Ideal gas results 331 Internal energy of an ideal gas

In addition to the ideal gas equation of state another property is necessary to define an ideal gas. We must specify its internal energy as a function of the thermodynamic coordinates. This can be accomplished by laboratory experiments to yield:

[internal energy of ideal gas] (3.7)

where N is the number of molecules in the system. There are many alternative forms of this relation because of the different ways we can describe an ideal gas:

where kB is Boltzmann's constant (Table 1.1), M is the mass of gas in the system, and f is a constant equal to 3 for an ideal monatomic gas (e.g., Ar), and f = 5 for most diatomic gases at room temperature (e.g., air). The internal energy can be determined by a series of experiments involving adiabatic processes in which AU = -Wa^b is easily measured. The constant f depends upon the internal structure of the molecules; it is known as the number of degrees of freedom in the molecule. For a diatomic molecule that is very stiff (does not stretch and contract under the temperature conditions being considered) such as O2 and N2 near room temperature, the value off is 5 because there are two more degrees of freedom due to the ability of the molecule to rotate in a two-dimensional plane, but not about the axis joining the two atomic constituents (its moment of inertia is too small about the axis joining the two atoms). At high temperatures (not naturally occurring in the lower atmosphere) the number of degrees of freedom goes up by two because of molecular vibration due to the spring-like binding. Hence, for normal atmospheric conditions such as encountered in the troposphere and stratosphere:

It is not difficult to see why the specific heat should be larger for molecules with larger f. Consider the constant volume heating case. If heating occurs in a box of monotonic gas, all the energy must go into increasing the linear (translational) kinetic energy of the molecules. If the molecule is spatially extended such as a diatomic molecule, it can rotate as well as translate. The added energy can go into rotational energy as well as translational energy. Hence, the heat capacity (amount of heat necessary to raise the system's temperature by 1K) will be larger. Basically the heat energy (that at the molecular level) must be shared among all the degrees of freedom, but only the linear kinetic energy goes into causing pressure since it carries momentum to the walls (or across boundaries).

Aremarkable theorem proved in the classical study of statistical mechanics shows that in equilibrium the energy will be shared equally between each of the rotational and translational modes (and vibrational modes when applicable): the principle of equipartition of energy. In the case of a diatomic molecule only two rotational modes are available, the rotation about the axis joining the two atoms does not count. In the case of a triatomic molecule in which the atoms are not in a straight line (e.g., CO2), all three rotational modes are involved: f = 6. Molecules actually can vibrate (stretching and contracting like masses joined by a spring) as well, and at sufficiently high temperatures these modes can enter and raise f even more, but as remarked above, this vibrational degree of freedom is not important in most f = 3 ideal monatomic gas (e.g., argon) f = 5 N2, O2 near STP f = 6 CO2, H2O and other nonlinear molecules.

applications of thermodynamics to the atmosphere. On the other hand, such modes of vibration and rotation play an important role in the absorption and emission of infrared radiation as it passes through air. In summary, each molecule on the average possesses 2kBT for each of its mechanical degrees of freedom. For 1 mol of such molecules at 300 K this is 1 R*T = 2.5 kJ for each degree of freedom. Hence for argon it is 7.5 kJ mol-1 and for O2 and N2 it is 12.5 kJ mol-1.

Example 3.6 Find the internal energy of a 1 kg mass of dry air at STP. Answer. We can use U = (f /2)pRdTV = (f /2)MRdT0, where M is the mass of the gas in the system (here 1 kg) and f is 5. Then U = 1.96 x 105 J = 196 kJ. □

Example 3.7 Compare the rise in potential energy due to lifting the 1 kg parcel to 9 km, approximately one scale height.

Answer. The change in potential energy is Mgh = 1.08 x 105 J = 108 kJ. Hence the gravitational potential change is comparable to the internal energy for a lift of one scale height. □