Physics 2C Summer Session II
Quiz #2
Multiple Choice. Choose the answer that best completes the
statement or answers the question.
1. A given volume of air is originally at T = 20 C with a pressure of
P = 105 P a: The air is is compressed to 25% of its original volume while its
temperature is increased to 45 C: What is the …nal pressure of the air?
A) 4:34 105 P a B) 2:71 104 P a C) 3:68 105 P a D) 4:00 105 P a
E) 2:30 104 P a
From the ideal gas law P V = nRT: Consider n to be the number of moles
of gas that is compressed then
nR
=
Pf
=
Pi V i
Vi Tf
Pf Vf
=
! Pf =
Pi
Tf
Ti
Vf Ti
Vi T f
273 + 45 5
10 = 4:34
Pi = 4
Vf T i
273 + 20
105 P a ! A
2. What is the thermal velocity, vrms ; of 1:5 moles helium (AU = 4)
atoms contained in a 40L volume at a pressure P = 3 104 P a?
A) 447m=s B) 949m=s C) 548m=s D) 775m=s E) 24; 500m=s
From the kinetic theory of gasses we …nd
p
p
p
vrms =
3kT =m = 3RT =mmol = 3P V = (nmmol )
p
vrms =
3 3 104 40 10 3 = (1:5 :004)
vrms = 774:6m=s ! D
3. The total kinetic energy of all the molecules in problem 2 is
A) 900J B) 1200J C) 3600J D) 600J
From the kinetic theory of gasses we …nd
KEtot =
3
3
3
N kT = P V =
2
2
2
3
104
E) 1800J
40
10
3
= 1800J ! E:
4. Astronomers observe a star whose black body spectra has a maximum
intensity at max = 524nm = 524 10 9 m: What is the total radiated power
of this star if it has a known radius of 6:4 105 km?
A) 8:65 1025 W
E) 2:70 1020 W
B) 6:75
1025 W
1
C) 2:70
1026 W
D) 5:66
1024 W
From Wien’s law the surface temperature of the star is
T =
2:89
10
3
=
max
2:89
524
10
10
3
9
= 5515K:
From the Stephan-Boltzman law
P
P
=
AT 4 = 4 R2 T 4
2
5:67
= 4 6:4 108
P
=
2:70
10
8
4
(5515)
1026 W ! C
5. How long does it take using a 220W heater to boil o¤ all the water
that originates from 750g of ice at 15 C?
A) 171 min B) 173 min C) 44 min D) 149 min E) 230 min
The speci…c heat of ice and water are 2050J=kg C and 4184J=kg C
respectively. Water has a latent heat of fusion and vaporization of Lf = 3:34
105 J=kg and Lv = 2:26 106 J=kg: Hence the total required heat transfer is
Q = :75 2050
Q =
2:282
105 + 4184
15 + 3:34
100 + 2:26
106
106 J
The time required to transfer this heat by a 220W heater is
t=
2:281 106
= 10368 sec = 172:85 min ! B
220
6. Consider the process shown in the …gure in which AB is a constant
volume process, AC is a constant pressure process, and BC lies on an isotherm.
Figure 1. Thermodynamic process.
2
The pressure and volume at point A are 100kP a and 5L respectively while the
volume at point C is 15L. The work done by the gas along the isotherm, BC is
A) 1648J B) 2414J
The work is given by
W
=
W
=
C) 804J
Z
D) 549J
E) 429J
VC
VC
= PB VB ln
VB
VB
1500 ln 3 = 1647:9J ! A
P dV = nRT ln
7. If the gas in the cycle in Figure 1 has a speci…c heat ratio of
how much heat much be transferred to the gas under process AC?
= 7=5;
A) 4500J B) 3500J C) 2500J D) 1400J E) 2800J
The heat transfer is during a constant pressure process is
Cp
Cp
nR T =
nR T =
R
Cp Cv
7=5
(P V ) =
P V =
105
1
2=5
Q = nCp T =
Q =
1
1
10
nR T
2
= 3500J ! B
8. For the same gas cycle in the cycle shown in Figure 1, how much
heat much be transferred to the gas under process AB?
A) 1800J B) 3750J C) 1500J D) 3500J E) 2500J
The heat transfer is during a constant volume process is
Cv
Cv
nR T =
nR T =
R
Cp Cv
1
5
(P V ) =
V P =
5 10
1
2
1
Q = nCv T =
Q =
1
1
1
3
nR T
2
105 = 2500J ! E
9. Consider the adiabatic process shown in Figure 2 where AB lies on
an adiabat with a speci…c heat ratio of = 5=3:.The pressure and volume at
point A are 250kP a and 1m3 respectively, while the volume at point B is 3m3 :
The pressure at point B is
3
Figure 2. Adiabatic Process
A) 120kP a B) 58kP a C) 32kP a D) 40kP a E) 83:3kP a
For an adiabatic process we have
PA VA = PB VB ! PB = PA (VA =VB ) = 250
5=3
103 (1=3)
= 40062P a ! D
10. What is the work done by this same gas in Figure 2 along adiabat
AB?
A) 1:48 105 J B) 1:72 105 J C) 2:75
E) 1:30 105 J
The work along the adiabat is found from
W
=
W
=
1
1
(Pi Vi
1:947
Pf V f ) =
103
40062
3
3
k = 1:38 10 23 J=K
7
R = 8:314J=Kmol
7
5
NA = 6:02 1023
= 5:67 10 8 W=m2 K 4
Some useful equations
PV
1
250
2=3
105 J
D) 1:95
105 J ! D
Some useful constants
2
Boltzmann’s constant
6 Universal gas constant
6
4
Avogadro’s number
Stephan-Boltzmann const.
PV
105 J
= N kT = nRT;
U =Q
W; W =
Z
P dV; T =
2:89
10
max
Cp
=
; P = e AT 4 ; Q = mc T; Q = mLf ; Q = mLv ;
Cv
p
p
3
= const; hKEi = kT; vrms = 3kT =m = 3RT =mmol
2
4
3