2015 USA Physics Olympiad Exam AAPT UNITED STATES PHYSICS TEAM AIP 2015 USA Physics Olympiad Exam DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor This examination consists of two parts. Part A has four questions and is allowed 90 minutes. Part B has two questions and is allowed 90 minutes. The first page that follows is a cover sheet.Examinees may keep the cover sheet for both parts of the exam. The parts are then identified by the center header on each page.Examinees are only allowed to do one part at a time,and may not work on other parts,even if they have time remaining. Allow 90 minutes to complete Part A.Do not let students look at Part B.Collect the answers to Part A before allowing the examinee to begin Part B.Examinees are allowed a 10 to 15 minutes break between parts A and B. Allow 90 minutes to complete Part B.Do not let students go back to Part A. Ideally the test supervisor will divide the question paper into 4 parts:the cover sheet (page 2), Part A (pages 3-7),Part B(pages 9-11),and several answer sheets for one of the questions in Part A(pages 13-13).Examinees should be provided parts A and B individually,although they may keep the cover sheet.The answer sheets should be printed single sided! The supervisor must collect all examination questions,including the cover sheet,at the end of the exam,as well as any scratch paper used by the examinees.Examinees may not take the exam questions.The examination questions may be returned to the students after April 15,2015. Examinees are allowed calculators,but they may not use symbolic math,programming,or graphic features of these calculators.Calculators may not be shared and their memory must be cleared of data and programs.Cell phones,PDA's or cameras may not be used during the exam or while the exam papers are present.Examinees may not use any tables,books, or collections of formulas. Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam 1 AAPT AIP 2015 UNITED STATES PHYSICS TEAM USA Physics Olympiad Exam DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor • This examination consists of two parts. • Part A has four questions and is allowed 90 minutes. • Part B has two questions and is allowed 90 minutes. • The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam. • The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining. • Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B. • Allow 90 minutes to complete Part B. Do not let students go back to Part A. • Ideally the test supervisor will divide the question paper into 4 parts: the cover sheet (page 2), Part A (pages 3-7), Part B (pages 9-11), and several answer sheets for one of the questions in Part A (pages 13-13). Examinees should be provided parts A and B individually, although they may keep the cover sheet. The answer sheets should be printed single sided! • The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after April 15, 2015. • Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas. Copyright c 2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Cover Sheet 2 AAPT UNITED STATES PHYSICS TEAM AIP 2015 USA Physics Olympiad Exam INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN Work Part A first.You have 90 minutes to complete all four problems.Each question is worth 25 points.Do not look at Part B during this time. After you have completed Part A you may take a break. Then work Part B.You have 90 minutes to complete both problems.Each question is worth 50 points.Do not look at Part A during this time. Show all your work.Partial credit will be given.Do not write on the back of any page.Do not write anything that you wish graded on the question sheets. Start each question on a new sheet of paper.Put your AAPT ID number,your name,the question number and the page number/total pages for this problem,in the upper right hand corner of each page.For example, AAPT ID# Doe,Jamie A1-1/3 A hand-held calculator may be used.Its memory must be cleared of data and programs.You may use only the basic functions found on a simple scientific calculator.Calculators may not be shared.Cell phones,PDA's or cameras may not be used during the exam or while the exam papers are present.You may not use any tables,books,or collections of formulas. Questions with the same point value are not necessarily of the same difficulty. In order to maintain exam security,do not communicate any information about the questions (or their answers/solutions)on this contest until after April 15, 2015. Possibly Useful Information.You may use this sheet for both parts of the exam. g=9.8 N/kg G=6.67×10-11N·m2/kg2 k=1/4r60=8.99×109N.m2/C2 km=4o/4r=10-7T.m/A c=3.00×108m/s k3=1.38×10-23J/K NA=6.02×1023(mol)-1 R=Na=8.31J/(mol·K) o=5.67×10-8J/(s·m2.K4) e=1.602×10-19C 1eV=1.602×10-19J h=6.63×10-34J.s=4.14×10-15eV,s me=9.109×10-31kg=0.511MeV/c2(1+x)n≈1+nx for<1 sin0≈0-ae3 for0l<1 cos0≈1-02forl0l<1 Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Cover Sheet 2 AAPT AIP 2015 UNITED STATES PHYSICS TEAM USA Physics Olympiad Exam INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Part B during this time. • After you have completed Part A you may take a break. • Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Part A during this time. • Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets. • Start each question on a new sheet of paper. Put your AAPT ID number, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example, AAPT ID # Doe, Jamie A1 - 1/3 • A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas. • Questions with the same point value are not necessarily of the same difficulty. • In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after April 15, 2015. Possibly Useful Information. You may use this sheet for both parts of the exam. g = 9.8 N/kg G = 6.67 × 10−11 N · m2/kg2 k = 1/4π0 = 8.99 × 109 N · m2/C 2 km = µ0/4π = 10−7 T · m/A c = 3.00 × 108 m/s kB = 1.38 × 10−23 J/K NA = 6.02 × 1023 (mol)−1 R = NAkB = 8.31 J/(mol · K) σ = 5.67 × 10−8 J/(s · m2 · K4 ) e = 1.602 × 10−19 C 1eV = 1.602 × 10−19 J h = 6.63 × 10−34 J · s = 4.14 × 10−15 eV · s me = 9.109 × 10−31 kg = 0.511 MeV/c 2 (1 + x) n ≈ 1 + nx for |x| 1 sin θ ≈ θ − 1 6 θ 3 for |θ| 1 cos θ ≈ 1 − 1 2 θ 2 for |θ| 1 Copyright c 2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 2 Part A Question Al Consider a particle of mass m that elastically bounces off of an infinitely hard horizontal surface under the influence of gravity.The total mechanical energy of the particle is E and the acceleration of free fall is g.Treat the particle as a point mass and assume the motion is non-relativistic. a.An estimate for the regime where quantum effects become important can be found by simply considering when the deBroglie wavelength of the particle is on the same order as the height of a bounce.Assuming that the deBroglie wavelength is defined by the maximum momentum of the bouncing particle,determine the value of the energy Ea where quantum effects become important.Write your answer in terms of some or all of g,m,and Planck's constant h. Solution Full points will only be awarded if it is clear that the examinee knew the deBroglie wavelength expression. The deBroglie wavelength is p=h/ so if the height H of the bounce is given by E mgH 2m and入=H,then h2 mgH= 2mH2 or H3= h2 2m2g or 瓦=5mg22 An answer of mg2h2 is acceptable,but will not receive full points if it was derived by dimensional analysis alone. b.A second approach allows us to develop an estimate for the actual allowed energy levels of a bouncing particle.Assuming that the particle rises to a height H,we can write -(+ H where p is the momentum as a function of height r above the ground,n is a non-negative integer,and h is Planck's constant. i.Determine the allowed energies En as a function of the integer n,and some or all of g, m,and Planck's constant h. Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 3 Part A Question A1 Consider a particle of mass m that elastically bounces off of an infinitely hard horizontal surface under the influence of gravity. The total mechanical energy of the particle is E and the acceleration of free fall is g. Treat the particle as a point mass and assume the motion is non-relativistic. a. An estimate for the regime where quantum effects become important can be found by simply considering when the deBroglie wavelength of the particle is on the same order as the height of a bounce. Assuming that the deBroglie wavelength is defined by the maximum momentum of the bouncing particle, determine the value of the energy Eq where quantum effects become important. Write your answer in terms of some or all of g, m, and Planck’s constant h. Solution Full points will only be awarded if it is clear that the examinee knew the deBroglie wavelength expression. The deBroglie wavelength is p = h/λ so if the height H of the bounce is given by E = mgH = p 2 2m and λ = H, then mgH = h 2 2mH2 or H3 = h 2 2m2g or Eq = 3 r 1 2 mg2h 2 An answer of p3 mg2h 2 is acceptable, but will not receive full points if it was derived by dimensional analysis alone. b. A second approach allows us to develop an estimate for the actual allowed energy levels of a bouncing particle. Assuming that the particle rises to a height H, we can write 2 Z H 0 p dx = n + 1 2 h where p is the momentum as a function of height x above the ground, n is a non-negative integer, and h is Planck’s constant. i. Determine the allowed energies En as a function of the integer n, and some or all of g, m, and Planck’s constant h. Copyright c 2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 4 ii.Numerically determine the minimum energy of a bouncing neutron.The mass of a neutron is mn=1.675x 10-27 kg=940 MeV/c2;you may express your answer in either Joules or eV. iii.Determine the bounce height of one of these minimum energy neutrons. Solution The integral is not particular difficult to solve, H 2v2m VE-mgx dx; 2V2mE V1-mgx/E da, Jo 2V2mE E V1-u du, mg Jo E 2V2mE vu dv, mg Jo =2v2 E3/22 mg3 So 9mg2h2 1 2/3 En= 32 n+2) Solving for the minimum energy we get E0= 9mg2h2 9(mc2)g2h2 =1.1×10-12eV 128 128c2 The bounce height is given by H= E0=10μm mg This is a very measurable distance! c.Let Eo be the minimum energy of the bouncing neutron and f be the frequency of the bounce. Determine an order of magnitude estimate for the ratio E/f.It only needs to be accurate to within an order of magnitude or so,but you do need to show work! Solution Asf=l/△t and Eo≈△E,Heisenberg's uncertainty relation yeilds △E△t≈h Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 4 ii. Numerically determine the minimum energy of a bouncing neutron. The mass of a neutron is mn = 1.675×10−27 kg = 940 MeV/c 2 ; you may express your answer in either Joules or eV. iii. Determine the bounce height of one of these minimum energy neutrons. Solution The integral is not particular difficult to solve, n + 1 2 h = 2 Z H 0 p dx = 2√ 2m Z H 0 p E − mgx dx, = 2√ 2mE Z H 0 p 1 − mgx/E dx, = 2√ 2mE E mg Z 1 0 √ 1 − u du, = 2√ 2mE E mg Z 1 0 √ v dv, = 2√ 2 E3/2 √ mg 2 3 so En = 3 r 9mg2h 2 32 n + 1 2 2/3 Solving for the minimum energy we get E0 = 3 r 9mg2h 2 128 = 3 r 9(mc2)g 2h 2 128c 2 = 1.1 × 10−12eV The bounce height is given by H = E0 mg = 10 µm This is a very measurable distance! c. Let E0 be the minimum energy of the bouncing neutron and f be the frequency of the bounce. Determine an order of magnitude estimate for the ratio E/f. It only needs to be accurate to within an order of magnitude or so, but you do need to show work! Solution As f = 1/∆t and E0 ≈ ∆E, Heisenberg’s uncertainty relation yeilds ∆E∆t ≈ h Copyright c 2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 5 Question A2 Consider the circuit shown below.Is is a constant current source,meaning that no matter what device is connected between points A and B,the current provided by the constant current source is the same. 4R 2R 2R ● ● 2R3 A B 4R a.Connect an ideal voltmeter between A and B.Determine the voltage reading in terms of any or all of R and Is. Solution An ideal voltmeter has infinite resistance,so no current flows between A and B. Out of symmetry,the same current must flow down each leg,so the current in a leg isIs/2. Assume the potential at the bottom is zero. The potential at A is the same as the junction to the left of A,or,by simple application of Ohm's law VA-52R-LR. The potential at B is found the same way,so R=2I,R. VB =2 The difference is VA-VB=-Is R. The negative is not important for scoring purposes. b.Connect instead an ideal ammeter between A and B.Determine the current in terms of any or all of R and Is. Solution An ideal ammeter has zero resistance.So the problem is simply finding the current through the effective 6R resistor that connects the two vertical branches.This current will flow to the left. Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 5 Question A2 Consider the circuit shown below. Is is a constant current source, meaning that no matter what device is connected between points A and B, the current provided by the constant current source is the same. 2R 4R 4R 2R Is 4R 2R A B a. Connect an ideal voltmeter between A and B. Determine the voltage reading in terms of any or all of R and Is. Solution An ideal voltmeter has infinite resistance, so no current flows between A and B. Out of symmetry, the same current must flow down each leg, so the current in a leg is Is/2. Assume the potential at the bottom is zero. The potential at A is the same as the junction to the left of A, or, by simple application of Ohm’s law VA = Is 2 2R = IsR. The potential at B is found the same way, so VB = Is 2 4R = 2IsR. The difference is VA − VB = −IsR. The negative is not important for scoring purposes. b. Connect instead an ideal ammeter between A and B. Determine the current in terms of any or all of R and Is. Solution An ideal ammeter has zero resistance. So the problem is simply finding the current through the effective 6R resistor that connects the two vertical branches. This current will flow to the left. Copyright c 2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 6 Out of symmetry,the current through each vertical resistance of 2R must be the same,as well as the currents through each vertical resistance of 4R.This will give us a few equations: Is 12+I4 I2=I6+I4 144R 122R+166R Solve, Is 16+214 414=2(I6+I4)+616 or I4=416 and Is=916 or 6= c.It turns out that it is possible to replace the above circuit with a new circuit as follows: B From the point of view of any passive resistance that is connected between A and B the circuits are identical.You don't need to prove this statement,but you do need to find It and Rt in terms of any or all of R and Is. Solution Use the previous results.For a short,all of the current will flow through AB,so It=I6=Is For an open circuit,the potential across AB will be IsR,so Rt=9R. Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 6 Out of symmetry, the current through each vertical resistance of 2R must be the same, as well as the currents through each vertical resistance of 4R. This will give us a few equations: Is = I2 + I4 I2 = I6 + I4 I44R = I22R + I66R Solve, Is = I6 + 2I4 4I4 = 2 (I6 + I4) + 6I6 or I4 = 4I6 and Is = 9I6 or I6 = 1 9 Is c. It turns out that it is possible to replace the above circuit with a new circuit as follows: It Rt A B From the point of view of any passive resistance that is connected between A and B the circuits are identical. You don’t need to prove this statement, but you do need to find It and Rt in terms of any or all of R and Is. Solution Use the previous results. For a short, all of the current will flow through AB, so It = I6 = 1 9 Is. For an open circuit, the potential across AB will be IsR, so Rt = 9R. Copyright c 2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A > Question A3 A large block of mass mo is located on a horizontal frictionless surface.A second block of mass mt is located on top of the first block;the coefficient of friction (both static and kinetic)between the two blocks is given by u.All surfaces are horizontal;all motion is effectively one dimensional. A spring with spring constant k is connected to the top block only;the spring obeys Hooke's Law equally in both extension and compression.Assume that the top block never falls off of the bottom block;you may assume that the bottom block is very,very long.The top block is moved a distance A away from the equilibrium position and then released from rest. WWWWMG mb a.Depending on the value of A,the motion can be divided into two types:motion that expe- riences no frictional energy losses and motion that does.Find the value Ac that divides the two motion types.Write your answer in terms of any or all of u,the acceleration of gravity g,the masses mt and mb,and the spring constant k. Solution The maximum possible acceleration of the top block without slipping is moamax=umtg.If the top block is not slipping then the angular frequency is given by k w2=1 mi+mb amax≥Aw22 or Ae= mt b.Consider now the scenario A>Ac.In this scenario the amplitude of the oscillation of the top block as measured against the original equilibrium position will change with time.Determine the magnitude of the change in amplitude,AA,after one complete oscillation,as a function of any or all of A,u,g,and the angular frequency of oscillation of the top block wt. Solution There are several ways to do this. The energy of an oscillation is approximately equal to B=kA2. Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 7 Question A3 A large block of mass mb is located on a horizontal frictionless surface. A second block of mass mt is located on top of the first block; the coefficient of friction (both static and kinetic) between the two blocks is given by µ. All surfaces are horizontal; all motion is effectively one dimensional. A spring with spring constant k is connected to the top block only; the spring obeys Hooke’s Law equally in both extension and compression. Assume that the top block never falls off of the bottom block; you may assume that the bottom block is very, very long. The top block is moved a distance A away from the equilibrium position and then released from rest. mb mt a. Depending on the value of A, the motion can be divided into two types: motion that experiences no frictional energy losses and motion that does. Find the value Ac that divides the two motion types. Write your answer in terms of any or all of µ, the acceleration of gravity g, the masses mt and mb, and the spring constant k. Solution The maximum possible acceleration of the top block without slipping is mbamax = µmtg. If the top block is not slipping then the angular frequency is given by ω2 = s k mt + mb , so amax ≥ Aω2 2 or Ac = µg mt k 1 + mt mb b. Consider now the scenario A Ac. In this scenario the amplitude of the oscillation of the top block as measured against the original equilibrium position will change with time. Determine the magnitude of the change in amplitude, ∆A, after one complete oscillation, as a function of any or all of A, µ, g, and the angular frequency of oscillation of the top block ωt . Solution There are several ways to do this. The energy of an oscillation is approximately equal to E = 1 2 kA2 . Copyright c 2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 8 Take the derivative,and △E=kA△A That energy is lost to friction.if AAc then the top block has almost completed a complete half cycle before the bottom block catches up with it (speed),so the energy lost in half a cycle is approximately 1 △E=2Af=2Aμmtg 2 where f is the frictional force. Combine, 4μmtg=k△A, or △A=4m9=49 42 c.Assume still that A>Ac.What is the maximum speed of the bottom block during the first complete oscillation cycle of the upper block? Solution The bottom block accelerates according to mt a=μ mb The upper block oscillates as if it is free,since the bottom block exerts a constant force on it,so wt=Vk/mt gives a half period of t=TVmt/k. The maximum speed is then mt b=μTVmt/R mh Note that the maximum speed of the top block is Ut Awt The ratio is =μ9mbkA mtmt t Remember that Ae=μ9K t mt 1+ b %=Am A mb+mt ←1 Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 8 Take the derivative, and ∆E = kA∆A. That energy is lost to friction. if A Ac then the top block has almost completed a complete half cycle before the bottom block catches up with it (speed), so the energy lost in half a cycle is approximately 1 2 ∆E = 2Af = 2Aµmtg where f is the frictional force. Combine, 4µmtg = k∆A, or ∆A = 4µmtg k = 4 µg ωt 2 c. Assume still that A Ac. What is the maximum speed of the bottom block during the first complete oscillation cycle of the upper block? Solution The bottom block accelerates according to a = µg mt mb The upper block oscillates as if it is free, since the bottom block exerts a constant force on it, so ωt = p k/mt gives a half period of t = π p mt/k. The maximum speed is then vb = µg mt mb π p mt/k Note that the maximum speed of the top block is vt = Aωt The ratio is vb vt = µg mt mb mt kA Remember that Ac = µg mt k 1 + mt mb , so vb vt = Ac A mt mb + mt 1 Copyright c 2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 9 Question A4 A heat engine consists of a moveable piston in a vertical cylinder.The piston is held in place by a removable weight placed on top of the piston,but piston stops prevent the piston from sinking below a certain point.The mass of the piston is m =40.0 kg,the cross sectional area of the piston is A 100 cm2,and the weight placed on the piston has a mass of m =120.0 kg. Assume that the region around the cylinder and piston is a vacuum,so you don't need to worry about external atmospheric pressure. At point A the cylinder volume Vo is completely filled with liquid water at a temperature To=320 K and a pressure Pmin that would be just sufficient to lift the piston alone,except the piston has the additional weight placed on top. Heat energy is added to the water by placing the entire cylinder in a hot bath. At point B the piston and weight begins to rise. At point C the volume of the cylinder reaches Viax and the temperature reaches Tmax.The heat source is removed;the piston stops rising and is locked in place. Heat energy is now removed from the water by placing the entire cylinder in a cold bath. At point D the pressure in the cylinder returns to Pmin.The added weight is removed;the piston is unlocked and begins to move down. The cylinder volume returns to Vo.The cylinder is removed from the cold bath,the weight is placed back on top of the piston.and the cycle repeats. Because the liquid water can change to gas,there are several important events that take place At point W the liquid begins changing to gas. At point X all of the liquid has changed to gas.This occurs at the same point as point C described above. At point Y the gas begins to change back into liquid. At point Z all of the gas has changed back into liquid. When in the liquid state you need to know that for water kept at constant volume,a change in temperature AT is related to a change in pressure AP according to △P≈(106Pa/K)△T When in the gas state you should assume that water behaves like an ideal gas. Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 9 Question A4 A heat engine consists of a moveable piston in a vertical cylinder. The piston is held in place by a removable weight placed on top of the piston, but piston stops prevent the piston from sinking below a certain point. The mass of the piston is m = 40.0 kg, the cross sectional area of the piston is A = 100 cm2 , and the weight placed on the piston has a mass of m = 120.0 kg. Assume that the region around the cylinder and piston is a vacuum, so you don’t need to worry about external atmospheric pressure. • At point A the cylinder volume V0 is completely filled with liquid water at a temperature T0 = 320 K and a pressure Pmin that would be just sufficient to lift the piston alone, except the piston has the additional weight placed on top. • Heat energy is added to the water by placing the entire cylinder in a hot bath. • At point B the piston and weight begins to rise. • At point C the volume of the cylinder reaches Vmax and the temperature reaches Tmax. The heat source is removed; the piston stops rising and is locked in place. • Heat energy is now removed from the water by placing the entire cylinder in a cold bath. • At point D the pressure in the cylinder returns to Pmin. The added weight is removed; the piston is unlocked and begins to move down. • The cylinder volume returns to V0. The cylinder is removed from the cold bath, the weight is placed back on top of the piston, and the cycle repeats. Because the liquid water can change to gas, there are several important events that take place • At point W the liquid begins changing to gas. • At point X all of the liquid has changed to gas. This occurs at the same point as point C described above. • At point Y the gas begins to change back into liquid. • At point Z all of the gas has changed back into liquid. When in the liquid state you need to know that for water kept at constant volume, a change in temperature ∆T is related to a change in pressure ∆P according to ∆P ≈ (106 Pa/K)∆T When in the gas state you should assume that water behaves like an ideal gas. Copyright c 2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 10 Of relevance to this question is the pressure/temperature phase plot for water,showing the re- gions where water exists in liquid form or gaseous form.The curve shows the coexistence condition, where water can exist simultaneously as gas or liquid. 240 200 160 ed) 120 Liquid Region 80 40 Gas Region 0 280 300 320 340 360 380 400 Temperature(K) The following graphs should be drawn on the answer sheet provided. a.Sketch a PT diagram for this cycle on the answer sheet.The coexistence curve for the liquid/gas state is shown.Clearly and accurately label the locations of points B through D and W through Z on this cycle. b.Sketch a PV diagram for this cycle on the answer sheet.You should estimate a reasonable value for Vmax,note the scale is logarithmic.Clearly and accurately label the locations of points B through D on this cycle.Provide reasonable approximate locations for points W through Z on this cycle. Solution We are using the Magnus form to approximate the coexistence curve: P=(610.94Pa)e17.625r+aao Note that T is measured in centigrade! We should compare this to the approximation that is valid for ideal gases at low temperatures compared to the critical temperature and constant latent heats: P=Pe ( Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 10 Of relevance to this question is the pressure/temperature phase plot for water, showing the regions where water exists in liquid form or gaseous form. The curve shows the coexistence condition, where water can exist simultaneously as gas or liquid. 280 300 320 340 360 380 400 0 40 80 120 160 200 240 Temperature (K) Pressure (kPa) Liquid Region Gas Region The following graphs should be drawn on the answer sheet provided. a. Sketch a PT diagram for this cycle on the answer sheet. The coexistence curve for the liquid/gas state is shown. Clearly and accurately label the locations of points B through D and W through Z on this cycle. b. Sketch a PV diagram for this cycle on the answer sheet. You should estimate a reasonable value for Vmax, note the scale is logarithmic. Clearly and accurately label the locations of points B through D on this cycle. Provide reasonable approximate locations for points W through Z on this cycle. Solution We are using the Magnus form to approximate the coexistence curve: P = (610.94 Pa)e 17.625 T T +(243.04 ◦C) Note that T is measured in centigrade! We should compare this to the approximation that is valid for ideal gases at low temperatures compared to the critical temperature and constant latent heats: P = P0e L R T −T0 T T0 Copyright c 2015 American Association of Physics Teachers