176-final-exam-solutions-05-06-10

# 176-final-exam-solutions-05-06-10 - Physics 176 Final Exam...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Physics 176 Final Exam: Solutions Professor Greenside May 2010 The following answers are much more detailed and pedagogical than what was needed to get full credit for any given problem. My hope is that the answers will be insightful, help you to solve related problems, and perhaps help you to appreciate some of the thermal physics. Problems That Require Writing Please write your answers to the following problems on extra blank sheets of paper. Also make sure to write your name and the problem number at the top of each sheet. Unless otherwise stated, you need to justify your answers to get full credit . 1. (10 points) Sketch qualitatively correct graphs of the Gibbs free energy G = U- T S + P V versus temperature T for the three phases of water (ice, water, and steam) at atmospheric pressure. You should draw your three graphs on the same set of axes so that you can see how they imply which phase is stable at a given temperature. Also make sure to indicate where on your temperature axis T = 0 ◦ C and T = 100 ◦ C. Answer: This was Problem 5.30 on page 172 of Schroeder and should have been straightforward to solve if you had read Section 5.3 and had thought about the homework problems related to this section. Two important insights from pages 170-171 were that the qualitative behavior of the Gibbs free en- ergy G ( T, P, N ) as a function of temperature T or of pressure P could be understood from the relations ∂G ∂P ¶ T,N = V, (1) and ∂G ∂T ¶ P,N =- S, (2) which both follow from the thermodynamic identity for G : dG =- SdT + V dP + X i μ i dN i . (3) This identity in turn could be obtained from the given definition G = U- T S + P V and the thermo- dynamic identity dU = T dS- P dV : dG = d ( U- T S + P V ) (4) = dU- SdT- T dS + V dP + P dV (5) = ( T dS- P dV )- SdT- T dS + V dP + P dV. (6) Since you are being asked to plot G versus T for a fixed pressure of one atmosphere (and for a fixed number of particles N ), Eq. (2) is the desired equation to get some insight. This equation implies the following, if one also uses some background knowledge that entropy generally increases with temperature, and also increases as phases change from solid to liquid to gas: (a) the slope of G ( T ) for fixed pressure is always negative, i.e., G ( T ) is a decreasing function of T . (b) the slope of G ( T ) becomes more negative with increasing T since, generally, the entropy of a solid, a liquid, or a gas increases with temperature T . Thus not only does G decrease with increasing T , it has to decrease faster than a straight line and so have some curvature downward. 1 (c) finally, since generally S solid < S liquid < S gas for three different phases of the same substance, the slopes of the corresponding G ( T ) curves must increase in steepness in going from solid to liquid to gas for a given temperature....
View Full Document

## This note was uploaded on 10/20/2011 for the course PHYSICS 176 taught by Professor Behringer during the Spring '08 term at Duke.

### Page1 / 28

176-final-exam-solutions-05-06-10 - Physics 176 Final Exam...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online