Fluorescent lamps provide around 80 % of the world’s needs in artificial lighting. They consume seve — Physical Chemistry — Thermodynamics Chemistry Question
Fluorescent Lamps: Heating Inert Gas Atoms by Electrons
Fluorescent lamps provide around 80 % of the world’s needs in artificial lighting. They consume several times less energy per light output than incandescent light bulbs, and hence are important in the fight to reduce world’s energy consumption. Fluorescent lamps are filled with low pressure noble gas, such as argon, and also mercury vapor at even lower pressure. Electrical discharge in fluorescent lamps causes partial ionization of Hg, resulting in emergence of electrons and equal number of ionized Hg atoms. Collisions of electrons with neutral Hg atoms lead to the electronic excitation of the latter atoms, which emit UV light when decaying back to the ground state. The UV light strikes the glass surface of the tube covered with a phosphor, which produces a glow of visible light that we can see.
The electric field between the tube cathode and the anode continuously transfers energy to the electrons. The electrons redistribute the energy among themselves, quickly reaching the temperature on the order of 11,000 K. Similarly, neutral Ar atoms also quickly equilibrate thermally among themselves. However, because of a very large mass mismatch between electrons and argon, collisions between electrons and Ar atoms are extremely inefficient in transferring the electrons’ energy to Ar atoms. Hence, the argon temperature in a tube is much lower than the electrons’ temperature.
Using the steady state approximation, find the steady state temperature of neutral Ar gas in middle column of the fluorescent lamp, given that electrons’ temperature, Te, is 11,000 K and the temperature of the outer tube wall, Twall, is 313 K.
In all calculations use the following specific parameters describing a typical fluorescent lamp having dimensions of 120 cm in length and 3.6 cm in diameter, and Ar pressure of 3 Torr (1 Torr = 1 mm Hg; 1/760th of 1 atm pressure).
What is the total frequency, ν, of electron-Ar collisions in the tube having volume of 4.9 · 10–3 m3 and concentration of free electrons ne = 5.0 · 1017 m–3, if the mean collision time of an electron with Ar atoms is τ = 7.9 · 10–10 s?
Model Answer
First, let’s compute first lamp’s volume and surface area: V = πr2h = 4.9 · 10–3 m3; S = 2πrh = 0.27 m2. Next, we are told that one electron collides with an Ar gas within τ seconds, at the concentration of nAr. In a tube with volume V there are a total of neV electrons. Hence, per given time duration, τ, we expect the number of electron-Ar collisions to be neV times larger, compared with the one electron case. In terms of the collision frequency, ν = ne V / τ = 7.9 · 1027 s–1.
What is total rate of energy transfer from electrons to Ar in the tube, Je→Ar, in J·s–1? Assume that only a small fraction of electron’s energy, fe→Ar = 2.5 · 10–5, is transferred to an Ar atom per single collision, and the average energy of electrons and Ar atoms is kBT, where kB is the Boltzmann constant and T is the corresponding temperature. Note that in a collision between an electron and Ar atom, the energy is transferred in both directions.
Model Answer
The average energy of an electron is kBTe and that of Ar is kBTAr. In each collision, the net energy transfer from e to Ar is f ( kB)(Te - TAr), because the transfer of energy occurs in both directions. Furthermore, the number of electron-Ar collisions per second is neV / τ , as worked out in 16.1. Hence, per unit time, the total energy transfer rate in the tube from e to Ar is Je→Ar = (ne*V/τ) * fe→Ar * 1.5 * kB * (Te - TAr) in units of [J s-1].
Assuming a linear drop of temperature from the tube center of the wall, the total thermal energy transfer rate from heated Ar gas in the middle to the tube wall is J_{Ar→Wall} = κ_Ar * ((T_Ar - T_{Wall}) / R_{tube}) * S_t, where κAr is the thermal conductivity of argon, κAr = 1.772 · 10–4 J s–1 m–1 K–1, Rtube is the tube radius, Rtube = 3.6 cm, and St indicates the total area of the tube whose length is 120 cm. At the steady state, derive an expression for the temperature of the neutral Ar gas in the fluorescent lamp tube, TAr.
Model Answer
At steady state, all energy which is transmitted from electrons to Ar should be further transmitted to the wall, so there is no energy accumulation or depletion in the Ar gas. Hence, we should request a balance of the in and out energy transfer rates, JAr→Wall = Je→Ar. After substituting the expressions for JAr→Wall and Je→Ar, we solve this equation for TAr, to obtain: TAr = (κAr * T_Wall * S_t + 1.5 * (ne * V / τ) * fe→Ar * kB * Te * R_tube) / (κAr * S_t + 1.5 * (ne * V / τ) * fe→Ar * kB * R_tube) in units of [K]. Plugging these and other given parameters into the expression derived, we obtain that TAr = 377 K in the center of the fluorescent lamp. Notice that this temperature is much lower than the temperature of electrons at 11,000 K. Electrons transfer heat to Ar, however, this process is inefficient, and this energy quickly flows out as heat to the outside, providing a stable steady state situation.
Compare the energy loss through the heat transfer by Ar atoms to the tube walls with the total energy input of a 40 W fluorescent lamp (1 W = J s–1 ).
Model Answer
The heat loss through argon can be calculated using JAr→Wall = κAr * ((TAr - TWall) / Rtube) * St, resulting in Je→Ar = 17.1 J s–1 = 17.1 W, which is approximately 50 % of the energy consumed by the lamp, 40 W. It turns out that the overall efficiency of electrical energy conversion to light is only 15 % (because of additional losses), which is still several fold higher than the 2.5 % efficiency of the regular light bulb.
Recalculate TAr for the Ar pressures of 1 and 10 atmospheres, respectively. The only change in the parameters above will be in τ , which is inversely proportional to the pressure, τ ~ p–1. The thermal conductivity of Ar, κAr, is independent of pressure in this regime of pressures.
Model Answer
Using the expression for TAr derived in 16.3, and proportionally decreasing τ because of higher pressures, we find TAr = 6785 K for pAr = 1 atm and TAr = 10346 K for pAr = 10 atm. Hence, at higher neutral gas pressures, neutral atoms are found much closer to thermal equilibrium with the hot electron gas at 11,000 K. The reason for this is more efficient energy transfer rate from electrons to atoms due to more frequent collisions (because of larger number of Ar atoms), while the energy dissipation due to heat transfer stays roughly the same. Note: This calculation does not take into account that the temperature of electrons will be around 5000 K to 6000 K in tubes at high Ar pressure, because of more efficient cooling since energy transfer to the Ar gas is enhanced. So to accurately solve the problem at high pressures, these new values for the electron gas temperature would need to be assumed and the calculation needs to be redone with this new Te.