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Условие Data Answer sheet Отговори Решение

IPhO 1999, Теория — Задача 1. Absorption of radiation by a gas

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Всички материали от IPhO 1999

30th International Physics Olympiad — Padua, Italy — Theoretical competition — Thursday, July 22nd, 1999 · 18 юли 1999 г. · 10 т.

Please read this first:

The time available is 5 hours for 3 problems.
Use only the pen provided.
Use only the front side of the provided sheets.
In addition to the problem texts, that contain the specific data for each problem, a sheet is provided containing a number of general physical constants that may be useful for the problem solutions.
Each problem should be answered on separate sheets.
In addition to "blank" sheets where you may write freely, for each problem there is an Answer sheet where you must summarize the results you have obtained. Numerical results must be written with as many digits as appropriate to the given data; don’t forget the units.
Please write on the "blank" sheets whatever you deem important for the solution of the problem, that you wish to be evaluated during the marking process. However, you should use mainly equations, numbers, symbols, figures, and use as little text as possible.
It's absolutely imperative that you write on top of each sheet that you'll use: your name (“NAME”), your country (“TEAM”), your student code (as shown on the identification tag, “CODE”), and additionally on the "blank" sheets: the problem number (“Problem”), the progressive number of each sheet (from 1 to N, “Page n.”) and the total number (N) of "blank" sheets that you use and wish to be evaluated for that problem (“Page total”). It is also useful to write the section you are answering at the beginning of each such section. If you use some sheets for notes that you don’t wish to be evaluated by the marking team, just put a large cross through the whole sheet, and don’t number it.
When you've finished, turn in all sheets in proper order (for each problem: answer sheet first, then used sheets in order; unused sheets and problem text at the bottom) and put them all inside the envelope where you found them; then leave everything on your desk. You are not allowed to take any sheets out of the room.

This set of problems consists of 13 pages (including this one, the answer sheets and the page with the physical constants)

These problems have been prepared by the Scientific Committee of the 30th IPhO, including professors at the Universities of Bologna, Naples, Turin and Trieste.

Условие

A cylindrical vessel, with its axis vertical, contains a molecular gas at thermodynamic equilibrium. The upper base of the cylinder can be displaced freely and is made out of a glass plate; let's assume that there is no gas leakage and that the friction between glass plate and cylinder walls is just sufficient to damp oscillations but doesn't involve any significant loss of energy with respect to the other energies involved. Initially the gas temperature is equal to that of the surrounding environment. The gas can be considered as perfect within a good approximation. Let's assume that the cylinder walls (including the bases) have a very low thermal conductivity and capacity, and therefore the heat transfer between gas and environment is very slow, and can be neglected in the solution of this problem.

Through the glass plate we send into the cylinder the light emitted by a constant power laser; this radiation is easily transmitted by air and glass but is completely absorbed by the gas inside the vessel. By absorbing this radiation the molecules reach excited states, where they quickly emit infrared radiation returning in steps to the molecular ground state; this infrared radiation, however, is further absorbed by other molecules and is reflected by the vessel walls, including the glass plate. The energy absorbed from the laser is therefore transferred in a very short time into thermal movement (molecular chaos) and thereafter stays in the gas for a sufficiently long time.

We observe that the glass plate moves upwards; after a certain irradiation time we switch the laser off and we measure this displacement.

1. Using the data below and - if necessary - those on the sheet with physical constants, compute the temperature and the pressure of the gas after the irradiation. [2 points]

2. Compute the mechanical work carried out by the gas as a consequence of the radiation absorption. [1 point]

3. Compute the radiant energy absorbed during the irradiation. [2 points]

4. Compute the power emitted by the laser that is absorbed by the gas, and the corresponding number of photons (and thus of elementary absorption processes) per unit time. [1.5 points]

5. Compute the efficiency of the conversion process of optical energy into a change of mechanical potential energy of the glass plate. [1 point]

Thereafter the cylinder axis is slowly rotated by 90°, bringing it into a horizontal direction. The heat exchanges between gas and vessel can still be neglected.

6. State whether the pressure and/or the temperature of the gas change as a consequence of such a rotation, and - if that is the case – what is its/their new value. [2.5 points]

Data

Room pressure: $p_0 = 101.3\ \mathrm{kPa}$
Room temperature: $T_0 = 20.0^{\circ}\mathrm{C}$
Inner diameter of the cylinder: $2r = 100\ \mathrm{mm}$
Mass of the glass plate: $m = 800\ \mathrm{g}$
Quantity of gas within the vessel: $n = 0.100\ \mathrm{mol}$
Molar specific heat at constant volume of the gas: $c_\mathrm{V} = 20.8\ \mathrm{J/(mol\cdot K)}$
Emission wavelength of the laser: $\lambda = 514\ \mathrm{nm}$
Irradiation time: $\Delta \mathrm{t} = 10.0\ \mathrm{s}$
Displacement of the movable plate after irradiation: $\Delta s = 30.0\ \mathrm{mm}$

Answer sheet

In this problem you are requested to give your results both as analytical expressions and with numerical data and units: write expressions first and then data (e.g. $A=bc=1.23\ \mathrm{m^2}$ ).

1. Gas temperature after the irradiation …
Gas pressure after the irradiation …

2. Mechanical work carried out …

3. Overall optical energy absorbed by the gas …

4. Optical laser power absorbed by the gas …
Absorption rate of photons (number of absorbed photons per unit time) …

5. Efficiency in the conversion of optical energy into change of mechanical potential energy of the glass plate …

6. Owing to the cylinder rotation, is there a pressure change? YES ☐ NO ☐
If yes, what is its new value? …
Owing to the cylinder rotation, is there a temperature change? YES ☐ NO ☐
If yes, what is its new value? …

Отговори

Решение

At equilibrium the pressure $p$ inside the vessel must be equal to the room pressure $p_0$ plus the pressure induced by the weight of the movable base: $p = p_0 + \dfrac{mg}{\pi r^2}$ . This is true before and after irradiation. Initially the gas temperature is room temperature. Owing to the state equation of perfect gases, the initial gas volume $V_1$ is $V_1 = \dfrac{nRT_0}{p}$ (where $R$ is the gas constant) and therefore the height $h_1$ of the cylinder which is occupied by the gas is $h_1 = \dfrac{V_1}{\pi r^2} = \dfrac{nRT_0}{p_0 \pi r^2 + mg}$ . After irradiation, this height becomes $h_2 = h_1 + \Delta s$ , and therefore the new temperature is

T_2 = T_0\left(1 + \frac{\Delta s}{h_1}\right) = T_0 + \frac{\Delta s(p_0 \pi r^2 + mg)}{nR}.

Numerical values: $p = 102.32\ \mathrm{kPa}$ ; $T_2 = 322\ \mathrm{K} = 49^{\circ}\mathrm{C}$

The mechanical work made by the gas against the plate weight is $mg\Delta s$ and against the room pressure is $p_0 \pi r^2 \Delta s$ , therefore the total work is $L = (mg + p_0 \pi r^2)\Delta s = 24.1\ \mathrm{J}$
The internal energy, owing to the temperature variation, varies by an amount $\Delta U = nc_\mathrm{V}(T_2 - T_0)$ . The heat introduced into the system during the irradiation time $\Delta t$ is

Q = \Delta U + L = nc_\mathrm{V}\frac{T_0 \Delta s}{h_1} + (mg + p_0 \pi r^2)\Delta s = \Delta s\left(p_0 \pi r^2 + mg\right)\left(\frac{c_\mathrm{V}}{R} + 1\right).

This heat comes exclusively from the absorption of optical radiation and coincides therefore with the absorbed optical energy, $Q = 84\ \mathrm{J}$ .

The same result can also be obtained by considering an isobaric transformation and remembering the relationship between molecular heats:

Q = nc_p(T_2 - T_0) = n(c_\mathrm{V} + R)\left[\frac{\Delta s(p_0 \pi r^2 + mg)}{nR}\right] = \Delta s(p_0 \pi r^2 + mg)\left(\frac{c_\mathrm{V}}{R} + 1\right)

Since the laser emits a constant power, the absorbed optical power is $W = \dfrac{Q}{\Delta t} = \left(\dfrac{c_\mathrm{V}}{R} + 1\right)\dfrac{\Delta s}{\Delta t}(p_0 \pi r^2 + mg) = 8.4\ \mathrm{W}$ . The energy of each photon is $hc/\lambda$ , and thus the number of photons absorbed per unit time is $\dfrac{W\lambda}{hc} = 2.2\cdot 10^{19}\ \mathrm{s^{-1}}$
The potential energy change is equal to the mechanical work made against the plate weight, therefore the efficiency $\eta$ of the energy transformation is

\frac{mg\Delta s}{Q} = \frac{1}{\left(1 + \dfrac{p_0 \pi r^2}{mg}\right)\left(1 + \dfrac{c_V}{R}\right)} = 2.8\cdot 10^{-3} \approx 0.3\%

When the cylinder is rotated and its axis becomes horizontal, we have an adiabatic transformation where the pressure changes from $p$ to $p_0$ , and the temperature changes therefore to a new value $T_3$ . The equation of the adiabatic transformation $pV^{\gamma} = \text{constant}$ may now be written in the form $T_3 = T_2\left(\dfrac{p_0}{p}\right)^{\frac{\gamma - 1}{\gamma}}$ , where $\gamma = \dfrac{c_\mathrm{p}}{c_\mathrm{V}} = \dfrac{c_\mathrm{V} + R}{c_\mathrm{V}} = 1 + \dfrac{R}{c_\mathrm{V}} = 1.399$ . Finally $T_3 = 321\ \mathrm{K} = 48^{\circ}\mathrm{C}$

Grading guidelines

1.	0.5	Understanding the relationship between inner and outer pressure
	0.7	Proper use of the plate displacement
	0.2+0.2	Correct results for final pressure
	0.2+0.2	Correct results for final temperature
2.	0.6	Understanding that the work is made both against plate weight and against atmospheric pressure
	0.2+0.2	Correct results for work
3.	1	Correct approach
	0.5	Correct equation for heat
	0.3	Understanding that the absorbed optical energy equals heat
	0.2	Correct numerical result for optical energy
4.	0.2+0.2	Correct results for optical power
	0.5	Einstein’s equation
	0.3+0.3	Correct results for number of photons
5.	0.6	Computation of the change in potential energy
	0.2+0.2	Correct results for efficiency
6.	0.8	Understanding that the pressure returns to room value
	0.4	Understanding that there is an adiabatic transformation
	0.4	Equation of adiabatic transformation
	0.5	Derivation of $\gamma$ from the relationship between specific heats
	0.2+0.2	Correct results for temperature

For “correct results” two possible marks are given: the first one is for the analytical equation and the second one for the numerical value.
For the numerical values a full score cannot be given if the number of digits is incorrect (more than one digit more or less than those given in the solution) or if the units are incorrect or missing.
No bonus can be given for taking into account the gas weight

Physical constants and general data

In addition to the numerical data given within the text of the individual problems, the knowledge of some general data and physical constants may be useful, and you may find them among the following ones. These are nearly the most accurate data currently available, and they have thus a large number of digits; you are expected, however, to write your results with a number of digits that must be appropriate for each problem.

Speed of light in vacuum: $c = 299792458\ \mathrm{m\cdot s^{-1}}$
Magnetic permeability of vacuum: $\mu_0 = 4\pi\cdot 10^{-7}\ \mathrm{H\cdot m^{-1}}$
Dielectric constant of vacuum: $\varepsilon_0 = 8.8541878\ \mathrm{pF\cdot m^{-1}}$
Gravitational constant: $G = 6.67259\cdot 10^{-11}\ \mathrm{m^3/(kg\cdot s\ )}$
Gas constant: $R = 8.314510\ \mathrm{J/(mol\cdot K)}$
Boltzmann's constant: $k = 1.380658\cdot 10^{-23}\ \mathrm{J\cdot K^{-1}}$
Stefan's constant: $\sigma = 56.703\ \mathrm{nW/(m\ \cdot K^4)}$
Proton charge: $e = 1.60217733\cdot 10^{-19}\ \mathrm{C}$
Electron mass: $m_\mathrm{e} = 9.1093897\cdot 10^{-31}\ \mathrm{kg}$
Planck's constant: $h = 6.6260755\cdot 10^{-34}\ \mathrm{J\cdot s}$
Base of centigrade scale: $T_\mathrm{K} = 273.15\ \mathrm{K}$
Sun mass: $M_\mathrm{S} = 1.991\cdot 10^{30}\ \mathrm{kg}$
Earth mass: $M_\mathrm{E} = 5.979\cdot 10^{24}\ \mathrm{kg}$
Mean radius of Earth: $r_\mathrm{E} = 6.373\ \mathrm{Mm}$
Major semiaxis of Earth orbit: $R_\mathrm{E} = 1.4957\cdot 10^{11}\ \mathrm{m}$
Sidereal day: $d_\mathrm{S} = 86.16406\ \mathrm{ks}$
Year: $y = 31.558150\ \mathrm{Ms}$
Standard value of the gravitational field at the Earth surface: $g = 9.80665\ \mathrm{m\cdot s^{-2}}$
Standard value of the atmospheric pressure at sea level: $p_0 = 101325\ \mathrm{Pa}$
Refractive index of air for visible light, at standard pressure and 15 °C: $n_\mathrm{air} = 1.000277$
Solar constant: $S = 1355\ \mathrm{W\cdot m^{-2}}$
Jupiter mass: $M = 1.901\cdot 10^{27}\ \mathrm{kg}$
Equatorial Jupiter radius: $R_\mathrm{B} = 69.8\ \mathrm{Mm}$
Average radius of Jupiter’s orbit: $R = 7.783\cdot 10^{11}\ \mathrm{m}$
Jovian day: $d_\mathrm{J} = 35.6\ \mathrm{ks}$
Jovian year: $y_\mathrm{J} = 374.32\ \mathrm{Ms}$
$\pi$ : 3.14159265

Alternate source edition — physical constants

The alternate original IPhO_1999_Theo Question 1.pdf, page 5, prints the following units:

Gravitational constant: $G = 6.67259\cdot10^{-11}\ \mathrm{m^3/(kg\cdot s^2)}$ .

Stefan’s constant: $\sigma = 56.703\ \mathrm{nW/(m^2\cdot K^4)}$ .

In the primary original 1.pdf, page 5, the corresponding denominators are printed without these squared exponents: $\mathrm{kg\cdot s}$ and $\mathrm{m\cdot K^4}$ . The constants sheet above preserves that primary printed variant; both complete original editions are provided in the sources. The task, data, answer sheet, worked solution and grading are the same in both editions.

Оригинал в Архива: 1.pdf · официални решения: 1_sol.pdf · допълнителни данни: IPhO_1999_Theo Question 1.pdf · допълнителни данни: IPhO_1999_Theo Solution 1.pdf