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USAPhO 2011, Полуфинал — A1

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Всички материали от USAPhO 2011

AAPT UNITED STATES PHYSICS TEAM AIP 2011 — Semifinal Exam · 25 т.

Внимание: Бележка към темата

Printed source errors are preserved: A1’s solution labels both quoted volumes V₀ and says “we used the fat that”; A2’s solution reverses its graph-variable description; A4 prints ω = f/(2π). The solutions repeat the question statements and common instructions; they are shown once here. No individual subpart point values are printed.

Important Instructions for the Exam Supervisor

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 3 parts: the cover sheet (page 2), Part A (pages 3-4), and Part B (pages 6-7). Examinees should be provided parts A and B individually, although they may keep the cover sheet.
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 March 31, 2011.
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.
Please provide the examinees with graph paper for Part A.

Указания

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 1, 2011.

Possibly Useful Information

Possibly Useful Information. You may use this sheet for both parts of the exam.

$g = 9.8\ \mathrm{N/kg}$	$G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}$
$k = 1/4\pi\epsilon_0 = 8.99 \times 10^{9}\ \mathrm{N \cdot m^2/C^2}$	$k_{\mathrm{m}} = \mu_0/4\pi = 10^{-7}\ \mathrm{T \cdot m/A}$
$c = 3.00 \times 10^{8}\ \mathrm{m/s}$	$k_{\mathrm{B}} = 1.38 \times 10^{-23}\ \mathrm{J/K}$
$N_{\mathrm{A}} = 6.02 \times 10^{23}\ (\mathrm{mol})^{-1}$	$R = N_{\mathrm{A}} k_{\mathrm{B}} = 8.31\ \mathrm{J/(mol \cdot K)}$
$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s \cdot m^2 \cdot K^4)}$	$e = 1.602 \times 10^{-19}\ \mathrm{C}$
$1\mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$	$h = 6.63 \times 10^{-34}\ \mathrm{J \cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV \cdot s}$
$m_e = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}$	$(1+x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$
$\sin\theta \approx \theta - \frac{1}{6}\theta^3$ for $\lvert\theta\rvert \ll 1$	$\cos\theta \approx 1 - \frac{1}{2}\theta^2$ for $\lvert\theta\rvert \ll 1$

Условие

Single bubble sonoluminescence occurs when sound waves cause a bubble suspended in a fluid to collapse so that the gas trapped inside increases in temperature enough to emit light. The bubble actually undergoes a series of expansions and collapses caused by the sound wave pressure variations.

We now consider a simplified model of a bubble undergoing sonoluminescence. Assume the bubble is originally at atmospheric pressure $P_0 = 101\ \mathrm{kPa}$ . When the pressure in the fluid surrounding the bubble is decreased, the bubble expands isothermally to a radius of $36.0\ \mu\mathrm{m}$ . When the pressure increases again, the bubble collapses to a radius of $4.50\ \mu\mathrm{m}$ so quickly that no heat can escape. Between the collapse and subsequent expansion, the bubble undergoes isochoric (constant volume) cooling back to its original pressure and temperature. For a bubble containing a monatomic gas, suspended in water of $T = 293\ \mathrm{K}$ , find

a. the number of moles of gas in the bubble,

b. the pressure after the expansion,

c. the pressure after collapse,

d. the temperature after the collapse, and

e. the total work done on the bubble during the whole process.

You may find the following useful: the specific heat capacity at constant volume is $C_V = 3R/2$ and the ratio of specific heat at constant pressure to constant volume is $\gamma = 5/3$ for a monatomic gas.

Отговори

Решение

a. The most important thing in problems like this is to keep track of all the variables carefully. Let the initial pressure, volume, and temperature be $P_0$ , $V_0$ , and $T_0 = T$ . The steps are:

i. Isothermal expansion, after which we have $P_1$ , $V_1$ , and $T_1 = T_0$ .

ii. Adiabatic collapse, after which we have $P_2$ , $V_2$ , and $T_2$ .

iii. Isochoric cooling, to return to the initial state. This implies $V_2 = V_0$ .

In all cases the bubble contains an ideal monatomic gas, so

P_i V_i = nRT_i.

We are also given the volumes

V_0 = \frac{4}{3}\pi(4.50\,\mu\mathrm{m})^3 = 3.82 \times 10^{-16}\,\mathrm{m}^3, \quad V_0 = \frac{4}{3}\pi(36.0\,\mu\mathrm{m})^3 = 1.95 \times 10^{-13}\,\mathrm{m}^3

In particular, we have

n = \frac{P_0 V_0}{R T_0} = 1.58 \times 10^{-14}\,\mathrm{mol}.

b. During an isothermal expansion, $PV$ is constant by the ideal gas law. Thus

P_1 = \frac{P_0 V_0}{V_1} = 197\,\mathrm{Pa}.

c. During an adiabatic process, $PV^\gamma$ is constant where here $\gamma = 5/3$ , so

P_2 = \frac{P_1 V_1^\gamma}{V_2^\gamma} = 6.46 \times 10^{6}\,\mathrm{Pa}.

d. To find the temperature after collapse, we can use the ideal gas law again,

T_2 = \frac{P_2 V_2}{nR} = 1.88 \times 10^{4}\,\mathrm{K}.

e. The work done on the bubble is $dW = -P\,dV$ . During an isothermal expansion, we have

\int P\,dV = nRT \int \frac{dV}{V} = nRT \log\frac{V_f}{V_i}

so the work done on the bubble during the isothermal expansion is

W_1 = -nRT \log\frac{V_1}{V_0} = -2.40 \times 10^{-10}\,\mathrm{J}.

During the adiabatic collapse, it is easiest to find the work done on the bubble using the first law of thermodynamics, $\Delta E = Q + W$ . Since the process is adiabatic, $Q = 0$ , so

W_2 = \Delta E = nC_V \Delta T = \frac{3nR}{2}(T_2 - T_1) = 3.64 \times 10^{-9}\,\mathrm{J}

where we used the fat that $C_V = 3R/2$ for a monatomic gas. There is no work done during the isochoric process, so the total work done on the bubble is

W = 3.40 \times 10^{-9}\,\mathrm{J}.

Since the process is cyclic, this energy must have been radiated away by the bubble, in a flash of light. Note that the final (positive) sign is important. Some textbooks define $dW$ to be the work done on the bubble, while some define it to be the work done by the bubble. Depending on the conventions used, there may be some extra signs in intermediate steps, but the final answer doesn’t depend on the convention.

Оригинал в Архива: USAPhO-2011.pdf · официални решения: USAPhO-2011-Solutions.pdf