Съдържание

Условие Решение

USAPhO 2005, Полуфинал — A2

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Всички материали от USAPhO 2005

2005 Semi-Final Exam · 25 т.

2005 Semi-Final Exam — Instructions

AAPT / AIP — United States Physics Team 2005.

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Work Part A first. You have 90 minutes to complete all four problems.
After you have completed Part A, you may take a break.
Then work Part B. You have 90 minutes to complete both problems.
Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example: 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.
Do not discuss the contents of this exam with anyone until after the submission deadline.
Good luck!

Possibly Useful Information

Quantity	Value
Gravitational field at the Earth’s surface	$g=9.8$ N/kg
Newton’s gravitational constant	$G=6.67\times10^{-11}$ N·m $^2$ /kg $^2$
Coulomb’s constant	$k=1/4\pi\epsilon_0=8.99\times10^9$ N·m $^2$ /C $^2$
Biot-Savart constant	$k_m=\mu_0/4\pi=10^{-7}$ T·m/A
Speed of light in a vacuum	$c=3.00\times10^8$ m/s
Boltzmann’s constant	$k_B=1.38\times10^{-23}$ J/K
Avogadro’s number	$N_A=6.02\times10^{23}$ (mol) $^{-1}$
Ideal gas constant	$R=N_Ak_B=8.31$ J/(mol·K)
Stefan-Boltzmann constant	$\sigma=5.67\times10^{-8}$ J/(s·m $^2$ ·K $^4$ )
Elementary charge	$e=1.602\times10^{-19}$ C
1 electron volt	$1\text{ eV}=1.602\times10^{-19}$ J
Planck’s constant	$h=6.63\times10^{-34}\text{ J}\cdot\text{s}=4.14\times10^{-15}\text{ eV}\cdot\text{s}$
Electron mass	$m=9.109\times10^{-31}\text{ kg}=0.511\text{ MeV}/c^2$
Binomial expansion	$(1+x)^n\approx1+nx$ for $\lvert x\rvert\ll1$
Small angle approximations	$\sin\theta\approx\theta$ , $\cos\theta\approx1-\tfrac12\theta^2$

Условие

A sample of water of mass 0.0360 kg and temperature $100^\circ$ C is added to an enclosed container of volume 2.00 m $^3$ that contains air at a pressure of $1.00\times10^5$ Pa and a temperature of 500 K. Assume both air and steam can be treated as ideal gases and that all of the water vaporizes. What is the final temperature and pressure inside the container?

Possibly useful information:

Molar Heat Capacities	Air	Steam
$C_V$ (at constant volume)	20.8 J/(mol·K)	27.9 J/(mol·K)
$C_P$ (at constant pressure)	29.1 J/(mol·K)	36.2 J/(mol·K)

$L_V$ (Latent heat of vaporization) $=2.26\times10^6$ J/kg.

Molar mass of water $=18.0$ g/mol.

Решение

Source note: the numerical substitution below interchanges the initial air and steam temperatures. The printed equation and the printed final answer are both retained.

Vaporizing the water will require energy:

\Delta U_{\rm vap}=mL_v=(0.0360\text{ kg})(2.26\times10^6\text{ J/kg})=81.4\text{ kJ}.\qquad (A2-1)

We find that we have $\frac{0.0360\text{ kg}}{0.018\text{ kg/mol}}=2.0\text{ mol}$ of water.

Using $pV=nRT$ , we find that we have 48.1 mol of air.

Since the volume is constant, the change in internal energy of the air and of the steam are given by

\Delta U=nC_V\Delta T.\qquad (A2-2)

The total change in internal energy of the system must be zero:

\Delta U_{\rm vap}+\Delta U_{\rm steam}+\Delta U_{\rm air}=0.\qquad (A2-3)

Substituting (A2-1) and (A2-2) into (A2-3):

81.4\text{ kJ}+(nC_V\Delta T)_{\rm steam}+(nC_V\Delta T)_{\rm air}=0.\qquad (A2-4)

81.4\text{ kJ}+(2.0\text{ mol})(27.9\text{ J/(mol}\cdot\text{K)})(T_f-500\text{ K})+(48.1\text{ mol})(20.8\text{ J/(mol}\cdot\text{K)})(T_f-373\text{ K})=0.

T_{\rm final}=416\text{ K}.\qquad (A2-5)

There are a total of 2.0 mol + 48.1 mol = 50.1 mol of gas in the container, so from the ideal gas law,

p=\frac{nRT}{V}=\frac{(50.1\text{ mol})(8.31\text{ J/mol}\cdot\text{K})(416\text{ K})}{2.00\text{ m}^3}=86.7\text{ kPa}.\qquad (A2-6)

Оригинал в Архива: Semifinal 2005.pdf · официални решения: Semifinal 2005 Solutions.pdf