PHYSICS 214A Midterm Exam February 10, 2009

Transcription

1 Clearly Print LAS NAME: FIRS NAME: SIGNAURE: I.D. # PHYSICS 2A Midterm Exam February 0, Do not open the exam until instructed to do so. 2. Write your answers in the spaces provided for each part of each problem. Place answers in the boxes if provided. Show your calculations in the available space or the blank facing page. Please write clearly. If we can t read your writing, you won t get credit. R 8.3 J/K-mol k B ergs/k atm dynes/cm K 0 o C DO NO WRIE BELOW HIS LINE A MAX POINS SCORE INIIALS Problem 20 Problem 2 20 Problem 3 20 Problem 20 Problem 5 20 Problem 6 20 otal 00

2 . (a) (5 points) If a three dimensional system has 0 6 particles, how many dimensions does its phase space have? You may neglect internal degrees of freedom of the particles. Answer: Number of dimensions of phase space 6N (b) (5 points) What is the most important fundamental assumption of statistical mechanics? Answer: An isolated system in equilibrium is equally likely to be in any of its accessible microstates. 2

3 (c) (5 points) Is anything wrong with the plot of entropy S versus energy E? If so, explain. S Answer: Yes, something is wrong. At high temperatures the temperature is negative because S E < 0 () S/ E is the slope of the curve. Also note that S/ E 0 which implies the temperature is infinity at the maximum. hen < 0 for E < E max which implies that the temperature is discontinuous. (d) (5 points) Is anything wrong with the plot of entropy S versus temperature? If so, explain. S E Answer: Yes, something is wrong. he specific heat at constant volume C ( ) S (2) he region with negative slope ( S/ < 0) has a negative specific heat which is unphysical. 3

4 2. (20 points) Use the Legendre transformation to derive the Gibbs free energy G(, p) from the Helmholtz free energy F (, ). In other words, derive the relation between G and F. Also derive the associated Maxwell relation. Answer: df Sd pd (3) Use a Legendre transformation to change the independent variable from p to. d(p ) pd + dp pd d(p ) dp df Sd [d(p ) dp] Sd d(p ) + dp d(f + p ) Sd + dp Since the Gibbs free energy is defined as G F + p, we have dg Sd + dp () From this we see that Using we have the Maxwell relation S G p ( ) S p ( ) G ( ) G p p G p ( ) p (5) (6)

5 3. (20 points) An ideal monatomic gas at 300K expands adiabatically and reversibly to twice its volume. What is its final temperature? Answer: For an adiabatic process p γ const where γ c P /c 5/3. Method I: p i γ i p f γ f f 2 i p i γ i p f (2 γ i ) p i 2 γ p f p f 2 γ p i p f f nr f p f f ( 2 γ p i ) (2i ) nr f 2 γ p i i nr f p i i nr i 2 γ (nr i ) nr f f 2 γ i i 300 K (300 K) (300 K) 89 K Method II: γ const γ i i γ f f f 2 i γ i i (2 i ) γ f i 2 γ f f i 2 γ (300 K) (300 K) K final 89 K 5

6 . (20 points) wo identical bodies, each with constant heat capacity C p, originally at temperatures and 2, respectively, are used as heat reservoirs for a Carnot engine operating with infinitesimal reversible cycles. If the bodies remain at constant pressure and undergo no phase changes, the engine will eventually come to rest with both bodies at a final temperature f. Find f in terms of and 2. q engine w q 2 2 Answer: A Carnot engine has S 0. his implies that dq S f C p d f C p d + 2 ( ) ( ) f f C p ln + C p ln ( ) 2 2 C p ln f f 2 f 2 2 f 2 f 2 6

7 5. (20 points) wo volumes of gas are separated by a partition. he system as whole is thermally isolated. In volume there is mole of helium gas. In volume 2 there are 2 moles of argon At time t 0 the partition is removed. After a while equilibrium is established. What is the change in entropy S S f S i between the initial and final states? (You may treat the gases as ideal gases.) N mole He N 2 moles 2 Ar 2 Answer: he change in entropy is due to the increase in the number of configurations of the gas particles when the partition is removed. Method I: Let Ω be the number of states. S k B ln Ω 2 2 N 2 2N Ω i N N 2 2 N (2 ) 2N 2 2N 3N Ω f (3 ) N +N 2 (3 ) N +N 2 (3 ) 3N 3 3N 3N S k B ln Ω f k B ln Ω i ( ) Ωf k B ln Ω ( i ) 3 3N k B ln N k B [ln ( ) 27 R ln ( J N ( )] K mole 5.8 J/K where N mole ) ( ( )) 27 ln Method II: he energy is constant. For systems with interacting particles, the temperature is not well defined. However, for an ideal gas, the particles are noninteracting, so the energy and temperature do not change when the volume changes as long as the system is thermally isolated. de 0 7

8 dq ds de + dw dw pd ds pd p nr ds nr d ( ) d S nr nr ln f S He n R ln(3) ( 3 S Ar n 2 R ln 2) i S S He + S Ar (( R ln (3 n 3 n2 ) ) + R ln 2) [ ( )] 3 n +n 2 R ln R [ ln 2 n 2 ( )] ) ( 27 R ln 5.8 J/K Method III: Start with the Maxwell relation: ( ) ( ) S p ( ) p ds d p nr ( p ) nr ds nr d ( ) d S nr nr ln f S He n R ln(3) ( 3 S Ar n 2 R ln 2) S S He + S Ar 8 i

9 (( R ln (3 n 3 n2 ) ) + R ln 2) [ ( )] 3 n +n 2 R ln R [ ln 2 n 2 ( )] ) ( 27 R ln 5.8 J/K 9