PHYSICS 221 SPRING 2014

FINAL EXAM: May 6, 2014 7:00pm - 9:00pm Name (printed): ______________________________________________ Recitation Instructor: _________________________ Section #_______ INSTRUCTIONS: This exam contains 25 multiple-choice questions plus 2 extra credit questions, each worth 4 points. Choose one answer only for each question. Choose the best answer to each question. Answer all questions. Allowed material: Before turning over this page, put away all materials except for pens, pencils, erasers, rulers and your calculator. There is a formula sheet attached at the end of the exam. Other copies of the formula sheet are not allowed. Calculator: In general, any calculator, including calculators that perform graphing, is permitted. Electronic devices that can store large amounts of text, data or equations (like laptops, palmtops, pocket computers, PDA or e-book readers) are NOT permitted. Wireless devices are NOT permitted. If you are unsure whether or not your calculator is allowed for the exam, ask your TA. How to fill in the bubble sheet:

Use a number 2 pencil. Do NOT use ink. If you did not bring a pencil, ask for one. You will continue to use the same bubble sheet that you already used for the first midterm exam. Bubble answers 55-81 on the bubble sheet for this exam.

Please turn over your bubble sheet when you are not writing on it. If you need to change any entry, you must completely erase your previous entry. Also, circle your answers on this exam. Before handing in your exam, be sure that your answers on your bubble sheet are what you intend them to be. You may also copy down your answers on a piece of paper to take with you and compare with the posted answers. You may use the table at the end of the exam for this. When you are finished with the exam, place all exam materials, including the bubble sheet, and the exam itself, in your folder and return the folder to your recitation instructor. No cell phone calls allowed. Either turn off your cell phone or leave it at home. Anyone answering a cell phone must hand in their work; their exam is over.

Best of luck,

Drs. Kai-Ming Ho, Eli Rosenberg, and Kerry Whisnant

55) In the figure on the right, a diatomic ideal gas is going clockwise through a cyclic process. Which of the following statements about the work done in the process is correct?

A) It is equal to the area under the curve abc. B) It is equal to the area under the curve adc. C) It is zero. D) It is equal to the area enclosed by the curve abcda. E) It is equal to the area under the curve ab minus the area under the curve dc. Solution: The work done for a cycle is always the area enclosed by the cycle on a P-V diagram. Answer: D. 56) A 20-g bullet has an initial speed of 200 m/s. After going through a 2.5 cm-

thick wall, it’s speed is 180 m/s. What is the average force on the bullet as it passes through the wall, in kN?

A) 11 B) 30 C) 19 D) 6.4 E) 3.0 Solution: We can use the work-kinetic energy theorem plus the fact that the work done is the average force times the displacement: !KE =W12m(vf

2 " vi2 ) = "Fd

F =m(vi

2 " vf2 )

2d=

(0.020)(1802 " 2002 )2(0.025)

= 3.0#103 N.

Answer: E.

57) An ice cube at 0°C is placed inside a large metal box at 70°C. Heat is

transferred and the ice melts and the box cools until both are at 40°C. Which statement is correct?

A. The entropy of the ice increases; the entropy of the metal box decreases; the total entropy is unchanged. B. The entropy of the ice increases; the entropy of the metal box decreases; the total entropy increases. C. The entropy of the ice increases; the entropy of the metal box decreases; the total entropy decreases. D. The entropy of the ice decreases; the entropy of the metal box increases; the total entropy is unchanged. E. The entropy of the ice decreases; the entropy of the metal box is unchanged; the total entropy deceases.

!

74) In the figure on the right, a diatomic ideal gas is going clockwise through a cyclic process. Which of the following is an accurate statement? A) The work done in the process is equal to the area under the curve abc. B) The work done in the process is equal to the area enclosed by the cyclic process. C) The work done in the process is equal to the area under the curve adc. D) The work done in the process is zero. E) The work done in the process is equal to the area under ab minus the area under dc. 75) A heat engine takes 9.0 moles of a diatomic ideal gas through the reversible cycle abca, on the p-V diagram, as shown. The path bc is an isothermal process. The temperature at c is 640 K, and the volumes at a and c are 0.03 m3 and 0.22 m 3, respectively. In the figure on the right, for the path ab, the heat absorbed by the gas, in kJ, is closest to: A) 145 B) !100 C) zero D) !145 E) 100

Solution: The entropy change is dS = dQ/t. For the ice, dQ is positive since it is warming up, so dS is positive. For the box, dQ is negative since it is giving up heat, so dS is negative. By the second law of thermodynamics the total entropy change cannot be negative; since the heat exchange is occurring at changing temperatures, entropy for the system as a whole is increasing. Answer: B. 58) A 73-kg mass is hung from a massless

horizontal beam. The beam is attached to a wall at the left end and supported by a cable on the right end (see diagram). The angle θ is 300 and the length of the beam is 2.0 m. If the maximum tension the cable can withstand without breaking is 1000 N, what is the maximum distance x from the wall the mass can be hung, in m, that will not break the cable?

A) 0.50 B) 2.0 C) 1.4 D) 1.1 E) 0.70 Solution: We can use the fact that the net torque about the left end of the beam must be zero. Taking CCW torque to be positive, mgxmax !TmaxLsin! = 0

xmax =TmaxLsin!

mg=

(1000)(2.0)sin30°(73)(9.8)

=1.4 m.Answer: C.

59) Two boxes are connected by a massless string and are pulled along a

frictionless horizontal surface by a horizontal force (see figure). If mA = 2.0 kg, mB = 3.0 kg, and F = 18 N, what is the tension in the string, in N?

A) 7.2 B) 18 C) 11 D) 9.0 E) 0 Solution: The acceleration of the system as a whole is F/(mA+mB) = (18)/(5.0)=3.6 m/s2. Then T = mA a = (2.0)(3.6) = 7.2 N. Answer: A.

60) Water at +50.0 °C with a mass of 0.231 kg is placed in a freezer whose temperature is kept at –50.0 °C. The water eventually turns to ice at –50.0 °C. How much heat is absorbed by the freezer, in kJ? The specific heat of ice is 2.10 kJ/kg·°C, the specific heat of liquid water is 4.19 kJ/kg·°C, and the latent heat of fusion of water is 334 kJ/kg.

A) 750 B) 600 C) 450 D) 300 E) 150

Solution: The water must first cool down to 00C, freeze, then cool down to -500C. The total heat removed from the ice (and therefore absorbed by the freezer) is (using units kJ for energy and kg for mass) Q =mcwater!T +mLfusion +mcice!T =m(cwater!T + Lfusion + cice!T ) = (0.231)[(4.19)(50.0)+334+ (2.10)(50.0)]=150 kJ.

Answer: E.

61) A Heat Pump with a coefficient of performance 4.0 deposits 8.0 J of heat into

a room. How much heat does it remove from the outside air, in J? A) 12 B) 10 C) 8.0 D) 6.0 E) 4.0 Solution: The coefficient of performance for a heat pump is K = |QH|/|W| = |QH|/(|QH| - |QC|). Therefore |QC| = |QH|(K-1)/K = (8.0)(4.0-1)/4.0 = 6.0. Answer: D. 62) A 1500-kg car moves eastward with speed 65 mph. It hits an 8000-kg truck

that is traveling at 20 mph towards the southwest. If they stick together, what is the speed of the two vehicles immediately after the collision, in mph?

A) 12 B) 16 C) 20 D) 24 E) 28 Solution: Momentum is conserved during the collision. In the x direction

mcvc !mtvt cos45° = (mc +mt )vx

vx =mcvc !mtvt cos45°

mc +mt

=(1500)(65)! (8000)(20)(0.707)

1500+8000= !1.6 mph.

In the y direction

0!mtvt cos45° = (mc +mt )vy

vy =0!mtvt sin 45°

mc +mt

=0! (8000)(20)(0.707)

1500+8000= !11.9 mph.

Then v = vx2 + vy

2 =12 m/s. Answer: A.

The following information is used in problems 63 and 64. The transverse displacement of wave traveling down a stretched string of mass per unit length 1.0 gm/cm is described by the function

y(x,t ) = (5.00 cm) cos[(5.00 rad/m)x + (30.0 rad/s)t ], where x is the position in m and t is the time is seconds. 63) What is the wavelength of this wave, in m?

A) 0.05 B) 1.26 C) 0.80 D) 4.80 E) 0.20 Solution: The standard form for a wave is y(x, t) = Acos(kx !!t) , where k = 2! / "Then ! = 2" / k = 2(3.14159) / (5.00) =1.26 m. Answer: B. 64) What is the tension on the string, in N? A) 0.0036 B) 0.020 C) 0.60 D) 3.6 E) 6.0 Solution: We have k = 5.00 and ω = 30.0. The velocity is ω/k and is related to the string tension by (everything is put into standard units) !k= v = T

µ or T = µ !

k!

"#

$

%&

2

=0.0010.01

30.05.00!

"#

$

%&

2

= 3.6 N. Answer: D.

65) An object moves in a circular path at constant speed. Compare the directions

of the object’s velocity and acceleration vectors.

A) They are in the same direction. B) They are in the opposite direction. C) They are perpendicular. D) The acceleration vector is zero and has no direction. E) The velocity and acceleration vectors for this motion have no fixed relationship. Solution: The acceleration for uniform circular motion is in the inward radial direction; the velocity is tangential. Therefore they are perpendicular. (Note that if the speed was changing, there would have been a tangential component to the acceleration and the acceleration would not be perpendicular to the velocity.) Answer: C.

66) An unknown substance is in a container where the pressure is below the triple-point pressure of the substance. The temperature in the container can be varied. Which states of the substance can be observed as you vary the temperature?

A) Either gas, liquid, or solid B) Gas or liquid, but not solid C) Gas or solid, but not liquid D) Liquid or solid, but not gas E) Only gas Solution: The triple point on a p-T diagram is the point where a substance can have all three forms – solid, liquid, and gas. Below the triple point, a liquid is not possible. Answer: C. 67) A baseball is hit with initial speed 40 m/s at an angle 300 above the

horizontal. Its initial height is 1.0 m above the ground. A 20-m high fence is 130 m away. How high up on the fence does the ball hit, in m? Neglect air resistance,

A) The baseball hits the ground before reaching the fence. B) 7.0 C) 12 D) 17 E) The baseball goes over the fence. Solution: The equations for projectile motion are:

y = y0 + v0 yt !12gt2 = y0 + (v0 sin! )t !

12gt2

x = x0 + v0 xt = 0+ (v0 cos! )t = v0t cos!

We can solve for t using the x equation (the time the ball reaches the fence), then plug that into the y equation (the height on the fence):

t = xv0 cos!

=130

40cos30°= 3.75 s

y = y0 + (v0 sin! )t ! 12gt2 =1.0+ (40)(sin30)(3.75)! 1

2(9.8)(3.75)2 = 7.0 m.

(You may have gotten a slightly different answer if you rounded off your result for the time.) Answer: B.

Triple point (water)

At triple point Tmelting = Tboiling

Liquid water does not exist for p < 610 Pa!!

T

p

solid liquid

gas

Triple point

Critical point

Triple point for water: p = 610 Pa (0.006 atm), T = 273.16K

68) A police car siren emits sound that has frequency 2000 Hz and wavelength 0.172 m. A stationary observer measures a frequency of 1854 Hz. Relative to the observer the police car is moving

A) away from the observer at 27 m/s. B) towards from the observer at 27 m/s. C) away from the observer at 25 m/s. D) towards from the observer at 25 m/s. E) in a circle with angular velocity 5825 rad/s. Solution: The formula for the Doppler shift if the source is moving is

f = f0v

v+ vs, where we use the plus sign in the denominator since the observed

frequency is lower, which means that the source is moving away from the observer. The sound wave speed is v = ! f = (0.172)(2000) = 344 m/s . Solving for vs,

vs = vf0f!1

"

#$

%

&'= 344 2000

1854!1

"

#$

%

&'= 27 m/s. Answer: A.

69) A bullet is fired horizontally, and at the same instant a second bullet is

dropped from rest from the same height. Compare the times it takes for the two bullets to hit the ground.

A) The fired bullet hits the ground first. B) The dropped bullet hits the ground first. C) The two bullets hit the ground at the same time. D) The relative times cannot be determined without knowing the masses of the bullets. E) The relative times cannot be determined without knowing the speed of the fired bullet. Solution: The initial heights, initial vertical velocities, and accelerations are the same for each bullet. Therefore the vertical motions are the same and they hit the ground at the same time. Answer: C. The following information is used in problems 70 and 71. Eight moles of a diatomic ideal gas are compressed adiabatically from an initial pressure of 4.0 atm and an initial volume of 50.0 L to a final volume of 25.0 L. 70) What is the initial temperature of the gas, in °C? A) 3.0 B) 300 C) 2800 D) 28 E) -3.0

Solution: The ideal gas law is pV = nRT. Putting everything in standard units (1 L = .001 m3, 1 atm = 1.0 x 105 N/m2)

T = PVnR

=(4.0!105)(0.050)

(8.0)(8.31)= 301 K = 28°C.

If you used 1.01 x 105, you get 304 K = 31 0C. Answer: D.

71) What is the final pressure of the gas, in atm? A) 2.00 B) 6.50 C) 10.6 D) 8.01 E) 1.50 Solution: The process is adiabatic, soP1V1

! = P2V2! , where γ = 1.4 for a diatomic

gas. Therefore p2 = p1(V1 /V2 )! = (4.0)(0.050 / 0.025)1.4 =10.6 atm. Answer: C. 72) A large pendulum consists of a massive bob suspended from the ceiling by a

very light steel wire. The coefficient of linear expansion of steel is 1.2 x 10-5/K. At -400C the period of the pendulum is 5.000 s. By how much does the period of the pendulum change when it is heated to 1000C, in ms? [You can ignore the mass of the wire.]

A) 1.7

B) 4.2

C) 5.0

D) 8.4 E) 0 Solution: The period of the pendulum is T = 2! L / g , so

TT0

=2! L / g2! L0 / g

=LL0

= 1+"!T

T = T0 1+"!T = (5.000) 1+ (1.2"10#5 )(140) = 5.0042 s.

The change in the period is therefore 0.0042 s, or 4.2 ms. Answer: B. 73) A 2.0-kg mass swings through an arc on the end of a

40-cm long massless string (see diagram). The speed of the mass is 4.0 m/s at the bottom of the arc. What is the tension in the string, in N, when the string is horizontal?

A) 41 B) 60 C) 33 D) 20 E) 10 Solution: Conservation of energy gives 12mvi

2 +mgyi =12mvf

2 +mgyf , or 12mvi

2 =12mvf

2 +mgR , so

vf2 = vi

2 ! 2gL = (4.0)2 ! 2(9.8)(0.40) = 8.16m2 /s2. When the string is horizontal, only the tension provides centripetal force, so T =mvf

2 / L = (2.0)(8.16) / (0.40) = 41 N. Answer: A. 74) A 2.5-kg particle is moving in a straight line with speed 2.0 m/s as

shown, where d = 0.60 m. The angular momentum of the particle about the point O has magnitude _________ kg-m2/s and direction _______

A) 1.2, parallel to the velocity. B) 1.2, out of the page. C) 1.5, opposite to the velocity. D) 3.0, to the left. E) 3.0, into the page. Solution: The angular momentum ismrvsin! = (2.5)(0.60)(2.0)sin90° = 3.0 kg-m2 /s. By the right-hand rule, the angular momentum is into the page. Answer: E. 75) Two masses are attached to each other by a long, massless string. The

string runs over a massless pulley (see diagram). The coefficient of kinetic friction between m1 and the incline is 0.10 and the coefficient of static friction is 0.20. If there is no slipping of the string on the pulley, m1 = 3.0 kg, m2 = 1.0 kg, and α = 350, what is the acceleration of m2, in m/s2?

A) 0 B) 2.4, down C) 9.8, down D) 2.4, up E) 1.2, up Solution: In this case we choose the string as the x-axis, with motion to the right and down as positive. Treating the two masses as one system, we can ignore the tension in the string, which is internal to the system. To determine which direction friction acts, we look at the other forces in the x direction: m2g!m1gsin! = (1.0)(9.8)! (3.0)(9.8)sin35° = !7.1 N. Therefore the system wants to move to the left (m1 down the incline, with m2 going up), and friction acts up the incline. The next question is whether or not static friction is enough to hold the objects in place; the static friction force is fS = µSN = µSm1gcos! = (0.20)(3.0)(9.8)cos35° = 4.8 N.

8

51. A cylinder is rotating at an angular speed of 4.00 rad/s. If the moment of inertia of the cylinder

about its axis of rotation is 2.00 kg !m2 , the rotational kinetic energy of the cylinder is ____ J.

A. 8 B. 10 C. 12 D. 14 E. 16

52. A massless ideal cable is wound around the outside of a uniform solid

cylinder with a mass M = 20.0 kg and radius R = 0.500 m. The

cylinder rotates on a frictionless axle pointing into the page, as shown

in the figure. A force with magnitude F pulls horizontally on the

cable, as shown. If the magnitude of the angular acceleration of the

cylinder is 5.00 rad/s2, then F = ____ N.

[The moment of inertia of a uniform solid cylinder about its

continuous rotation symmetry axis is Icm

= (1 / 2)MR2 ]

A. 5 B. 10 C. 15 D. 20 E. 25

53. A particle with mass m = 0.500 kg is moving in a straight line with speed v = 6.00 m/s as shown, where the distance d shown is d = 0.667 m. The angular momentum of the particle about the point O has a magnitude of

____ kg !m2 /s and has a direction pointing ____

A. 0, the direction is not defined. B. 2, out of the page. C. 2, into the page. D. 4, upwards on the page. E. 4, downwards on the page.

This is not enough to stop the motion, so the m1 slides down the incline and m2 moves up. Friction is therefore kinetic and has magnitude fk = µkN = µkm1gcos! = (0.10)(3.0)(9.8)cos35° = 2.4 N.

The net force on the system is therefore 2.4 - 7.1 = - 4.7 N (down the incline). The acceleration is then a = Fnet / (m1 +m2 ) = !4.7 / 4.0 = !1.2 m/s2. Answer: E. 76) A 0.500 kg mass on a spring is displaced 0.25 m from its equilibrium position

and begins to oscillate with simple harmonic motion about equilibrium. The spring constant of the spring is k = 200 N/m. What is the speed of the mass when it passes through its equilibrium position, in m/s?

A) 0.0125 B) 5.0 C) 0.50 D) 2.5 E) 0.25 Solution: The standard simple harmonic motion equation is x = Acos(!t +") , where ! = k /m = 200 / 0.500 = 20 rad/s , and so the velocity is v = dx / dt = !A! sin(!t +") . The maximum velocity is the constant in front of the sine function in the velocity, so vmax = A! = (0.25)(20) = 5.0 m/s. Answer: B. 77) Three forces act on a 2.5-kg, uniform, cylindrical wheel as shown, where r = 40 cm and R = 80 cm. If the forces are F1 = 20 N, F2 = 10 N, and F3 = 25 N, through how many complete revolutions does the wheel rotate in 1.7 s if the wheel is initially at rest? A) 1.4 B) 5.7 C) 8.2 D) 2.9 E) 4.1

Solution: The angle it rotates through is !! ="0t +12#t2 = 0+ 1

2#t2 = 1

2#t2 , so we

need the angular acceleration. We can find that using the torque and moment of

inertia, ! = " / I , where I = 12mR2 for a uniform cylinder. Choosing CCW torque

as positive, the net torque is ! net = riFi =! " (0.40)(20)+ (0.80)(10)" (0.80)(25) = 20 N-m where in each case the force is perpendicular to the radial vector, so all angles

are 900. Then ! = 2" netmR2 =

2(20)(2.5)(0.80)2 = 25 rad/s2 . Plugging into the angle equation,

!! =12"t2 =

12

(25)(1.7)2 = 36.1 rad. We just divide this by 2π to get the number of

revolutions, 5.7 rev. Answer: B.

Laboratory  final  exam    

78)  To  estimate  the  height  of  a  cliff,  Sam  drops  a  stone  that  can  be  clearly  heard  as  it  hits  the  water  at  the  bottom  of  the  cliff.  Using  a  stopwatch,  he  measures  the  time  that  the  stone  spends  in  the  air.    

After  several  attempts,  he  computes  the  average  value  and  standard  uncertainty  of  his  measurements:    

  ( )2.3 0.1  st = ±      Thus,  the  estimated  height  of  the  cliff  is:      

( )( )22 21 19.81  m/s 2.3  s 26  m

2 2h gt= = =  

 What  is  the  uncertainty  in  the  height  due  to  the  uncertainty  in  the  time?  

 A. ±  0.5  m  B. ±  1  m  C. ±  2  m  D. ±  3  m  E. ±  4  m  

   The  minimum  and  maximum  values  of  the  time  are  tmin  =  2.2  s  and  tmax  =  2.4  s.  With  these  values,  we  obtain  hmin  =  24  m  and  hmin  =  28  m.  Thus,  h  =  (26±2)  m.    Alternatively,    

        ( )2

2 21 1 1 12

2 2 2 2h g t t gt gt t g t⎛ ⎞= ±Δ = ± Δ + Δ⎜ ⎟

⎝ ⎠  

The  last  term  is  negligible  if  Δt  <<  t,  so 212

h gt gt t= ± Δ  

Plugging  in  the  given  values,   ( )( )( )29.81  m/s 2.3  s 0.1  s 2  mgt tΔ = =  

   Answer: C.    

79)  In  one  of  this  semester’s  experiments,  a  cart  was  pulled  on  a  track  as  shown  in  the  figure  below.  The  force  probe  was  fastened  to  the  body  of  the  cart,  and  the  whole  system  (cart  +  probe)  was  pulled  through  a  string  attached  to  the  probe’s  hook.  

   In  any  experiment,  it  is  very  important  to  know  exactly  what  each  sensor  measures.  In  this  setup,  which  of  the  following  is  the  best  description  of  the  force  measured  by  the  probe  when  the  system  is  in  motion?    A. The  net  force  on  the  cart  B. The  net  force  on  the  cart  and  probe  system  C. The  weight  of  the  hanging  mass  D. The  tension  in  the  string  E. The  friction  between  the  cart  and  the  track      The  probe  measures  the  force  exerted  on  it,  and  it  is  the  string  that  is  attached  to  it.      Note  that  this  tension  is  smaller  than  the  weight  of  the  hanging  mass  when  the  system  is  in  motion  and  has  an  acceleration.  For  a  hanging  mass  of  mass  m,  Newton’s  second  law  gives    

 mg T ma

T mg ma mg

− =

= − <

   

 Answer: D.        

   

Cart  

Force  probe  Pulley  

Hanging  mass  

80)  A  motion  detector  like  the  ones  we  used  in  the  labs  is  placed  on  the  floor,  facing  up.  A  ball  is  thrown  vertically  up  right  above  the  detector,  and  its  position  is  measured  while  it  is  in  the  air  (trajectory  shown  in  the  figure  to  the  right).  Remember  that  these  detectors  measure  the  distance  of  the  object  to  the  detector,  so  the  positive  x  axis  is  as  shown  in  the  figure.  

 The  detector  sends  the  position  and  time  data  to  the  computer,  where  the  software  uses  a  numerical  derivative  to  obtain  the  corresponding  velocity  vx  as  a  function  of  time.  Which  of  the  plots  below  is  closest  to  the  graph  you  will  obtain  for  the  motion  of  this  ball?      

 Answer: A.

vx  

t  0  

A  

vx  

t  0  

B  

vx  

t  0  

C  

vx  

t  0  

D  

vx  

t  0  

E  

Motion  detector  

+x  

81)  In  the  setup  shown  in  the  figure  to  the  right,  the  syringe  is  slowly  pushed  in  while  the  pressure  in  the  sample  of  gas  is  measured  with  a  pressure  sensor.  The  whole  system  is  submerged  in  icy  water,  so  the  temperature  in  the  gas  remains  constant  at  273K.    

 The  graphs  below  shows  the  volume  markings  in  the  syringe  as  the  piston  is  pushed  versus  the  pressure,  and  versus  the  inverse  of  the  pressure.  The  latter  is  a  linear  graph,  as  expected.    How  many  moles  of  gas  are  there  in  the  sample?    A. 0.2  moles  B. 0.4  moles  C. 2  moles  D. 4  moles  E. It  cannot  be  determined  unless  we  

know  which  gas  this  is.      The  observed  dependence  seems  to  indicate  that  this  gas  behaves  as  an  ideal  gas  for  this  range  of  pressure,  volume  and  temperature.      For  an  ideal  gas.  pV  =  nRT  should  be  verified.  Note  that  V  is  the  entire  volume  of  the  sample,  V  =  Vsyringe    +  Vbottle  and  tubes    

( )                  

   

syr b t

syr b t

p V V nRT

nRTV V

p

+

+

+ =

⇓

= −

   

 The  last  equation  corresponds  to  the  linear  behavior  observed  in  the  second  graph.  The  slope  of  the  linear  fit  (m)  should  thus  be  equal  to  nRT    

5 39.08 10  ml  kPa 908  m  Pa 908  Jm= × = =    

( )

908  J0.4  mol

J8.31   273  K

mol  K

nRT m

mn

RT

=

= = =⎛ ⎞⎜ ⎟⎝ ⎠

   

Answer: B.

55    D     64    D     73    A  

56    E     65    C     74    E    

57    B     66    C     75    E      

58    C     67    B     76    B      

59    A     68    A     77    B    

60    E     69    C     78    C  

61    D     70    D     79    D  

62    A     71    C     80    A  

63    B     72    B     81    B