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Условие Отговори Решение

USAPhO 2012, Полуфинал — Задача 2. Question A2

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Всички материали от USAPhO 2012

UNITED STATES PHYSICS TEAM 2012 — Semifinal Exam · 25 т.

Important Instructions for the Exam Supervisor

DO NOT DISTRIBUTE THIS PAGE

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 minute 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 April 1, 2012.
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, 2012.

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

An ideal (but not necessarily perfect monatomic) gas undergoes the following cycle.

The gas starts at pressure $P_0$ , volume $V_0$ and temperature $T_0$ .
The gas is heated at constant volume to a pressure $\alpha P_0$ , where $\alpha > 1$ .
The gas is then allowed to expand adiabatically (no heat is transferred to or from the gas) to pressure $P_0$
The gas is cooled at constant pressure back to the original state.

The adiabatic constant $\gamma$ is defined in terms of the specific heat at constant pressure $C_p$ and the specific heat at constant volume $C_v$ by the ratio $\gamma = C_p/C_v$ .

a. Determine the efficiency of this cycle in terms of $\alpha$ and the adiabatic constant $\gamma$ . As a reminder, efficiency is defined as the ratio of work out divided by heat in.

b. A lab worker makes measurements of the temperature and pressure of the gas during the adiabatic process. The results, in terms of $T_0$ and $P_0$ are

Pressure	units of $P_0$	1.21	1.41	1.59	1.73	2.14
Temperature	units of $T_0$	2.11	2.21	2.28	2.34	2.49

Plot an appropriate graph from this data that can be used to determine the adiabatic constant.

c. What is $\gamma$ for this gas?

Отговори

Решение

a. Label the end points as 0, 1, and 2. The ideal gas law yields $T_1 = \alpha T_0$ . The quantity $PV^\gamma$ is conserved along the adiabatic process $1 \to 2$ , so

P_1V_1^\gamma = P_2V_2^\gamma = \frac{1}{\alpha}P_1V_2^\gamma \quad\Rightarrow\quad V_2 = V_1\alpha^{1/\gamma} = V_0\alpha^{1/\gamma}.

Again using the ideal gas law

T_2 = \alpha^{1/\gamma}T_0.

Now, heat enters the gas during the isochoric process $0 \to 1$ , so

Q_{\mathrm{in}} = nC_v\Delta T = nC_v(\alpha - 1)T_0.

Heat exits the system during the process $2 \to 0$ , so

Q_{\mathrm{out}} = nC_p\Delta T = nC_p(\alpha^{1/\gamma} - 1)T_0.

The work done is the difference,

W = Q_{\mathrm{in}} - Q_{\mathrm{out}} = nC_v(\alpha - 1)T_0 - nC_p(\alpha^{1/\gamma} - 1)T_0

and the efficiency is then

\eta = \frac{W}{Q_{\mathrm{in}}} = \frac{C_v(\alpha - 1) - C_p(\alpha^{1/\gamma} - 1)}{C_v(\alpha - 1)} = 1 - \gamma\frac{\alpha^{1/\gamma} - 1}{\alpha - 1}

where we used the definition $\gamma = C_p/C_v$ .

b. For an adiabatic process, the quantity

PV^\gamma\left(\frac{T}{PV}\right)^\gamma = P^{1-\gamma}T^\gamma

is constant, which implies that

P \propto T^{\gamma/(\gamma-1)}.

Thus, if we make a graph with $\log T$ on the horizontal axis and $\log P$ on the vertical, we will find a line with slope $\gamma/(\gamma - 1)$ . To get full credit, all five data points should be plotted and used, though one can get an approximate result with just two.

c. Using the data given, we find a slope of about 3.5, giving $\gamma \approx 1.4$ , as expected for a diatomic gas. Note that credit will not be given for simply writing this number down; some degree of data analysis is necessary.

STOP: Do Not Continue to Part B

If there is still time remaining for Part A, you should review your work for Part A, but do not continue to Part B until instructed by your exam supervisor.

Оригинал в Архива: USAPhO-2012.pdf · официални решения: USAPhO-2012-Solutions.pdf