{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw3_solutions

# hw3_solutions - MS&E 303 Fall 2003 Homework#3 Due Wednesday...

This preview shows pages 1–3. Sign up to view the full content.

MS&E 303 – Fall 2003 Homework #3 – Due Wednesday, September 24 1. Yet more practice. Reduce the following thermodynamic questions to their simplest forms involving only co- ordinates P, V, T, S , physical parameters α and β (expansion coefficient and the isothermal compressibility), and the constant pressure specific heat c P . (a) ∂V ∂T S ∂V ∂T S = - ∂S ∂T V ∂S ∂V T = - ∂U ∂T V ` ∂U ∂S V ∂P ∂T V = - C v /T ∂T ∂P V = C v T ∂V ∂P T ∂V ∂T P = C v T - βV αV = - C v β αT = - ( C p - α 2 V T β ) β αT = αV - C p β αT (b) ∂H ∂S V ∂H ∂S V = ∂H ∂S P + ∂H ∂P S ∂P ∂S V = T - V ∂T ∂V S = T + V ∂S ∂V T ∂S ∂T V = T + V ∂P ∂T V ∂U ∂T V ∂U ∂S V = T - V ∂V ∂T P ∂V ∂P T C v /T = T + αV T βC v = T + αV T ( βC p - α 2 V T ) = T + 1 β αV T · C p - α (c) At constant temperature, show that c p varies with pressure proportional to the curvature of the volume with temperature 2 V ∂T 2 P . Find the constant of proportionality (ie. solve the problem exactly). 1

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

View Full Document
The question is just English for the derivative ∂C p ∂P T . ∂C p ∂P T = ∂H ∂T P ∂P T = 2 H ∂T∂P = ∂H ∂P T ∂T P = ∂H ∂P S + ∂H ∂S P ∂S ∂P T ∂T P = V - T ∂V ∂T P ∂T P = ∂V ∂T P - ∂T ∂T P ∂V ∂T P - T 2 V ∂T 2 P = - T 2 V ∂T 2 P It is much less “obvious”, but turns out easier to use the definition of C p related to the derivative of S . ∂C p ∂P T = T ∂S ∂T P ∂P T = T ∂S ∂T P ∂P T = T 2 S ∂T∂P = T ∂S ∂P T ∂T P = - T ∂V ∂T P ∂T P = - T 2 V ∂T 2 P (d) Under adiabatic and reversible conditions, the pressure of a system is increased. Obtain an expression for the rate that the temperature increases with pressure. Use this result to determine the temperature increase for adiabatically and reversibly compressing water from 1 atm to 10 atm at 3.98 o C. Adiabatic reversible is just a euphanism for constant entropy (second law) – no heat flow under re- versible equals no entropy flow. Thus the appropriate partial derivative has constant entropy as the condition. So the mathematical translation is ∂T ∂P S . ∂T ∂P S = - ∂S ∂P T ∂S ∂T P = - - ∂V ∂T P C p /T = αV T C p . For water over a finite pressure change, we just have to integrate Δ T = Z 10 atm 1 atm ∂T ∂P S dP ∂T ∂P S Δ P = αV T C p Δ P.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern