Съдържание

Условие Решение

USAPhO 2000, Полуфинал — A1

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Всички материали от USAPhO 2000

2000 Semi-Final Exam · 20 т.

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.
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. It may not be used in graphing mode. Calculators may not be shared. You may not use any tables, books, or collections of formulas. You may use a ruler or straight edge.
Questions with the same point value are not necessarily of the same difficulty.
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	$k=1/4\pi\epsilon_0=9.0\times10^9$ 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	$c=3.0\times10^8$ 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
Elementary charge	$e=1.6\times10^{-19}$ C
1 electron volt	$1\text{ eV}=1.6\times10^{-19}$ 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	$m=9.1\times10^{-31}\text{ kg}=0.51\text{ MeV}/c^2$
Binomial expansion	$(1+x)^n\simeq1+nx\quad\text{for }\lvert x\rvert\ll1$

Условие

An empty 4.2-g balloon is filled with air to a pressure of 2.0 atm at the ambient temperature of $20.0^\circ$ C. When filled it is spherical with a radius of 15.0 cm. The mean molecular mass of air is 28.8 g/mole and the atmospheric pressure is $1.01\times10^5$ Pa.

a. The filled balloon is weighed by placing it on a scale. What is the scale reading? [15 т.]

b. The filled balloon is now gently heated until its radius increases by 10%. What is the scale reading now? [5 т.]

Решение

Let $R=0.15\text{ m}=$ the radius of the balloon

$p=2\times1.01\times10^5\text{ Pa}=$ The pressure inside the balloon

$m_b=0.0042\text{ kg}=$ The mass of the empty balloon

$T=(20+273)\text{ K}=$ The temperature of the air

a. (15 т.)

The volume of the balloon is $V=\frac43\pi R^3=\frac43\pi(0.15\text{ m})^3=0.0141\text{ m}^3$ .

Using the ideal gas law

n=\frac{pV}{RT}=\frac{(2.02\times10^5\text{ Pa})(0.0141\text{ m}^3)}{(8.31\text{ J/mole}\cdot\text{K})(293\text{ K})}=1.17\text{ moles}.

Therefore the mass of the air inside the balloon is

m=nM=(1.17\text{ moles})(28.8\text{ g/mole})=33.7\text{ g}=0.0337\text{ kg}.

The scale reading will be equal to the force of gravitational attraction $(m+m_b)g$ on the balloon and the enclosed air minus the buoyant force $B=m_{dis}g=\rho gV$ acting on the balloon, where $m_{dis}$ is the mass of the air displaced by the balloon.

F_s=(m+m_b)g-m_{dis}g=(m+m_b-m_{dis})g.

The temperature and volume of the air displaced is the same as that of the air within the balloon but the pressure is only half as much.

m_{dis}=n_{dis}M=\tfrac12nM=\tfrac12m=0.0168\text{ kg}.

F_s=(0.0337\text{ kg}+0.0042\text{ kg}-0.0168\text{ kg})(9.8\text{ m/s}^2)=0.21\text{ N}.

b. (5 т.)

The amount of air inside the balloon and its mass do not change. The volume of air displaced and its mass both change by the amount

m'_{dis}=\rho V'=\rho\tfrac43\pi(R')^3=\rho\tfrac43\pi(1.1R)^3=(1.1)^3\rho\tfrac43\pi R^3=(1.1)^3m_{dis}=(1.1)^3(0.0168\text{ kg})=0.0224\text{ kg}.

F_s=(m+m_b-m'_{dis})g=(0.0337\text{ kg}+0.0042\text{ kg}-0.0224\text{ kg})(9.8\text{ m/s}^2)=0.15\text{ N}.

Оригинал в Архива: Semifinal 2000.pdf · официални решения: Semifinal 2000 Solutions.pdf