Tutorial 4 Notes

# We are given the following data n 15 mol t 1 15 o c

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and final states that we have enough information to calculate. We are given the following data; n = 1.5 mol, T 1 = 15 o C p1 = 9.0 atm, A = 100 cm 2 pext=1.5 atm, adiabatic process (q = 0), h=15 cm and C v,m = 28.8 J K -1 mol -1 . And we are asked to calculate: q , w , Δ U , Δ T , Δ S a) q = 0 b) For work the pressure is constant and so: 2 1 V ext V ext w pdV p dV p V = ! = ! = ! " # # The change in volume is given by area*height so: ext w p Ah = ! We will need the conversion factors: 1 m 3 = 10 6 cm 3 and 1 atm = 101325 Pa ( ) ( )( ) ( ) ( ) -1 2 6 3 -3 1.5 atm 101325 Pa atm 100.0 cm 15 cm 10 m cm 230 J w ! = ! = ! c) Since q = 0, Δ U = w = -230 J d) Since this process is adiabatic: V V ext ext V dU w C dT pdV C dT p dV p dT dV C ! = = " = " = " Since p ext and C v is constant we can integrate straightforwardly and substitute in the values:

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( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 1 , -1 2 6 3 -3 -1 -1 1.5 atm 101325 Pa atm 100.0 cm 15 cm 10 m cm 1.5 mol 28.8 J K mol 5.3 K T V ext V T V ext V ext V m p dT dV C p T V C p Ah nC ! = ! " = ! " = ! = ! = ! # # e) To calculate entropy we must find the equations for the heat of the two reversible processes we have chosen to connect our initial state to our final state: Constant volume cooling: We determined this equation in last question: 2 1 2 , 1 ln T V cooling T V m C S dT T T nC T ! = " # = \$ % & ( isothermal expansion 2 1 ln V S nR V ! " # = \$ % & And so the total entropy will be: cooling isothermal expansion 2 2 , 1 1 ln ln V m S S S T V nC nR T V ! = ! + ! " # " # = + \$ % \$ % & & We don’t know the volume, but the ideal gas equation can be used to calculate the volume of the initial state: ( ) ( ) ( ) ( ) ( ) 1 1 1 -1 -1 -1 3 3 1.5 mol 8.314 J K
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