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CHEM 477: Physical Chemistry Lab II Heat Capacity Ratio by Kundtβs Method Used and adapted with permission from Paul Ayers (McMaster University) Background By definition, πΆπΆπ£π£Μ (or πΆπΆππΜ ) denotes the amount of heat energy that must be absorbed by one mole of a gas at constant volume (or pressure) to raise the temperature of the gas by one degree. The absorbed heat energy causes the molecules to move faster (increase in translational energy), to rotate faster (increase in rotational energy) and to vibrate faster (increase in vibrational energy). Thus, the knowledge of heat capacities plays a role in understanding the complexity of gaseous molecules. Unfortunately, the easiest method for determining the individual heat capacities of gases is beyond the scope of an undergraduate laboratory class. The heat capacity ratio, πΎπΎ = πΆπΆππΜ βπΆπΆπ£π£Μ (1), is just as useful in understanding the structure of gaseous molecules and is more accessible experimentally. The equipartition theorem states that πΆπΆπ£π£Μ of an ideal gas depends on the types of motion the molecule is able to perform. The different types of motion are called degrees of freedom. The more degrees of freedom a molecule possesses, the more energy a molecule can absorb before its temperature increases (πΆπΆπ£π£Μ is larger). Theoretical calculations, experimental measurements, and common sense suggest that monatomic gases have the smallest heat capacities, while complex molecules have larger heat capacities. Each type of molecular motion β translations, rotations, and vibrations β contribute to the heat capacity. As per classical statistical mechanics, translational and rotational degrees of freedom contribute ½RT per mole, while vibrations contribute RT. The number of translational and rotational degrees of freedom available to a molecule depends on the moleculeβs shape (linear or non-linear) and the number of atoms in the molecule. Diatomic molecules have two rotational degrees of freedom and one vibrational degree of freedom. Linear polyatomic molecules have 3N β 5 vibrational degrees of freedom, where N is the number of atoms that make up the molecule. Non-linear polyatomic molecules have 3N β 6 vibrational degrees of freedom. For example, an ideal monatomic gas can move along the x-, y-, and z-axes, and therefore has three translational degrees of freedom. Because it has no rotational or vibrational degrees of freedom, its πΆπΆπ£π£Μ is 3/2R. A diatomic gas has three translational degrees of freedom, two rotational degrees of freedom, and one vibration. It has a πΆπΆπ£π£Μ of 7/2R. The energy level spacing for translations and rotations are very close together, and at room temperatures, many translational and rotational energy levels are available to gas molecules. Vibrational energy levels are spaced further apart, and these levels are not as likely to be populated at room temperature. While classical theory predicts a 1R contribution to πΆπΆπ£π£Μ for vibrations, quantum calculations and experimental measurements show much smaller contributions. Spring 2009 Revised 01/27/2009 Page 1 of 4 Heat Capacity Ratio by Kundtβs Method CHEM 477: Physical Chemistry Lab II Using Speed of Sound to Measure Heat Capacity Ratio Sound waves require elastic media for their propagation, as a train of sound waves consists essentially of a sequence of elastic displacements of elements of the medium. These displacements take place in the direction of the propagation of the wave: sound waves are longitudinal in nature. A stimulus, when applied to the medium at rest, causes the elements to vibrate about their mean positions and, because they do not vibrate in phase, this causes pressure (and hence density) differences in the medium, resulting in compressions and rarefactions along the wavefront. These compressions and rarefactions take place with such rapidity that the process is β for all intents and purposes β adiabatic. That is, there is no time for heat exchange between the vibrating elements and their surroundings, and ππ = 0. The adiabatic compressions and expansions of ideal gases are governed by ππππ πΎπΎ = constant (2). The speed of sound in any medium depends on the elasticity of the medium and its density. It can be shown that the speed of sound ππ in any medium is given by: πΎπΎ ππ = οΏ½ (3) ππ where πΎπΎ is the bulk modulus of elasticity of the medium and ππ its density. For a gas undergoing abiabatic pressure changes, the bulk modulus of elasticity is simply πΎπΎπΎπΎ, where ππ is the absolute pressure of the gas. Thus: πΎπΎπΎπΎ ππ = οΏ½ (4) ππ If an assumption is made that the gas behaves ideally, then ππππ = ππππππ (5) also applies, which upon rearrangement gives: ππ ππ = π π π π (6) ππ where ππ is the molar mass of the gas. Substitution of Equation 6 into Equation 4 and rearrangement gives πΎπΎ = ππππ 2 (7) π π π π If the distance between corresponding points on the wave front (e.g. maximum compressions) is called the wavelength ππ, and the rate at which such compressions pass a given section is called the frequency ππ, then the speed of sound is also given by and (8) ππ = ππππ πΎπΎ = Spring 2009 ππππ 2 ππ 2 (9) π π π π Revised 01/27/2009 Page 2 of 4 Heat Capacity Ratio by Kundtβs Method CHEM 477: Physical Chemistry Lab II In this experiment, which is based on a modified Kundt's tube (Figure 1), a sound-wave of known frequency is sent down a long plastic tube and is made to reflect from a movable piston, creating a standing wave with a series of maxima, called antinodes, and minima in between, called nodes. When the standing waveform is plotted on the oscilloscope, it appears as a diagonal line, which is a result of interference between two waves of the same frequency travelling with the same speed in opposite directions. The distance between nearest nodes (or anti-nodes) is equal to ππ/2. The successive nodes (or anti-nodes) are in opposite phase, i.e., they differ in phase by 180° (a phenomenon of which we will take advantage in this experiment), with maximum sound intensity occurring at the nodes and minimum at the anti-nodes. Figure 1. Diagram of a Kundtβs tube similar to that used in this experiment. Experimental Fill the heat exchanger with water, turn it on, and set the temperature to 25° C. Allow the temperature to equilibrate before beginning the experiment. Move the piston as far from the speaker as possible and purge the Kundtβs tube with one of the gases to be studied (N2, CO2, or Ar). Flow the gas for 15 β 30 sec to purge the tube completely. Reduce the flow of the gas to a slow stream that is still sufficient to prevent air from diffusing into the tube during the experiment. Note: be careful while purging with CO2, as it cools considerably upon expansion from the bulk cylinder. Connect the output of the signal generator to the speaker leads and channel 1 (X) of the oscilloscope. Connect the microphone leads to channel 2 (Y) of the oscilloscope. Connect the battery to the microphone power leads. Select an appropriate frequency on the signal generator (3000 Hz for CO2, air, and N2; 4000 Hz for Ar). Finally, switch the signal generator on. A high frequency noise should be heard from the speaker and a waveform pattern will appear on the oscilloscope. Details on adjusting the oscilloscope will be provided by the T.A. Ideally, the components should be adjusted so that the two signals to the oscilloscope are about equal in magnitude. Spring 2009 Revised 01/27/2009 Page 3 of 4 Heat Capacity Ratio by Kundtβs Method CHEM 477: Physical Chemistry Lab II Slowly move the piston toward the speaker until a diagonal line appears on the screen. Record the position of the piston. Continue to move the piston inward until you see another diagonal line that is perpendicular to the first diagonal line. Again, note the position of the piston. Continue taking as many readings as the geometry of the tube allows. If necessary, repeat the entire procedure until reproducible results are obtained. Repeat with the other gases provided (CO2, N2, air, and Ar). Repeat the measurements at a minimum of 3 different frequencies for each gas. Waste Disposal There are no waste concerns for this experiment. Calculations Given the values of ππ/2 and using the known frequency calculate πΎπΎ for each gas from Equation 9. Using πΆπΆππΜ β πΆπΆπ£π£Μ = π π and Equation 1 determine both πΆπΆπ£π£Μ and πΆπΆππΜ for each of the gases (you have two equations and two unknowns). Discussion Compare your experimental πΆπΆπ£π£Μ values, the literature values, and the theoretical values and discuss the role of vibrational energy levels in explaining any difference. Discuss the relationship of the value of πΎπΎ with respect to the structure of the gas molecules. Spring 2009 Revised 01/27/2009 Page 4 of 4