Exercise 1 Five moles of an ideal monatomic gas are at an initial temperature of 395K . Then, 1650J of heat is added to the gas and the work done on the gas is 925J . a) What is the change in internal energy during this process? (5 points) Following the first principle of thermodynamics, the internal energy changes by U =QW . We have Q=1650J and W =−925J since the work is done on the gas and not by the gas. In total U =1650J−−925J=2575J b) What is the final temperature of the gas? (6 points) For an mono­atomic ideal gas, a change in internal energy corresponds to a change in temperature 3 2U according to U = n⋅R T −T  from which T =T  . Numerically we obtain 2 f i f i 3n⋅R 2×2575J T f =395K =436.3K 3×5mol×8.31 J⋅mol−1⋅K −1

Exercise 2 A thermal engine produces 1600J of work while releasing 2000J of heat to the air at a temperature of 300K. a) What is the amount of heat provided by the hot reservoir during the operation of this engine? (5 points)

We have W =1600J and Qc=2000J . For energy to be conserved we must have Q H=QCW so

Q H=2000J1600J=3600J b) What is the efficiency of the engine? (5 points)

By definition the efficiency e=W /Q H . Here e=1600J/3600J=0.444 or 44.4 % c) For a Carnot engine with the same efficiency and cold reservoir temperature, what is the temperature of the hot reservoir? (5 points)

The largest possible efficiency for given hot and cold reservoir temperatures is eCarnot=1−T C /T H which T H=T C /1−eCarnot . By setting eCarnot=e=0.444 we obtain the smallest temperature for the hot reservoir allowing in principle the given engine performance: T H=300K/1−0.444=540K . Exercise 3 A loudspeaker produces a spherical isotropic harmonic wave with a power of 1W and a frequency of 25kHz . The speed of sound is c=343m⋅s−1 . a) What are the period and the wavelength of the wave detected by a stationary observer located at a distance of 250m ? (5 points) The period of a frequency f wave is T =1/ f . In this case, T =1/2.5×104 Hz=4×10−5 s=40 s . The wave length is =c⋅T =c/ f =343 m⋅s−1 /2.5×104 Hz=0.0137m=1.37cm b) What is the intensity of the wave 250m from the speaker? (5 points) We have a power P=1W . At a distance r from the source, the wave being spherical and isotropic, the intensity is I =P /4 r2 with r=250m we obtain I =1W/ 4××250m2 =1.27×10−6 W⋅m−2 . c) At a distance of 10km the intensity of the wave is 7.8×10−10 W⋅m−2 In decibels (dB), how does −12 −2 this compare to the hearing threshold I 0=1.0×10 W⋅m ? (5 points) −10 −2 An intensity of I =7.8×10 W⋅m is greater than the hearing threshold I 0 by −10 −12 =10×log I / I 0=10×log 7.8×10 /1×10 =28.9dB d) What frequency would be measured by an observer moving away from the loud speaker at a speed of 150km/h ? (6 points) The frequency observed by an observer moving away from the source at speed v is −1 f o= f 1−v/c . Here we have v=150km/h=150000/3600s=41.66m⋅s so, numerically, 4 f o=2.5×10 Hz×1−41.66/343=21963Hz Exercise 4 Five moles of an ideal monatomic gas are going through the process cycle ABCA shown on the figure. The process A­B is an isothermal 5 compression with T =350K , P A=1.2×10 Pa and 5 P B=2.5×10 Pa .

a) What are the volumes V A and V B occupied by the gas in the states A and B respectively? (5 points)

We have T A=T B=350K and we can use the ideal gas law to compute V A and V B : 5 3 V A=n⋅R⋅T A/ PA=5×8.31×350/1.2×10 Pa=0.1212m 5 3 V B=n⋅R⋅T B/ P B=5×8.31×350/2.5×10 Pa=0.0582m b) What is the temperature of the gas in state C ? (5 points)

From the graph, PC= PB and V C =V A . With this we can use the ideal gas law to obtain 5 PC⋅V C 2.5×10 ×0.1212 T = = =729K C n⋅R 5×8.31 c) Complete the following table. Detail your work. (3 points for each quantity, 36 points total)

● Since AB is isothermal, U AB=0

● W AB=n⋅R⋅T AB ln V B /V A=5×8.31×350×ln0.0582/0.1212=−10668J

● Since U AB=0 , Q AB=W AB =−10668J 3 3 U = n⋅RT −T = ×5×8.31729−350=23636J ● BC 2 C B 2 5 ● W BC =P BV C −V B=P B V A−V B=2.5×10 0.1212−0.0582=15750J

● Q BC=U BCW BC=2363615750=39386J

● Since CA is an isochore W CA=0 3 3 U = n⋅RT −T = n⋅RT −T =−23636J ● CA 2 A C 2 B C

● Since W CA=0 , U CA=Q CA=−23636J

U Q W A  B 0 ­10668J ­10668J B C 23636J 39386J 15750J C  A ­23636J ­23636J 0 TOTAL 0 5082J 5082J d) What is the efficiency of this engine and how does it compare to the Carnot efficiency corresponding to temperatures T A and T C ? (5 points) The total amount of work produced by the engine is W =5082J while the heat flowing to the engine from the hot reservoir is Q H=39386J . The efficiency is e=W /Q H =5082/39386=0.129 or

12.9%. The Carnot efficiency between the extreme temperatures T A and T C is

eCarnot=1−T C /T H=1−350/729=0.52 so it is possible to quadruple the engine efficiency without violating the second principle of thermodynamics.