CHE222_solutions3

# CHE222_solutions3 - ChE 222: Problem Set #3 Due: February 1...

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ChE 222: Problem Set #3 Due: February 1 I, 2008 (8:OO am) lay) 1 .) Koretsky, Problem 2.1 1 ((y) 2.) Koretsky, Problem 2.13 1 9 ) 3.) Koretsky, Problem 2.16 1 7 ) 4.) Koretsky, Problem 2.17

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Path B (ii) . Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the process is isothermal Equation 2.48 states that enthalpy is a function of temperature only for an ideal gas. Therefore, Performing an energy balance and neglecting potential and kinetic energy produces For an isothermal, adiabatic process, Equation 2.77 states Substituting the values from the problem statement gives
1 bar w =[8.311[m0i K]]((88+ 273.15) K)ln(-) 3 bar Using the energy balance above + I (4 (i). See path on diagram in part (a) f a (ii). Since the overall process is isothermal and u and h are state functions The definition of work is w=- PEdv I +( During the constant volume part of the process, no work is done. The work must be solved for the constant pressure step. Since it is constant pressure, the above equation simplifies to

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## This homework help was uploaded on 04/18/2008 for the course CHE 222 taught by Professor Benjamin during the Spring '08 term at SDSMT.

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CHE222_solutions3 - ChE 222: Problem Set #3 Due: February 1...

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