) = 6754 = 0 = = = 6754 = + = + = = = = = = = = = =  = ln = ln  = 22 5 dU w Q dU w pdV p RT V w RT V dV RT V V RT p p w RT J dU Q w Q J H U P V H U P V T T U U T T P V P V H H dS Q T pdV T nR V dV S R V V R P P S . This is an ideal gas so along an isothermal path. (a) , (b) (c) For an ideal gas at constant temperature ( ) Also, when Therefore (d) 6 ≈ 3 5 ∫ ∫ ∫








vapor liq vapor liq Problem 1.7 Problem 1.8 final initial V V V V vap liq vap vapor kg m J g J g J g, w pdV w pdV w p dV p w p V V w pV p # Pa . J kg H U pV H U pV U p V U H p V U J g J g = = = = ( ) = = 10 0 592 = 167504 = 167 = +  =  +( ) =  +   =    = 2261 167 = 2094 (a) Work done in the process is: Define water as the system Initial State = Liquid Final State = Vapor So we write (Since is constant) Since the specific volume of a vapor is much larger than for a liquid This is about 7.5% of the heat of evaportation (b) Conclusion: The heat one has to transfer to water to evaporate it is partly used for increasing the internal energy of water (2094 breaking bonds) but aslo for the work required by the vapor expansion. 7 2 7 ′ ′ R cp     ′ ′ ′ ′ ′ 2 2 2 2 1 2 1 2 2 Problem 2.3

   




I II I II sys sys I II I II I II II p p p II II II total I II total w w Q Q U U U w w Q Q w Q Q Q Q nc T nc T T n . c R J mole K T T T P P T K Q Q . J mole K K Q kJ w Q Q kJ w kJ U Ap V A dU w Q ∂W pdV ∂Q dU w ( + ) + ( + ) =   = 0 ( + ) = ( + ) = 0 =  = ( ) = 613 5 = 7 2 = 29 = = 298 2 15 = 168 = 613 3 29 (298 168 ) = 2313 = ( + ) = (0 + 2313 ) = 2313 = = = + = = 0 = Two parts: (I) Adiabatic expansion from 1 to 2 (II) Heating at constant P from 2 to 2 Since state 1 and 2 are at the same temperature and for an ideal gas is only a function of temperature, and So we can calculate ’s or ’s. Since (adiabatic process) let’s calculat moles, Need to find Now for moles So, ( positive constant) (First Law) Along a reverisble path, when only p-V work is possible. Along a reversible adiabatic path, and therefore 10 2 2
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Problem 2.4         [ ] V p V p machine machine J K s dU ∂U ∂p dp ∂U ∂V dV ∂U ∂p ApV ∂U ∂V Ap ApV dp Ap dV pdV AV dp Ap dV dp Ap dV AV Ap A V A Ap V Ap V Usys Q C Q C W Q C . MW . MW MW Q C Q C MW dS Q T dS Q C Q C Q C . MW K . MW K MW K . J K s . = + = 2 = (2 ) + ( ) = 2 = ( + 1) ( + 1) = 2 ln( + 1) = ln 2 + ln ( + 1) = ( + 1)  = (500 ) + (300 ) + + (25 ) = 0 1 5 + 0 5 1 + (25 ) = 0 (25 ) = 1 (500 ) 773 + (300 ) 573 + (25 ) 298 0 1 5 773 + 0 5 573 1 298 = 542 6 542 6 0 (1) Going back to equation 1 After intergration constant new constant = even newer constant (a) Using the first law (b) Chack to see if the second law is satisfied But has to be zero since we are operating at steady state 11 2 7 R cp → → → → → → → v f ∂S ∂T p p b a  

→             ∫   1 3 1 2 2 3 1 2 2 3 2 2 1 2 1 2 3 388 253 2 3 Problem 2.5 Problem 2.6 Problem 2.7 U Q w Q U nc T w moles . J mole K T J T K T K S S S S S is T T T p p K S c dT T . S . J mole K T > T p  = + = 0  =  = (10 ) 5 2 8 314  = 80000  = 385 = 388  =  +  = 0  = = 773 2 100 = 253  = = 7 2 8 314 ln 388 253  = 12 4 (a) Since the process is adiabatic, (b) since this process is reversible and adiabatic at constant pressure, temperatre varies What is though? The temperatre that would be reached if expansion was reversible. In case (b) extra work is done on the gas so that its internal energy will remain higher than in (a) (a) Total amount of energy required Necessary heat input is the total enthalpy change of the material (since is constat) Determined by the final and initial state 12 ( 1) ′ ′ ′ ′ ′ ′  ′ Problem 3.1 o o   o   o  o  o LEVEL 3 PROBLEMS   ∫ ∫        ( 1) ( 1) ( 1) 1 ( 1) 1 3 2 2 →

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′ ′ ′ ′                  o o o o o o o o o o o o o o o o V V o o V V V V V V o o PV V V tot o o o PV V V nRT V f i f =  = ( ) ( ) = = ( ) ) ( ) ) ) = = ( ) ( ) = ( ) = = 0  = ( ) ( ) = ( ) = =  =  = ( 1) 1 = ( 1) 1 1  = ( ) ( ) = 1 =  = ( ) ( ) = ( ) + 1 = 2 + 3 = = 2 = 20 = 0 = 0  = = V , P V , P dU Q U U P , V U P , V Q A P P P V , P V, P V , P V, P P V PV P P V V U P , V U P , V A Pr P r V , P V, P Q U U P, V U P , V P dV P V V, P . P V PV P PV V U PV dV V U PV  V PV  V V U U P, V U P , V r r U U P, V U P , V A Pr P r r NH N H P , n P P P atm dV W U Q U H U H PV First go from (i.e. constant volume) A constant colume, if only P-V work is possible is unknown, but we do know because of the path we chose from ( that ( and ( are connected by a reversible adiabat. Therefore, or So, (with ) Second go from Along the reversible adiabat and therefore with and representing values along the adiabat going through To calaculate this integral use So, (with ) Combining the first and second step (with ) The reaction at 1200K is T and V are constant but increases by a factor of 2 in the reaction. Therefore (b) Heat flow The system is underconstant volume ( ) so and We can compute from because 15  a   ∫ 1200 2 2 2 2 Problem 3.2



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J mole reaction T v N v H a reaction v N v H p v a J mole J mole K a o c o f o f f o o o o c f f c c o f o c v f o f c c o U H PV H nRT H RT n H kJ mole n moles U J mole . J mole K K U Q U U c , c , dT T U c , c , c c R T . T K n n P P T T V V n P V RT n n P V RT V n RT P U n n c T T U Q W W Q W P V n RT  =  ( ) =  ( ) =    = 87  = 2  = 87000 (2) 8 314 (1200 )  = = 67046 =  = 0  + ( +3 ) = 0 = 1200  +3 = = 1200 67046 4 24 686 = 521 = + = =  = ( + ) ( )  = + = = 0 = = and (c) System is adiabtic and under constant volume So, use The system cools down which is expected given that the reaction is endothermic. We can define the system as the gas in the tank ( moles) and the gas that will be squeezed into the tank ( moles). It will helpful to define some variables = The initial pressure in the tank = The external pressure pushing gas in = The initial temperature of the gas = The final temperature of the gas (what we want to solve for) = The volume of the tank = The volume of the gas pushed in We can write the following relationships Since this is an ideal gas we know that the internal energy change is only a function of temperature, given by Givem that the process is adiabatic (isolated), ( ) But for this case we know that 16    





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5 2 7 2 2 1 2 1 1 1 2 1 2 1 1 1 1 Problem 3.3 Problem 3.4 c o o c v f o c o f f v f o o f f o o f f v f o f o o f f v f o f o v f o f v f f o v o f v v f f v f o o o o o o o v = ( + ) ( ) = ( ) = ( ) = ( ) ( + ) = ( + ) = ( + ) + = = 298 1 ( + ) (0 5 + 1 ) = 7 6 298 = 347 6 +  =   =  =  +  = =  +( ) =   = + = = 0 = = ( ) = n RT n n c T T n RT P V RT c T T RT P V RT P V RT P V RT c T T P RT P RT P c T T P T R c T P R P c T P T R c P R P c c R T T R R . R c T . K A B AB V Q. P . P w P V U Q P V Q U P V P P P P Q U PV Q H H U U Q w w Q U U P V U U nc T T P V nRT or the external pressure (which is constant) times the volume of gas pushed in. Now we can just work through eliminating variables to find the final temperature. Treating air as a diatomic ideal gas, we can use and solve for Assume that the volume change resulting from the reaction is and the heat of reaction is During the reaction, the system will perform work against the environment which has a constant pressure Since the pressure of the environment is constant, the external work done by the system is Using the first law In the initial and final states of the reaction (both equilibrium states), the pressure of the gas has to equal (otherwise they wouldn’t be equilibrium states). Therefore if is the pressure of the gas or where is the difference of enthalpies in the initial and final states. Closed system solution System is gas flowing into the tank ( (adiabatic)) For an ideal gas, 17