CHE222_solutions3

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 8

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online