For a non-rigid diatomic molecule, the degrees of freedom \( f \) are given by:
\[f = 5 + 2(3N - 5)\]
Since \( N = 2 \) (for diatomic molecules):
\[f = 5 + 2(3 \times 2 - 5) = 7\]
The energy of one molecule is:
\[\text{Energy} = \frac{f}{2} k_B T = \frac{7}{2} k_B T\]
For 10 molecules, the total energy is:
\[10 \times \frac{7}{2} k_B T = 35 k_B T\]
To determine the energy of 10 non-rigid diatomic molecules at temperature \( T \), we need to consider the degrees of freedom available to a non-rigid diatomic molecule and apply the principles of statistical mechanics.
Diatomic molecules, unlike monatomic molecules, have additional degrees of freedom due to their rotational and vibrational motions. In the case of non-rigid diatomic molecules:
Total degrees of freedom for a non-rigid diatomic molecule = 3 (translational) + 2 (rotational) + 2 (vibrational) = 7.
According to the equipartition theorem, each degree of freedom contributes \(\frac{1}{2} k_B T\) to the energy at thermal equilibrium, where \( k_B \) is the Boltzmann constant.
Hence, the energy per molecule is:
\(E = \frac{7}{2} k_B T\)
For 10 molecules, the total energy will be:
\(E_{\text{total}} = 10 \times \frac{7}{2} k_B T = 35 k_B T\)
Thus, the total energy for 10 non-rigid diatomic molecules at temperature \( T \) is:
\(35 \, K_B T\)
Therefore, the correct answer is:
$35 \, K_B T$

Which one of the following graphs accurately represents the plot of partial pressure of CS₂ vs its mole fraction in a mixture of acetone and CS₂ at constant temperature?
