Съдържание

Условие Отговори Решение

USAPhO 2003, Полуфинал — Задача 4. A4

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Всички материали от USAPhO 2003

AAPT / AIP United States Physics Team 2003 — 2003 Semi-Final Exam

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

Gravitational field at the Earth’s surface	$g = 9.8\ \mathrm{N/kg}$
Newton’s gravitational constant	$G = 6.67 \times 10^{-11}\ \mathrm{N{\cdot}m^2/kg^2}$
Coulomb’s constant	$k = 1/4\pi\varepsilon_0 = 9.0 \times 10^{9}\ \mathrm{N{\cdot}m^2/C^2}$
Biot-Savart constant	$k_\mathrm{m} = \mu_0/4\pi = 10^{-7}\ \mathrm{T{\cdot}m/A}$
Speed of light in a vacuum	$c = 3.0 \times 10^{8}\ \mathrm{m/s}$
Boltzmann’s constant	$k_\mathrm{B} = 1.38 \times 10^{-23}\ \mathrm{J/K}$
Avogadro’s number	$N_\mathrm{A} = 6.02 \times 10^{23}\ \mathrm{(mol)^{-1}}$
Ideal gas constant	$R = N_\mathrm{A}k_\mathrm{B} = 8.31\ \mathrm{J/(mol{\cdot}K)}$
Stefan-Boltzmann constant	$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s{\cdot}m^2{\cdot}K^4)}$
Elementary charge	$e = 1.6 \times 10^{-19}\ \mathrm{C}$
1 electron volt	$1\ \mathrm{eV} = 1.6 \times 10^{-19}\ \mathrm{J}$
Planck’s constant	$h = 6.63 \times 10^{-34}\ \mathrm{J{\cdot}s} = 4.14 \times 10^{-15}\ \mathrm{eV{\cdot}s}$
Electron mass	$m = 9.1 \times 10^{-31}\ \mathrm{kg} = 0.51\mathrm{MeV/c^2}$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$
Small angle approximations	$\sin\theta \approx \theta$ ; $\cos\theta \approx 1 - \tfrac{1}{2}\theta^2$

Условие

An ideal gas has a heat capacity per unit mole at constant pressure of $C_\mathrm{P} = 41.9$ J/(mol·K) and a heat capacity per unit mole at constant volume of $C_\mathrm{V} = 33.5$ J/(mol·K). An 0.100-mole sample of this gas goes through the accompanying cycle. Process 1–>2 is isobaric (constant pressure), 2–>3 is adiabatic (no heat enters or leaves), and 3–>1 is isothermal (constant temperature).

P–V diagram (P vertical, V horizontal) of a cycle: horizontal isobar from point 1 to point 2 (arrow to the right), a steep adiabat from 2 down to point 3 (arrow downward), and an isotherm from 3 back up to 1 (arrow toward 1).

$T_1 = 300$ K	$T_2 = 400$ K
$V_1 = 0.0118\ \mathrm{m^3}$	$V_2 = 0.0157\ \mathrm{m^3}$

a. Find the work done by the gas during each process ( $W_{1\text{–>}2}$ , $W_{2\text{–>}3}$ , and $W_{3\text{–>}1}$ ). [15 т.]

b. What is the efficiency of this cycle? [10 т.]

Отговори

Решение

a. Process $1\to 2$ takes place at constant pressure. The work done is

W_{1\to 2} = p\Delta V.

For a constant pressure process, the ideal gas law yields $p\Delta V = nR\Delta T$
and

W_{1\to 2} = nR\Delta T = (0.10\ \mathrm{mol})(8.31\ \mathrm{J/mol\cdot K})(400\ \mathrm{K} - 300\ \mathrm{K}) = +83.1\ \mathrm{J}.

Process $2\to 3$ is adiabatic, $Q = 0$ . Combining this with the first law of thermodynamics,

Q = W + \Delta U.

The work done is the negative of the change in internal energy.

W_{2\to 3} = -\Delta U = -nC_V\Delta T = -nC_V\left(T_3 - T_2\right).

Since process $3\to 1$ is isothermal, $T_3 = T_1 = 300\ \mathrm{K}$ .
And

W_{2\to 3} = -(0.10\ \mathrm{mol})(33.5\ \mathrm{J/mol\cdot K})(300\ \mathrm{K} - 400\ \mathrm{K}) = +335\ \mathrm{J}.

For the isothermal process $3\to 1$ , the work done is

W_{3\to 1} = nRT\ln\left(\frac{V_1}{V_3}\right).

Since $2\to 3$ is adiabatic,

p_2V_2^{\gamma} = p_3V_3^{\gamma}

with $\gamma = \frac{C_P}{C_V} = \frac{41.9}{33.5} = \frac{5}{4}$ . Combining this with the ideal gas law $pV = nRT$ ,

nRT_2V_2^{\gamma-1} = nRT_3V_3^{\gamma-1}.

Solving for $V_3$ ,

V_3^{\gamma-1} = \left(\frac{T_2}{T_3}\right)V_2^{\gamma-1}

V_3 = \left(\frac{T_2}{T_3}\right)^{1/(\gamma-1)}V_2 = \left(\frac{T_2}{T_3}\right)^4V_2 = \left(\frac{400}{300}\right)^4\left(0.0157\ \mathrm{m^3}\right) = 0.0496\ \mathrm{m^3}

Finally,

W_{3\to 1} = (0.10\ \mathrm{mol})(8.31\ \mathrm{J/mol\cdot K})(300\ \mathrm{K})\ln\left(\frac{0.0118}{0.0496}\right) = -358\ \mathrm{J}.

b. The efficiency is the total work done divided by the heat input. Heat is input during process $1\to 2$ and leaves during $3\to 1$ .

Q_{in} = Q_{1\to 2} = nC_p\Delta T = (0.10\ \mathrm{mol})(41.9\ \mathrm{J/mol\cdot K})(100\ \mathrm{K}) = +419\ \mathrm{J}

W_{net} = W_{1\to 2} + W_{2\to 3} + W_{3\to 1} = +83\ \mathrm{J} + 335\ \mathrm{J} - 358\ \mathrm{J} = 60\ \mathrm{J}

e = \frac{60}{419} = 0.14

Оригинал в Архива: Semifinal 2003.pdf · официални решения: Semifinal 2003 Solutions.pdf