Thermodynamics & Statistical Mechanics Exam - Spring 2006, Exams of Thermodynamics

Download Thermodynamics & Statistical Mechanics Exam - Spring 2006 and more Exams Thermodynamics in PDF only on Docsity! 1 PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS SPRING 2006 NAME: _________ANSWERS______________________ There are six pages to this examination. Check to see that you have them all. CREDIT GRADE Problem 1 20% Problem 2 30% Problem 3 25% Problem 4 15% Problem 5 10% TOTAL 100% 1. (20%) In winter, around Troy, a layer of ice usually forms on the surface of a pond or lake, but the water remains liquid below this layer. One possible explanation for this phenomenon is based on the variation of pressure with depth in the water. Deeper in the water, the pressure is greater, so the freezing point is lower. Do you think that this is the entire explanation? Support your answer by calculating the thickness of the ice layer that would form on a lake at a uniform temperature of – 1.0ºC, if this explanation is correct. Useful information: v′ = 1.09 ×10-3 m3/kg v ′′ = 1.00 ×10-3 m3/kg 12 =3.34 ×10 5 J/kg )( 12 vvTT P ′−′′ = ∆ ∆  , so T vvT P ∆ ′−′′ =∆ )( 12 . ∆P due to a layer of water of depth y is ρgy. T vvT gy ∆ ′−′′ = )( 12ρ , so g T v vTg Tv vvTg T vvT y ∆       − ′ ′′ = ∆′ ′−′′ = ∆ ′−′′ = 1)()( 121212  ρ 2 33 33 5 m/s 8.9 K) 0.1( 1 /kgm 1009.1 /kgm 1000.1K) 273( J/kg) 1034.3( −       − × × × = − − y Thickness = 1510 m units Too thick! Not the whole story. Docsity.com 2 2. (30%) A quantity of gas is taken reversibly around the cycle a-b-c-d-a shown on the T-S diagram shown in the figure below. a) (4%) The system goes around the cycle in the direction a-b-c-d-a. Is it operating as a heat engine or a refrigerator? (Circle the correct answer.) HEAT ENGINE REFRIGERATOR b) (12%) Calculate the heat transferred in each step of the cycle. The sign of the heat transferred in each step is important. You may leave your answers in terms of R. Qa-b = T(Sb – S a) = (600 K)(2R – R/2) = (600 K)(1.5 R) Qa-b = (900 K)R For b-c, ∆S , so Qb-c = 0 Qc-d = T(Sd – S c) = (200 K)( R/2 – 2R) = (200 K)(– 1.5 R) Qc-d = – (300 K)R For d-a, ∆S , so Qd-a = 0 Docsity.com 5 c) (5%) Find the Helmholtz function F for the N molecules.         +−                    −=+−−= 1ln222ln)1ln(ln 22/3 2 NIkThh mkTVNkTNZNkTF ππ         +−                    −= 1ln222ln 22/3 2 NIkThh mkTVNkTF ππ d) (7%) Use the Helmholtz function to calculate the pressure of the gas as a function of temperature and volume. This should give the same ideal gas formula as a monatomic gas. ( ) V NkTP NIkT hh mkTVNkT V P NIkT hh mkTVNkT VV FP =                 +−                    +− ∂ ∂ −=                 +−                    − ∂ ∂ −= ∂ ∂ −= 1ln222lnln 1ln222ln 22/3 2 22/3 2 ππ ππ V nRT V NkTP == Docsity.com 6 4. (15%) An astronomer observes the light from a distant gas cloud in space. This is the same astronomer that appeared in the second quiz, but it is a different gas cloud. A wavelength analysis of the light reveals two sharp lines which are characteristic of two energy transitions in a particular molecule. The lower energy line (2.1 × 10-3 eV) corresponds to the energy for a transition between the first excited state and the ground state while the higher energy line (3.2 × 10-3 eV) corresponds to the transition between the second excited state and the ground state. (As before, these energies are determined in the rest frame of the cloud; all relativistic corrections due to motion of the cloud with respect to earth have already been included. Please, do not worry about this complication here.) Based on the measured intensities of the two lines, the astronomer determines that there are 4 times more molecules in the first excited state than there are in the second excited state, and neither state is degenerate. Compute the temperature of the gas cloud based on this information. Z eNf kT i iε− = , so kT kT kT e e e f f 12 2 1 2 1 εε ε ε − − − == . Then, kTf f 12 2 1ln εε −=      , so       − = 2 1 12 ln f f k T εε ( )4lneV/K) 1062.8( eV 101.2eV 102.3 5 33 − −− × ×−× =T T = 9.2 K 5. (10%) This question deals with the laws of thermodynamics. You need not write them out, just answer the questions by circling the correct choices. Your instructor trusts you. a) (2%) Are you able to state the Zeroth Law of Thermodynamics in some reasonable form? YES NO b) (2%) Are you able to state the First Law of Thermodynamics in some reasonable form? YES NO c) (2%) Are you able to state the Second Law of Thermodynamics in some reasonable form? YES NO d) (2%) Are you able to state the Third Law of Thermodynamics in some reasonable form? YES NO e) (2%) Are you able to state the Fourth Law of Thermodynamics in some reasonable form? YES NO Docsity.com