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14.5 Distribution of molecular speeds. For a continuum of energy levels, where and. Combining the above equations, one has ε is a kinetics energy calculated through (1/2)mv 2, thus d ε = mv d v The above equation can be transformed into (in class demonstration).
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14.5 Distribution of molecular speeds • For a continuum of energy levels, where and
Combining the above equations, one has • ε is a kinetics energy calculated through (1/2)mv2, thus dε = mvdv • The above equation can be transformed into (in class demonstration)
14.6 Equipartition of energy • From Kinetics theory of gases showing that the average energy of a molecule is the number of degrees of freedom (f) of its motion. • For a monatomic gas, there are three degrees of freedom, one for each direction of the molecule’s translational motion. • The average energy for a single monatomic gas molecule is (3/2)kT (in class derivation). • The principle of the equipartition of energy states that for every degree of freedom for which the energy is a quadratic function, the mean energy per particle of a system in equilibrium at temperature T is (1/2)kT.
14.7 Entropy change of mixing revisited • From classical thermodynamics Δs = - nR (x1lnx1 + x2lnx2) where x1 = N1/N and x2 = N2/N • Now consider mixing two different gases with the same T and P, the increase in the total number of configurations available to the system can be calculated with
From Boltzmann relationship • Using Stirling’s approximation (see white board for details) we get Δs = - nR (x1lnx1 + x2lnx2) • From statistical point of view, when mixing two of the same type of gases under the same T and V (i.e. non-distinguishable particles with the same Ej), there is no change in the total number of available microstates, thus Δs equals 0
Demon (II) Figure 14.4 Maxwell’s demon in action. In this version the demon operates a valve, allowing one species of a two-component gas (hot or cold) through a partition separating the gas from an initially evacuated chamber. Only fast molecules are allowed through, resulting in a cold gas in one chamber and a hot gas in the other.
Problem 14.2: Show that for an assembly of N particles that obeys Maxwell-Boltzmann statistics, the occupation numbers for the most probable distribution are given by: • Solution
14.3a) Show that for an ideal gas of N molecules, • Solution:
14-3(b) For calculate • Solution:
