Question

Consider a classical ideal gas of N molecules in equilibrium at temperature T. Each molecule has two energy levels, -∈ and ∈. The mean energy of the gas is

• A. 0
• B. \(N\isin\tan h(\frac{\isin}{k_BT})\)
• C. \(-N\isin\tan h(\frac{\isin}{k_BT})\)
• D. \(\frac{\isin}{2}\)

Answer 1

The Correct Option is C

To find the mean energy of a classical ideal gas with N molecules, each having two energy levels, we use the concept of statistical mechanics. The possible energy levels for each molecule are given as \(-\varepsilon\) and \(\varepsilon\).

In statistical mechanics, the probability of a system being in a particular energy state is given by the Boltzmann factor. This factor is given by:

\(P(E) = \frac{e^{-E/k_BT}}{Z}\)

where \(Z\) is the partition function, \(k_B\) is the Boltzmann constant, and \(T\) is the temperature. The partition function, \(Z\), for a single molecule is calculated as follows:

\(Z = e^{-\varepsilon/k_BT} + e^{\varepsilon/k_BT}\)

The probability of a molecule being in the state \(-\varepsilon\) is:

\(P(-\varepsilon) = \frac{e^{-\varepsilon/k_BT}}{Z}\)

The probability of a molecule being in the state \(\varepsilon\) is:

\(P(\varepsilon) = \frac{e^{\varepsilon/k_BT}}{Z}\)

The mean energy of one molecule, \(\bar{E}\), is calculated by averaging over these two states:

\(\bar{E} = -\varepsilon \cdot P(-\varepsilon) + \varepsilon \cdot P(\varepsilon)\)

Substituting in the expressions for \(P(-\varepsilon)\) and \(P(\varepsilon)\):

• \(\bar{E} = -\varepsilon \frac{e^{-\varepsilon/k_BT}}{e^{-\varepsilon/k_BT} + e^{\varepsilon/k_BT}} + \varepsilon \frac{e^{\varepsilon/k_BT}}{e^{-\varepsilon/k_BT} + e^{\varepsilon/k_BT}}\)
• \(\bar{E} = \frac{\varepsilon (e^{\varepsilon/k_BT} - e^{-\varepsilon/k_BT})}{e^{\varepsilon/k_BT} + e^{-\varepsilon/k_BT}}\)

This simplifies to:

\(\bar{E} = \varepsilon \tanh(\frac{\varepsilon}{k_BT})\)

Thus, for N molecules, the total mean energy of the gas is:

\(E = N \varepsilon \tanh(\frac{\varepsilon}{k_BT})\)

The correct answer is selected based on the fact that the mean energy of the gas would result in net energy direction considering the dual levels, leading to:

\(E = -N \varepsilon \tanh(\frac{\varepsilon}{k_BT})\)

Therefore, the answer is \(-N\varepsilon \tanh(\frac{\varepsilon}{k_BT})\).