MidtermExam_260_Spring_2024_solutions

© Mark Schlossman – do not copy or post without permission 1 PHYSICS 260 NAME SOLUTIONS Midterm Exam September 29, 2023 All work must be shown to receive full credit. Work must be legible and comprehensible. Answers, including units , must be clearly indicated. Equation Sheet is on the last page. PROBLEM POINTS SCORE 1 16 2 20 3 26 4 20 5 18 TOTAL 100

© Mark Schlossman – do not copy or post without permission 2 1. (16 points) Multiple Choice – Circle the correct answers (for example circle the label “i”). More than one choice may be correct for each problem. Note that points will be taken off for wrong answers. (a) (4 points) Write a mathematical expression for the First Law of Thermodynamics. ___ D U = Q – W by __________________________________ (b) (4 points) Equipartition i. refers to the distribution of energy in a system at thermal equilibrium. ii. refers to energy units, independent of the temperature, to a system’s degrees of freedom. iii. refers to assigning energy units to translational motion, then if any energy remains, assigning energy units to rotational and vibrational motion. iv. can be used to relate the internal energy U of a system to its temperature. v. says that two systems placed into thermal contact reach equilibrium by making equal changes in temperature. (c) (4 points) Heat capacity i. is the amount of heat in a system. ii. is proportional to the number of degrees of freedom in a system. iii. is used to determine the amount of heat required to change the temperature of a system. iv. has units of joules/(area second). v. is relevant only for processes at constant volume. (d) (4 points) For a monatomic ideal gas in thermal equilibrium at high temperature (> 1000 K) , i. degrees of freedom include translational, rotational, and vibrational motions. ii. U = 3/2 NkT . iii. U = 7/2 NkT . iv. U = 7/2 pV .

© Mark Schlossman – do not copy or post without permission 4 3. (26 points) One mole of a monatomic ideal gas undergoes the Stirling engine cycle shown in the figure. Show your work to earn full or partial credit. (a) (6 points) Calculate the work done by the gas in process 1. Provide a numerical value with units. Answer : W = 0 Joules because it is a constant volume process. This can be determined from ࠵? !" = ∫ ࠵? ࠵?࠵? . (b) (14 points) How much heat, ࠵? , is transferred to the gas from the hot reservoir? Mark the sign (+ or –) of the heat transfer and provide a numerical value with units. Answer : Heat is transferred from the hot reservoir during the two processes, 1 and 2, that have the gas in contact with the hot reservoir. In process 1, Q = a nR ( T h - T c ), which can be derived from Q = C V D T or from the First Law, D U = Q – W by , noting that W by = 0 and D U = a nR D T . Here, a = 3/2 because the gas is monatomic, n = 1 mole, and R = 8.31 J/(mol K). ࠵? # = a nR ( T h - T c ) = (3/2) (1) (8.31) (60) J = +748 J. Energy is transferred to the gas. In process 2, Q = nRT h ln( V final / V initial ), which can be derived from the First Law, D U = Q – W by , noting that D U = 0 because the temperature is constant, so that ࠵? = ࠵? !" = ∫ ࠵? ࠵?࠵? = ∫ $%& ! ’ ࠵?࠵? = ࠵?࠵?࠵? ( ∫ )’ ’ ’ "#$%& ’ #$#’#%& = ࠵?࠵?࠵? ( ln ’ "#$%& ’ #$#’#%& = ࠵?࠵?࠵? ( ln 2 ࠵? * = ࠵?࠵?࠵? ( ln 2 = (1) (8.31) (320) ln 2 = +1843 J. Energy is transferred to the gas. So, ࠵? +,+-. = ࠵? # + ࠵? * = 748 + 1843 = +2591 J. Heat energy is transferred to the gas. V p T h = 320 K V a 2 V a 1 2 3 4 T c = 260 K

© Mark Schlossman – do not copy or post without permission 5 (c) (6 points) What is the change in total entropy, D S tot , during process 2? Provide a numerical value with units. Answer : There are two ways to calculate this. First, this isothermal process is reversible so that D S tot = 0 J/K. Second, calculate the entropy changes of the system and the reservoir and add them. ∆࠵? /010/2,3/ = ࠵? ࠵? = −࠵? * 320 where Q 2 is from part (b). ∆࠵? 1"1+04 = ࠵? ࠵? = ࠵? * 320 ∆࠵? +,+ = ∆࠵? /010/2,3/ + ∆࠵? 1"1+04 = −࠵? * 320 + ࠵? * 320 = 0 J/K

© Mark Schlossman – do not copy or post without permission 7 5. (18 points) More Multiple Choice – Circle the correct answers (for example, circle the label “i”). More than one choice may be correct for each problem. Note that points will be taken off for wrong answers. (a) (6 points) How much heat is absorbed by the gas when one mole of an ideal gas of molecules with 16 degrees of freedom is heated from 300 to 400 K at a fixed volume? i. 200 k Joules. ii. 400 k Joules. iii. 800 k Joules. iv. 200 R Joules. v. 400 R Joules. vi. 800 R Joules. b) (6 points) Select the true statements about entropy. i. Changes in the entropy of an object must always be zero or positive. ii. Entropy is just the heat Q transferred in a process. iii. Total entropy remains unchanged during a reversible process. iv. The change in entropy of an object undergoing an adiabatic process can be positive or negative. v. The change in total entropy is positive when a cold object is placed in thermal contact with a hot object. c) (6 points) A heat engine consists of a thermodynamic cycle. The engine starts in an initial thermodynamic state (for example, an initial value of pressure and volume), then undergoes a series of processes that return the engine to its initial state. So, the final state and the initial state of the cycle are the same thermodynamic state. Circle the correct statements below. i. The change in internal energy, Δ U , for the heat engine cycle is zero because U is a state function. ii. The energy that goes into the engine during the cycle is equal to the energy that leaves the engine during the cycle, so that Q h = Q c . iii. The energy that goes into the engine during the cycle is equal to the energy that leaves the engine during the cycle, so that Q h = Q c + W by . iv. The efficiency of any heat engine depends only on the temperatures of the hot and cold reservoirs, so that it can be calculated using ࠵? = 1 − (࠵? 6 ࠵? ( ⁄ ) .

© Mark Schlossman – do not copy or post without permission 8 Equation Sheet ࠵? = 1.38 × 10 7*8 J K ⁄ ; ࠵? 9 = 6.02 × 10 *8 ; ࠵? = 8.31 J mol ∙ K; 1 atm = 1.01 × ⁄ 10 : N m * ⁄ ࠵? ; = 1.66 × 10 7*< kg (atomic mass unit); ࠵? =-+0/ = 4.18 kJ kg ∙ K ⁄ ࠵?࠵? = ࠵?࠵?࠵? = ࠵?࠵?࠵?; ∆࠵? = ࠵? − ࠵? !" ; ࠵?(K) = ࠵?(°C) + 273.15 ࠵? ,$ = −࠵? !" ; ࠵? !" = ࠵?∆࠵? or ࠵? !" = \ ࠵? ࠵?࠵? ࠵? = 2 3 ࠵? ࠵? 〈࠵?࠵?〉 ࠵? = ࠵?࠵?࠵?࠵?; 〈thermal energy per quadratic energy term or "degree of freedom"〉 = 1 2 ࠵?࠵? ࠵?࠵? > = constant; ࠵? >7# ࠵? = constant; ࠵? = ࠵? + 1 ࠵? ࠵? = ࠵? ∆࠵? ; ࠵? = ࠵?∆࠵? = ࠵?࠵?∆࠵?; ࠵? = \ ࠵? ࠵?࠵? ࠵? ’ = l ∆࠵? ∆࠵? m ’ ; ࠵? ? = l ∆࠵? ∆࠵? m ? + ࠵? l ∆࠵? ∆࠵? m ? ; ࠵? ? ࠵? ’ = ࠵? ∆࠵? = ࠵? /02 ࠵? = \ ࠵?࠵? /02 ࠵? ; ∆࠵? +,+-. ≥ 0 ࠵? = ࠵? ࠵? ( = 1 − ࠵? 6 ࠵? ( = 1 − ࠵? 6 ࠵? ( ࠵? = −࠵? ࠵?࠵? ࠵?࠵? ; ࠵? = ࠵? ∙ ࠵?; ࠵? 63/6.0 = ࠵?࠵? * ; ࠵? 6".3$)0/ = 2࠵?࠵? * + 2࠵?࠵?࠵?; ࠵? 1?(0/0 = 4࠵?࠵? *