EXPERIMENT 3. HEAT-CAPACITY RATIOS FOR GASES

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1 EXERIMENT 3. HEAT-CAACITY RATIOS FOR GASES The ratio Cp/Cv of the heat capacity of a gas at constant pressure to that at constant volume will be determined by either the method of adiabatic expansion. Several gases will be studied, and the results interpreted in terms of the contribution made to the specific heat by various molecular degrees of freedom. THEORY In considering the theoretical calculation of the heat capacities of gases, we shall only be concerned with perfect gases. Since Ĉp = Ĉv + R for an ideal gas (where Ĉp and Ĉv are the molar quantities Cp/N and Cv/N), our discussion can be restricted to Cv. The number of "degrees of freedom" for a molecule is the number of independent coordinates needed to specify its position and configuration. Hence a molecule of n atoms has 3n degrees of freedom. These could be taken as the three cartesian coordinates of the n individual atoms, but it is more convenient to classify them as follows: Translational degrees of freedom: Three independent coordinates are needed to specify the position of the center of mass of the molecule. Rotational degrees of freedom: All molecules containing more than one atom require a specification of their orientation in space. As an example, consider a rigid diatomic molecule; such a model consists of two point masses (the atoms) connected by a rigid massless bar (the chemical bond). Through the center of mass, which lies on the rigid bar, independent rotation can take place about two axes mutually perpendicular to each other and to the rigid bar. (The rigid bar itself does not constitute a third axis of rotation under ordinary circumstances for reasons based on quantum theory, there being no appreciable moment of inertia about this axis.) Rotation of a diatomic molecule or any linear molecule can thus be described in terms of two rotational degrees of freedom. Nonlinear molecules for which the third axis has a moment of inertia of appreciable magnitude and constitutes another axis of rotation require three rotational degrees of freedom. 3 ibrational degrees of freedom: One must also specify the displacements of the atoms from their equilibrium positions (vibrations). The number of vibrational degrees of freedom is 3n - 5 for linear molecules and 3n - 6 for non- linear molecules. These values are determined by the fact that the total number of degrees of freedom must be 3n. For each vibrational degree of freedom there is a "normal mode" of vibration of the molecule, with characteristic symmetry properties and a characteristic harmonic frequency. The vibrational normal modes for CO and H O are illustrated schematically in Fig.. Using classical statistical mechanics one can derive the theorem of the equipartition of energy. According to this theorem kt / of energy is associated with each quadratic term in the expression for the energy. Thus there is associated with each translational or rotational degree of freedom for a molecule a contribution to the energy of kt/ of kinetic energy and for each

2 vibrational degree of freedom a contribution of kt/ of kinetic energy and kt/ of potential energy. (The corresponding contributions to the energy per mole of gas are RT/.) Clearly a monatomic gas has no rotational or vibrational energy but does have a translational energy of 3/RT per mole. The constant-volume heat capacity of a monatomic perfect gas is thus E T For diatomic or polyatomic molecules, we can write: Ĉv = = 3/ R () Ē = Ē(trans) + Ē(rot) + Ē(vib) () In Eq. (), contributions from electronic states have been neglected, since they are not significant at room temperature for most molecules. Also any small intermolecular energies which occur for imperfect gases are not considered. The equipartition theorem is based on classical mechanics. Its application to translational motion is in accord with quantum mechanics as well. At ordinary temperatures the rotational results are also in accord with quantum mechanics. (The greatest deviation from the classical result is in the case of hydrogen, H ; at temperatures below 00 K the rotational energy of H is significantly below the equipartition value, as predicted by quantum mechanics.)

3 The vibrational energy is, however, highly quantized and depends strongly on temperature; the various vibrational modes are at ordinary temperatures only partially "active," and the degree of activity depends strongly on the temperature. As a general rule, the heavier the atoms or the smaller the force constant of the bond (i.e., the lower the vibrational frequency), the more "active" is a given degree of freedom at a given temperature and the greater is the contribution to the heat capacity. Moreover, the frequencies of modes that are predominantly bending of bonds tend to be much lower than those that are predominantly stretching of bonds. In the case of most gaseous diatomic molecules (where the one vibrational mode is a pure stretch) the vibrational contribution to Ĉv is very small; for example, N would have its classical equipartition value for Ĉv only above about 4000 K. Many polyatomic molecules, especially those containing heavy atoms, will at room temperature have significant partial vibrational contributions to Ĉv. We are now in a position to calculate for polyatomic molecules approximate or at least limiting values for Ĉv and for the ratio Ĉp/Ĉv. For monatomic gases and all ordinary diatomic molecules (where vibration is not important at room temperature) definite values can be calculated. For a brief discussion of the calculation of Ĉv (vib) for polyatomic molecules, see Exp. 4. A. ADIABATIC EXANSION METHOD For the reversible adiabatic expansion of a perfect gas, the change in energy content is related to the change in volume by de = pd = NRT d = NRTd(ln( )) (3) Moreover, since E for a perfect gas is a function of temperature only, we can also write de = C v dt. Substituting this expression into Eq. (3) and integrating, we find that ˆ C v T = ln T where Ĉ and are molar quantities (that is, C v /N, /N). It has been assumed that Ĉ is constant over the temperature range involved. This equation predicts the decrease in temperature resulting from a reversible adiabatic expansion of a perfect gas. Consider the following two-step process involving a perfect gas denoted by A: Step I: Allow the gas to expand adiabatically and reversibly until the pressure has dropped from to. A(,,T ) A(,,T ) (5) Step II: At constant volume, restore the temperature of the gas to T. A(,,T ) A( 3,,T ) (6) For step I, we can use the perfect-gas law to obtain T = T = Rln (4) (7)

4 Substituting Eq. (7) into Eq. (4) and combining terms in /, we write since for a perfect gas Ĉ p = Ĉ + R (9) For step II, which restores the temperature to T l, Thus ln ( Cˆ + R) Cˆ = ln ˆ = C Cˆ = Cˆ ln = ln Cˆ 3 3 ln (8) (0) () This can be rewritten in the form Cˆ Cˆ log log = log log 3 () This method, due to Clement and Desormes, uses the very simple apparatus shown in Fig.. The change in state (5) is carried out by quickly removing and replacing the stopper of a large carboy containing the desired gas at a pressure initially somewhat higher than atm pressure, so that the pressure of gas in the carboy momentarily drops to atmospheric pressure. The change in state (6) consists of allowing the gas remaining in the carboy to return to its initial temperature. The initial pressure and the final pressure 3 are read from an open-tube manometer. The thermodynamic equations (3) to () apply only to that part of the gas that remains in the carboy after the stopper is replaced. We may imagine the gas initially in the carboy to be divided into two parts by an imaginary surface; the part above the surface leaves the carboy when the stopper is removed and presumably interacts irreversibly with the surroundings, but the part below the surface expands reversibly against this imaginary surface, doing work in pushing the upper gas out. The process is approximately adiabatic only because it is rapid; within a few seconds the gas near the walls will have received an appreciable quantity of heat by direct conduction from the walls, and the pressure can be seen to rise almost as soon as the stopper is replaced.

5 The experiment will presumably give a somewhat low result (low 3 and therefore low Ĉ ratio) if the expansion is to an appreciable degree irreversible, a somewhat low result if the stopper is left open so long that the conditions are not sufficiently adiabatic, and a somewhat high result if the stopper has not been removed for a long enough time to permit the pressure to drop momentarily to atmospheric. There should be no significant irreversibility if during the expansion there are no significant pressure gradients in the gas below the imaginary surface mentioned above, and such pressure gradients should not be expected if the throat area is small in comparison with the effective area of this imaginary surface. With an 8-liter carboy and a about 50 Torr above atm, the volume of air forced out should be about liter, which should provide a large enough surface area to fulfill this condition approximately if the throat diameter is not more than about or 3 cm. The effect of the length of time the stopper is removed is much more difficult to estimate by calculation; some idea can be obtained from the duration of the sound produced when the stopper is removed and from the rate of rise of the manometer reading immediately after the stopper is replaced. For the purpose of this experiment the student may assume that, if the stopper is removed completely from the carboy to a distance of or 3 in. away and replaced tightly as soon as physically possible, the desired experimental conditions are approximately fulfilled. Some additional assurance may be gained from the reproducibility obtained in duplicate runs. The adiabatic expansion method is not the best method of determining the heat- capacity ratio. Much better methods are based on measurements of the velocity of sound in gases. One such

6 method, described in art B of this experiment, consists of measuring the wavelength of sound of an accurately known frequency by measuring the distance between nodes in a sonic resonance set up in a Kundt's tube. Methods also exist for determining the heat capacities directly, although the measurements are not easy. Experimental The apparatus should be assembled as shown in Fig.. If desired, the carboy may be mounted in a thermostat bath; if so, it must be clamped securely to overcome buoyancy. The manometer is an open-tube manometer, one side of which is open to the atmosphere; the pressure that it measures, therefore, is the difference of pressure from atmospheric pressure. A suitable liquid for the manometer is dibutyl phthalate, which has a density of.046 g/cm 3 at room temperature (0 C), To convert manometer readings (millimeters of dibutyl phthalate) to equivalent readings in millimeters of mercury, multiply by the ratio of this density to the density of mercury, which is 3.55 g/cm 3 at 0 C. To find the total pressure in the carboy, the converted manometer readings should be added to atmospheric pressure as given by a barometer. It is unnecessary to correct all readings to 0 C, as all pressures enter the calculations as ratios. Seat the rubber stopper firmly in the carboy and open the clamps on tubes a and b. Clamp off the tube c. The connections shown in Fig. are based on the assumption that the gas to be studied is heavier than air (or the previous gas in the carboy) and therefore should be introduced at the bottom in order to force the lighter gas out at the top; in the event that the gas to be studied is lighter, the connections a and b should be reversed. Allow the gas to be studied to sweep through the carboy for 5 min. The rate of gas flow should be about 6 liters min -, or 00 ml sec - (measure roughly in an inverted beaker held under water in the thermostat bath). Thus, 5 volumes of gas (90 liters) will pass through the carboy. Retard the gas flow to a fraction of the flushing rate by partly closing the clamp on tube a. Carefully open the clamp on tube c, and then cautiously (to avoid blowing liquid out of the manometer) clamp off the exit tube b, keeping a close watch on the manometer. When the manometer has attained a reading of about 600 mm of oil, clamp off tube a. Allow the gas to come to the temperature of the thermostat bath (about 5 min), as shown by a constant manometer reading. Record this reading; when it is converted to an equivalent mercury reading and added to the barometer reading, is obtained. Remove the stopper entirely (a distance of or 3 in.) from the carboy, and replace it in the shortest possible time, making sure that it is tight. As the gas warms back up to the bath temperature, the pressure will increase and finally (in about 5 min) reach a new constant value 3, which can be determined from the manometer reading and the barometer reading. At some point in the procedure, a barometer reading ( ) should be taken. Repeat the steps above to obtain two more determinations with the same gas. For these repeat runs, long flushing is not necessary; one additional volume of gas should suffice to check the effectiveness of the original flushing. Measurements are to be made on both helium and nitrogen. If there is sufficient time, study carbon dioxide also.

7 CALCULATIONS For each of the three runs on He, N and CO, calculate Ĉ /Ĉ using Eq. (). Also calculate the theoretical value of Ĉ /Ĉ predicted by the equipartition theorem. In the case of N and CO, calculate the ratio both with and without vibrational contribution to Ĉ. DISCUSSION Compare your experimental ratios with those calculated theoretically, and make any deductions you can about the presence or absence of rotational and vibrational contributions, taking due account of the uncertainties in the experimental values. For CO, how would the theoretical ratio be affected if the molecule were nonlinear (like SO ) instead of linear? Could you decide between these two structures from the Ĉ /Ĉ ratio alone? References ) Lord Rayleigh, The Theory of Sound, nd Ed., vol. II, pp. 5-3, Dover, NY, 945. ) K. Schell and W. Heuse, Ann. hysik, 37, 79, 9; 40, 473, 93; 59, 86, 99.