Physics 2C Summer Session II
Quiz #2
Multiple Choice. Choose the answer that best completes the
statement or answers the question.
1. Astronomers observe a star whose black body spectra has a maximum
intensity at max = 489nm = 489 10 9 m: What is the radius of this star if its
total radiated power is P = 4:48 1026 W ?
A) 790; 000km B) 767; 000km C) 714; 000km D) 695; 000km
E) 802; 000km
From Wien’s law the surface temperature of the star is found from
T =
2:898
10
3
=
2:898 10 3
= 5926K:
489 10 9
From the Stephan-Boltzmann law
P
r
r
r
P 1
= e AT = 4 r T ! r =
4 T2
r
4:48 1026
1
=
= 7:14 108 m
4
5:67 10 8 59262
= 714; 000km ! C
4
2
4
2. A vacuum chamber is pumped down to a pressure of 10 7 torr; where
1torr = 1mm of mercury which is equivalent to 1atm=760 where 1atm = 1:013
105 P a: Assuming that nitrogen makes up 79% of the air left in the chamber,
how many molecules of nitrogen are contained in one liter at 20 C?
A) 6:9 1011 B) 2:8 1015
The pressure in SI units is
P = 1:013
105
C) 3:3
10
7
1012
D) 2:6
=760 = 1:333
10
1012
5
E) 2:8
P a:
From the ideal gas law the total number of molecules in one liter, 1L = 10
is
PV
1:333 10 5 10 3
N=
=
= 3:295 1012 :
kT
1:38 10 23 293:15
The total number of nitrogen molecules is
NN2 = :79
3:295
1012 = 2:60
1012 .
3
m3 ;
1012 ! D
3. What is the total translational kinetic energy of all of the air molecules
in the chamber of problem 2?
1
A) 1:8 10 6 J B) 2:0 10 8 J C)2:0 10 5 J
E) 2:67 10 7 J
The total translational kinetic energy is given by
KE
KE
D) 1:5
1
3
3
3
= N m v 2 = N kT = nRT = P V
2
2
2
2
7
3
10
=
1:013 105
10 3 = 2:0 10
2
760
8
10
5
J
J !B
One could also calculate
3
1
KE = N m v 2 = N kT
2
2
directly from N obtained in problem 2 with the same result.
4. What is the root mean square velocity, vrms ; of carbon atoms at a
pressure P = :1atm = 104 P a and temperature, T = 150 C?
A) 938m=s B) 29:7km=s C) 558m=s D) 763m=s E) 1356m=s
From the expression for vrms we …nd
p
p
3kT =m = 3RT =mmol
vrms =
p
vrms =
3 8:314 (273 + 150) = (12 10 3 )
vrms = 937:66m=s: ! A
5. A freezer extracts energy from its contents at the rate of 100W . How
long will it take in minutes to freeze 1:5kg of water whose original temperature
was 25 C ?
A) 62:6 min
B) 17:5 min
C) 73:1 min
D) 26:3 min
The total heat extracted to freeze the water is given by
Q = mc T + mLf = 1:5 4184
25 + 334
103 = 6:58
E) 109:7 min
105 J
Since the heat is being extracted at a rate of 100W = Q= t; the total time is
found from
t = Q=100 = 6580 sec = 109:7 min ! E
6. Consider a tightly sealed ‡ask with a volume of 8:0L at 0 C of He gas at
a pressure of 100kP a. He gas has a speci…c heat ratio of = 5=3: How much
heat is required to heat the He to 50 C?
A) 366J
B) 186J
C) 264J
D) 318J
2
E) 220J
Since the spec…c heat ratio is = Cp =Cv = 5=3 and Cp = Cv + R; we …nd
that the speci…c heat at constant volume is
Cv + R
Cv
=
Cv
=
5
! 3 (Cv + R) = 5Cv
3
3
R:
2
The heat transfer is found from
3
3
Tf Ti
nR (Tf Ti ) = nRTi
2
2
Ti
3
Tf Ti
3 5
50
Pi Vi
= 10
8 10 3
= 219:8J ! E
2
Ti
2
273
Q = nCv T =
Q =
7. Consider the process shown in the …gure where BA lies on an isotherm.
The pressure and volume at point A are 50kP a and 5L respectively while
the volume at point B is 1L. The work done by the gas along the isotherm, BA
is
A) 442J
B) 402J
C) 1250J
D) 322J
E) 529J
The expression for the work done along an isotherm yields
Z
Z
dV
Vf
W =
P dV = nRT
= P V ln
V
Vi
W = 50 103 5 10 3 ln 5 = 402J ! B
8. What is the net work done by the gas in the cyclic process ACBA?
A) 202J
B) 152J
C) 242J
D) 182J
The total work done by the gas in the cycle is
W = 402
P
V = 402
50
3
103
4
E) 402J
10
3
= 202J ! A
9. Consider the process shown in the …gure where BA lies on an adiabat.
The gas in this process has a speci…c heat at constant volume of Cv = 5R=2: For
this problem the pressure and volume at point A are 40kP a and 5L respectively.
The pressure at point B is
A) 707kP a
B) 585kP a
C) 381kP a
D) 403kP a
E) 571kP a
In an adiabatic process P V = const where = Cp =Cv = (Cv + R) =Cv =
(7=2) = (5=2) = 1:4: This means that the pressure at B is found from
1:4
PB = PA (VA =VB )
= 40
51:4 = 380:7kP a ! C
10. For this case what is the net work done by the gas along adiabat BA?
A) 571J
B) 678J
C) 902J
D) 452J
E) 438J
The work done by a gas along an adiabat is given by
W
=
W
=
380:7
Pf Vf
=
1
451:75J ! D
Pi Vi
103
10
3
40
1:4 1
103
5
10
Some useful constants
Boltzmann’s constant
Universal gas constant
Avogadro’s number
spec…c heat (water)
heat of fusion (water)
Stephan-Boltzmann const
k = 1:38 10 23 J=K
R = 8:314J=Kmol
NA = 6:02 1023
c = 4184J=kgK
Lf = 334kJ=kg
= 5:67 10 8
Some useful equations
P V = N kT = nRT;
U =Q
W; W =
Z
P dV;
P = e AT 4 ; Q = mc T; Q = mLf ;
= 2:898
p
p
3
v 2 = 3kT =m;
P V = const; hKEi = kT;
2
Vf
Pi Vi Pf Vf
W = nRT ln
; W =
Vi
1
4
cp
cv
10 3 =T;
=
3