CHEM
Lect+17+08

# Lect+17+08 - Lecture 17 Gibbs Energies Standard State...

• Notes
• nmgarof
• 25

This preview shows pages 1–7. Sign up to view the full content.

Lecture 17 Gibbs Energies Standard State Entropies Clausius inequality Gibbs Free Energy is a state function which determines if a system at constant T and P is at equilibrium. G < 0 spontaneous process G = 0 reversible process, at equilibrium G > 0 non-spontaneous process 2200 G ° = H ° - T S ° Examples

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

Entropy Changes with Changing Temperature Adiabatic process (Isentropic) Isochoric processes (dV=0) If C v constant over T i to T f then 2200 S = nC v lnT f /T i (dV=0) Isobaric process if C p constant T i to T f 2200 S = nC p lnT f /T i (dP=0) S = dq rev /T i f = 0.0 S = dq rev /T Ti Tf = (nC v )dT/T Ti Tf
S for isobaric heating of a Liquid or Solid 2.00 moles of water are warmed from 0°C to 100°C. Calculate S warm for this process. Use: For liquid water, c P = 4.184 J g -1 K -1 = 75.35 J mol -1 K -1 S = nc P dT T T 1 T 2 = nc P ln e ( T 2 / T 1 ) S = (2.00 mol) (75.35 J mol -1 K -1 ) ln e (373/273) = 47.03 J mol -1 K -1 Note that c P for anything other than an ideal gas must be measured and is usually found in a thermodynamic table (eg Appendix D)

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

Conceptual Problems 2.0 moles of ideal gas have Cv =3/2R and Cp = 5/2R. Assume both are constant over the temperature range of interest. 298 K isothermal expansion from 24.2 L to 48.4 L. What are q, w, E, S for the system? Isothermal expansion so E = 0, S greater than zero, and q=-w How do you calculate S? 2.0 moles of ideal gas with Cv =3/2R and Cp = 5/2 R. Assume both are constant over the temperature range of interest. 48.4 L constant volume heating from 298K to 596K. What are q, w, E, S for the system? Constant volume heating so E = q, w = 0, S greater than zero How do you calculate S?
Clausius Inequality Irreversible Expansion of an Ideal Gas Constant T in a large heat bath (surroundings) E = 0, So q = -w Consider the rapid expansion of an ideal gas w = - P ext dV but during an irreversible expansion P ext must be less than P int ; - w irrev = P ext dV < P int dV = - w rev But for a reversible expansion the external and internal pressure must remain equal through the expansion SO….The work done during an irreversible expansion is always less than the work done during a reversible expansion

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

Reversible and irreversible expansion between identical initial and final states Area under the curve is the work Reversible process always provides more work than irreversible process
This is the end of the preview. Sign up to access the rest of the document.
• Spring '08
• SHARP

{[ 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