Physics 2020Spring 2009Stephan LeBohec

COVER SHEET - EXAM 1

Constants:

k=1.38×10−23 J⋅K−1 R=8.31 J⋅mol−1⋅K −1 N A=6.02×1023

Thermodynamics:T K =T C 273

P⋅V =N⋅k⋅T =n⋅R⋅T First law of thermodynamics: U=Q−W

n= NN A

= mM

&  R=N A⋅k Ideal mono­atomic gas: U=32

n⋅R⋅T

Isochoric: W =0 Isobaric: W =P V f −V i

Isothermal W =n⋅R⋅T ln V f

V i

Adiabatic: W =−32

n⋅RT f−T i &

P⋅V =Constant

with = 53

for mono­atomic gases

e Actual=W Cycle

Q H

=1−QC

QH

eCarnot=1−T C

T H

Waves:

=c⋅T = cf

as  f = 1T cString= F

with =m

L

In a gas cSound= P

with  = 5

3for a mono­atomic ideal gas

Isotropic sound intensity: I= P

4⋅r 2 Comparing intensities =10 log10

I 1

I 2

in dB

Doppler effect:  f o= f s

c±vo

c∓vs

SEAT #

A. A 10−3 m3 flask contains one gram of H 2 ( 2g⋅mol−1 ) and one gram of O2 ( 32g⋅mol−1 ) gas in thermal equilibrium. For each statement, circle the option you find appropriate. You do not need to show your work.

1. [2 points] Each gas exerts an equal pressure.TRUE or FALSE ?

Because PO V =nO R T & PH V =nH R T so PH /PO=nH /nO .

2. [2 points] The root mean square (RMS) velocity of H 2 molecules is the same as the RMS velocityof O2 molecules.TRUE or FALSE ?

Because32

k T=12

mO v RMSO

2 = 12

mH vRMS H

2so vRMS O

≠v RMSH

3. [2 points] The average energy of an H 2 molecule is the same as the average energy of an O2

molecule.TRUE or FALSE ?

Because32

k T=12

mO v RMSO

2 = 12

mH vRMS H

2

4. [3 points] The RMS velocity of H 2 molecules is greater than the RMS velocity of O2 molecules bya factor of1, 4, 16 or 256 ?

32

k T=12

mO v RMSO

2 =12

mH vRMS H

2so vRMS H

/v RMS O=mO /mH= 32/2=4

5. [2 points] If the average temperature of the mixture is T , the average temperature of one species of molecules is greater than T and the average temperature of the other species is less than T .TRUE or FALSE ?

The system is in thermal equilibrium, all its parts are at the same temperature.

6. [3 points] When the temperature is doubled, the RMS velocities of both species of molecules are multiplied by a factor of1, 2 , 2 or 4

32

k T= 12

m v2so v=3kT /m , when the temperature doubles the RMS velocity is multiplied by 2

7. [2 points] When the temperature of the mixture is increased the amount of work done by the gas isNEGATIVE, ZERO, or POSITIVE

V =0 so no work is done

8. [2 points] It is possible to change the temperature of the gas in the flask via an adiabatic process.TRUE or FALSE ?

Adiabatic implies Q=0 and we also have W =0 , so U=0 and T=0

Physics 2020Spring 2009Stephan LeBohec

EXAM 1Name:_____________________________________ Student ID #:___________________________

TA (circle one): Michael Sarah Adam Isaac

1

B The intensity of the 2kHz harmonic wave generated by a 200m distant loudspeaker is 10−6 W⋅m−2

1. [6 points] What are the wavelength and period of the harmonic wave (speed of sound c=343m⋅s−1 )?

The wave length =c / f =343/2×103=0.1715m .

The period T=1 / f =1/2×103=5×10−4 s .

2. [8 points] Assuming power is distributed uniformly in all directions, what is the total power emitted by the loudspeaker?

The power P=4r 2 I=4×2002×10−6=0.5026W .

3. [8 points] How far from the loudspeaker must one go to find the harmonic wave intensity to be just 20dB above the human hearing threshold ( I 0=10−12W⋅m−2 )?

By definition of dB and using the square law on intensities: =10 log I 2/ I 0=10 log II 0

⋅2002

d 2 so2002 I

d2 I 0

=10/10

and d= II 0

×2002×10−/10= 10−6

10−12×2002×10−20/10=2×104 m=20km

C [14 points] There are 2 moles of a monatomic ideal gas with an initial temperature of 300K and an initialvolume of 0.1m3 . How much heat must be added to the gas to increase its volume to 0.4m3 while keepingthe pressure constant?

First principle: Q=U W with U= 32

nRT f −T i . Using the ideal gas law T f=P V f

nRand

P=nRT i /V i so T f=T iV f /V i and U= 32

nRT iV f /V i−1 . Then

W =P V f −V i=nRT i V f /V i−1 so Q= 52

nRT iV f /V i−1=2.5×2×8.31×300×4−1=37395J .

Physics 2020Spring 2009Stephan LeBohec

EXAM 1Name:_____________________________________ Student ID #:___________________________

TA (circle one): Michael Sarah Adam Isaac

2

D Two moles of a monatomic ideal gas goes through the cycle represented in Figure 1. Process A-B is an isothermal expansion with temperature

T A=400K from V A=0.02m 3 to V B=0.04m3 . Process B-C is an isobaric process and process C-A is an isochoric process.

1. [9 points] Calculate P A , PB and T C , the pressures andtemperature reached in A,B and C.

P A=nRT A/V A=2×8.31×400/0.02=3.324×105 Pa

PB=nRT B/V B=2×8.31×400/0.04=1.662×105 Pa

T C=PC V C

nR=

PB V A

nR=

nRT B V A

V B nR=200K

2. [30 points] Complete the following table. Detail your work.

U AB=0 so QAB=W AB=nRT Aln V B/V A=2×8.31×400 ln 0.04 /0.02=4608J

W BC=PBV C−V B=1.662×1050.02−0.04=−3324J

U BC=32

nT T C−T B=1.5×n×8.31200−400=−4986J

QBC=U BCW BC=−8310J

W CA=0 so QCA=U CA=32

nR T A−T C =1.5×2×8.31400−200=4986J

Process ΔU Q W

A → B 0 4608J 4608J

B → C ­4986J ­8310J ­3324J

C → A 4986J 4986J 0

Total 0 1284J 1284J

3. [7 points] What is the efficiency of this engine and how does it compare to the Carnot efficiency corresponding to temperatures T A and T C ?

During one cycle the total amount of work done is W =1284J while the total amount of heat flowing from the heat source is QH=4608J4986J=9594J so the efficiency of this thermal engine is

e=W /QH=1284 /9594=0.1338 or 13.38%.The smallest possible Carnot efficiency in comparison would be eCarnot=1−T C /T H=1−200/ 400=0.5 or 50%.

Physics 2020Spring 2009Stephan LeBohec

EXAM 1Name:_____________________________________ Student ID #:___________________________

TA (circle one): Michael Sarah Adam Isaac

3