IPhO 1999 · Theory problem 1

Absorption of radiation by a gas

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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.

  1. 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: \(c_V = 20.8\ \mathrm{J/(mol\cdot K)}\)
Emission wavelength of the laser: \(\lambda = 514\ \mathrm{nm}\)
Irradiation time: \(\Delta t = 10.0\ \mathrm{s}\)
Displacement of the movable plate after irradiation: \(\Delta s = 30.0\ \mathrm{mm}\)

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: \(\mu_0 = 4\pi\cdot10^{-7}\ \mathrm{H\cdot m^{-1}}\)
Dielectric constant of vacuum: \(\varepsilon_0 = 8.8541878\ \mathrm{pF\cdot m^{-1}}\)
Gravitational constant: \(G = 6.67259\cdot10^{-11}\ \mathrm{m^3/(kg\cdot s^2)}\)
Gas constant: \(R = 8.314510\ \mathrm{J/(mol\cdot K)}\)
Boltzmann's constant: \(k = 1.380658\cdot10^{-23}\ \mathrm{J\cdot K^{-1}}\)
Stefan's constant: \(\sigma = 56.703\ \mathrm{nW/(m^2\cdot K^4)}\)
Proton charge: \(e = 1.60217733\cdot10^{-19}\ \mathrm{C}\)
Electron mass: \(m_e = 9.1093897\cdot10^{-31}\ \mathrm{kg}\)
Planck's constant: \(h = 6.6260755\cdot10^{-34}\ \mathrm{J\cdot s}\)
Base of centigrade scale: \(T_K = 273.15\ \mathrm{K}\)
Sun mass: \(M_S = 1.991\cdot10^{30}\ \mathrm{kg}\)
Earth mass: \(M_E = 5.979\cdot10^{24}\ \mathrm{kg}\)
Mean radius of Earth: \(r_E = 6.373\ \mathrm{Mm}\)
Major semiaxis of Earth orbit: \(R_E = 1.4957\cdot10^{11}\ \mathrm{m}\)
Sidereal day: \(d_S = 86.16406\ \mathrm{ks}\)
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 visibile light, at standard pressure and \(15^\circ\mathrm{C}\): \(n_{\mathrm{air}} = 1.000277\)
Solar constant: \(S = 1355\ \mathrm{W\cdot m^{-2}}\)
Jupiter mass: \(M = 1.901\cdot10^{27}\ \mathrm{kg}\)
Equatorial Jupiter radius: \(R_B = 69.8\ \mathrm{Mm}\)
Average radius of Jupiter’s orbit: \(R = 7.783\cdot10^{11}\ \mathrm{m}\)
Jovian day: \(d_J = 35.6\ \mathrm{ks}\)
Jovian year: \(y_J = 374.32\ \mathrm{Ms}\)
\(\pi\): \(3.14159265\)

Official English text of the problem.

Answers

  1. After irradiation: \[ \boxed{T_1=322.1\ \mathrm{K}=49.0^\circ\mathrm{C},\qquad p_1=102.3\ \mathrm{kPa}.} \]

  2. Mechanical work done by the gas: \[ \boxed{W=24.1\ \mathrm{J}.} \]

  3. Radiant energy absorbed: \[ \boxed{E_{\mathrm{abs}}=84.4\ \mathrm{J}.} \]

  4. Absorbed laser power and photon absorption rate: \[ \boxed{P_{\mathrm{abs}}=8.44\ \mathrm{W},\qquad \dot N=2.18\times10^{19}\ \mathrm{s^{-1}}.} \]

  5. Efficiency for increasing the plate’s gravitational potential energy: \[ \boxed{\eta=2.79\times10^{-3}=0.279\%.} \]

  6. Both pressure and temperature decrease. After rotation: \[ \boxed{p_2=101.3\ \mathrm{kPa},\qquad T_2=321.2\ \mathrm{K}=48.1^\circ\mathrm{C}.} \]

Solution

We use the ideal-gas approximation, the given constant molar heat capacity, and negligible heat leakage and frictional energy loss. Initial and final states have no macroscopic kinetic energy. The room pressure used below is the problem-specific value \(p_0=101.3\ \mathrm{kPa}\).

The cylinder cross-sectional area and the measured volume increase are \[ A=\pi r^2=\pi(0.0500)^2 =7.85398\times10^{-3}\ \mathrm{m^2}, \] \[ \Delta V=A\Delta s =(7.85398\times10^{-3})(0.0300) =2.35619\times10^{-4}\ \mathrm{m^3}. \]

1. Temperature and pressure after irradiation

In vertical mechanical equilibrium, the upward force of the gas balances atmospheric pressure and the plate’s weight: \[ p_1A=p_0A+mg. \] Thus \[ p_1=p_0+\frac{mg}{A} =101300+\frac{(0.800)(9.80665)}{7.85398\times10^{-3}} =102298.9\ \mathrm{Pa}. \] The same balance applies before irradiation, so the initial and final gas pressures are equal: \[ \boxed{p_1=102.3\ \mathrm{kPa}.} \]

Let \(V_0\) and \(V_1\) be the gas volumes before and after irradiation. Applying the ideal-gas law to both states, \[ p_1V_0=nRT_0,\qquad p_1V_1=nRT_1. \] Subtracting gives \[ p_1\Delta V=nR(T_1-T_0). \] With \(T_0=293.15\ \mathrm{K}\), \[ T_1-T_0 =\frac{(102298.9)(2.35619\times10^{-4})} {(0.100)(8.314510)} =28.9898\ \mathrm{K}. \] Therefore \[ \boxed{T_1=322.1\ \mathrm{K}=49.0^\circ\mathrm{C}.} \]

2. Mechanical work done by the gas

The opposing force \(p_0A+mg\) is constant. With negligible frictional loss and zero initial and final plate kinetic energy, the gas does work \[ W=(p_0A+mg)\Delta s =p_0\Delta V+mg\Delta s =p_1\Delta V. \] Numerically, \[ W=(102298.9)(2.35619\times10^{-4}) =24.1036\ \mathrm{J}, \] so \[ \boxed{W=24.1\ \mathrm{J}.} \]

This consists of \(23.8683\ \mathrm{J}\) of work against atmospheric pressure and \(0.235360\ \mathrm{J}\) of work lifting the plate.

3. Radiant energy absorbed

The absorbed radiation rapidly becomes thermal energy. Energy conservation, with work positive when done by the gas, gives \[ E_{\mathrm{abs}}=\Delta U+W. \] For the ideal gas, \[ \Delta U=nc_V(T_1-T_0) =(0.100)(20.8)(28.9898) =60.2988\ \mathrm{J}. \] Consequently, \[ E_{\mathrm{abs}}=60.2988+24.1036 =84.4024\ \mathrm{J}, \] or \[ \boxed{E_{\mathrm{abs}}=84.4\ \mathrm{J}.} \]

Equivalently, because \(W=nR(T_1-T_0)\), \[ E_{\mathrm{abs}}=n(c_V+R)(T_1-T_0)=nc_P(T_1-T_0), \] where \[ c_P=c_V+R=29.114510\ \mathrm{J\,mol^{-1}K^{-1}}. \]

4. Absorbed power and photon absorption rate

The laser power is constant, hence \[ P_{\mathrm{abs}}=\frac{E_{\mathrm{abs}}}{\Delta t} =\frac{84.4024}{10.0} =\boxed{8.44\ \mathrm{W}}. \]

Each incident laser photon has energy \[ \varepsilon_\gamma=\frac{hc}{\lambda} =\frac{(6.6260755\times10^{-34})(299792458)} {514\times10^{-9}} =3.86468\times10^{-19}\ \mathrm{J}. \] All incident laser photons entering the gas are absorbed, so their absorption rate is \[ \dot N=\frac{P_{\mathrm{abs}}}{\varepsilon_\gamma} =\frac{8.44024}{3.86468\times10^{-19}} =\boxed{2.18\times10^{19}\ \mathrm{s^{-1}}}. \] This is also the rate of elementary absorption events for the laser photons.

5. Efficiency of conversion into the plate’s potential energy

The plate’s gravitational potential energy increases by \[ \Delta E_{\mathrm{pot}}=mg\Delta s =(0.800)(9.80665)(0.0300) =0.235360\ \mathrm{J}. \] Therefore the requested efficiency is \[ \eta=\frac{\Delta E_{\mathrm{pot}}}{E_{\mathrm{abs}}} =\frac{0.235360}{84.4024} =2.78854\times10^{-3}. \] Thus \[ \boxed{\eta=2.79\times10^{-3}=0.279\%.} \] Its small value reflects the energy used to heat the gas and to displace the surrounding atmosphere.

6. Pressure and temperature after rotation

Let \(\theta\) be the angle between the cylinder axis and the upward vertical. During the slow rotation, axial mechanical equilibrium gives \[ p(\theta)A=p_0A+mg\cos\theta, \] so \[ p(\theta)=p_0+\frac{mg}{A}\cos\theta. \] At \(\theta=90^\circ\), the plate’s weight has no axial component. Hence \[ \boxed{p_2=p_0=101.3\ \mathrm{kPa}.} \]

The rotation is slow, and heat exchange and frictional dissipation are negligible. The gas therefore undergoes a quasistatic adiabatic expansion. The first law gives \[ nc_V\,dT=-p\,dV. \] Using \(p=nRT/V\), \[ c_V\frac{dT}{T}=-R\frac{dV}{V}. \] Integration yields \[ TV^{R/c_V}=\text{constant}, \] or, defining \[ \gamma=\frac{c_P}{c_V} =\frac{c_V+R}{c_V} =1.399736, \] \[ TV^{\gamma-1}=\text{constant}. \]

Eliminating volume with the ideal-gas law gives \[ \frac{T_2}{T_1} =\left(\frac{p_2}{p_1}\right)^{(\gamma-1)/\gamma} =\left(\frac{p_0}{p_1}\right)^{R/(c_V+R)}. \] Therefore \[ T_2 =322.1398 \left(\frac{101300}{102298.9}\right)^{8.314510/29.114510} =321.2384\ \mathrm{K}. \] Thus \[ \boxed{T_2=321.2\ \mathrm{K}=48.1^\circ\mathrm{C}.} \]

Both pressure and temperature decrease; the temperature falls by approximately \(0.90\ \mathrm{K}\) because the expanding gas supplies work from its internal energy.

The solution was checked against the official solution.