THE SCIENCE AND APPLICATIONS OF ACOUSTICS - H. H. Arnold ...

16.2 Relaxation Processes 447expression for relaxation:dE v= 1 dt τ [E v − E v (T tr )] (16.1)This reversion to equilibrium occurs through individual molecular collisions inwhich energy transfers result.Let k 10 be defined as the rate at which the molecules descend from the firstexcited state to the ground state, owing to collisions at 1 atm pressure. This issimply the collision frequency M multiplied by the probability of energy transferP 1→0 multiplied by the mole fraction x 1 of the molecules in the first excited state.The reverse process will also occur, i.e., some molecules in the ground state willbecome excited at a rate k 01 . In equilibrium both rates are equal, and we can writek 10 x 1 = k 01 x 0The energy is quickly shared from the first excited level of the vibrational mode tohigh-level modes through vibrational exchanges. On the basis of quantum mechanicsgoverning the probabilities of energy exchanges between vibrational levels ofa harmonic oscillator, Landau and Teller (1936)derived the following expression:− dE v (= k 10 1 − e−hυ/kT ) [E v − E v (T tr )] (16.2)dtwhere h is the Planck constant (6.626 × 10 –34 J s), v is the vibrational frequencyof the relaxing mode, k is the Boltzmann constant (1.38 × 10 16 ergs K −1 ), and Tis the absolute temperature in K. Through comparison with Equation (16.1), therelaxation time τ for Equation (16.2) is found to be1τ =(16.3)k 10 (1 − e −hν/kT )Relaxation time and ultrasonic absorption and dispersion are interlinked. Therelaxation process causes the specific heat of the gas to be frequency-dependent.The specific heat of a simple gas can be traced to translational, vibratory, androtational contributions. Let us now consider the situation where any acousticallyimposed temperature variation is followed by both translational and rotationalenergy equilibrating extremely quickly. The specific heat C ′ is deemed independentof temperature or the temperature never deviates appreciably from the equilibriumvalue T 0 ; so we can write[E v − E v (T tr )] = C ′ (T ′ − T 0 ) (16.4)and insert this last expression into Equation (16.1) to yielddT ′= 1 (T ′ )− T tr (16.5)dt τNow consider the case when the translational temperature is suddenly raisedat time t = 0 from T 0 to a value T 1 . Two cases can be ascertained: The externaltemperature is kept constant at T 1 after t = 0, the energy seeping from the outsideinto the internal (rotational and vibrational) degrees of freedom. The solution of