Simple Nature - Light and Matter

If the gas is monoatomic, then we know what to expect for thisrelationship between energy and temperature: ∆E = (3/2)nk∆T ,where the factor of 3 came ultimately from the fact that the gaswas in a three-dimensional space, k/1. Moving in this space, eachmolecule can have momentum in the x, y, and z directions. It hasthree degrees of freedom. What if the gas is not monoatomic?Air, for example, is made of diatomic molecules, k/2. There is asubtle difference between the two cases. An individual atom of amonoatomic gas is a perfect sphere, so it is exactly the same nomatter how it is oriented. Because of this perfect symmetry, thereis thus no way to tell whether it is spinning or not, and in fact wefind that it can’t rotate. The diatomic gas, on the other hand, canrotate end over end about the x or y axis, but cannot rotate aboutthe z axis, which is its axis of symmetry. It has a total of five degreesof freedom. A polyatomic molecule with a more complicated,asymmetric shape, k/3, can rotate about all three axis, so it has atotal of six degrees of freedom.Because a polyatomic molecule has more degrees of freedom thana monoatomic one, it has more possible states for a given amount ofenergy. That is, its entropy is higher for the same energy. From thedefinition of temperature, 1/T = dS/ dE, we conclude that it has alower temperature for the same energy. In other words, it is moredifficult to heat n molecules of difluoroethane than it is to heat natoms of helium. When the Clément-Desormes experiment is carriedout, the result b therefore depends on the shape of the molecule!Who would have dreamed that such simple observations, correctlyinterpreted, could give us this kind of glimpse of the microcosm?Lets go ahead and calculate how this works. Suppose a gas isallowed to expand without being able to exchange heat with therest of the universe. The loss of thermal energy from the gas equalsthe work it does as it expands, and using the result of homeworkproblem 2 on page 335, the work done in an infinitesimal expansionequals P dV , sodE + P dV = 0 .(If the gas had not been insulated, then there would have been athird term for the heat gained or lost by heat conduction.)From section 5.2 we have E = (3/2)P V for a monoatomic idealgas. More generally, the equipartition theorem tells us that the 3simply needs to be replaced with the number of degrees of freedom α,so dE = (α/2)P dV + (α/2)V dP , and the equation above becomesRearranging, we have0 = α + 2 P dV + α 2 2 V dP .(α + 2) dV V = −αdP P.324 Chapter 5 Thermodynamics