Saved Bookmarks
| 1. |
State and prove law of equipartition of energy. |
|
Answer» Law of equipartition of energy : The energy of the molecules of a gas, in thermal equilibrium at a temperature T and containing large number of molecules, is equally divided among their available degrees of freedom, with the energy per molecule for each degree of freedom equal to `(1)/(2)k_(B)`T, where `k_(B)` is the Boltzmann constant. OR The average energy associated with each quadratic term in the energy of a molecule in a large sample of gas in thermal equilibrium at a temperature T is `(1)/(2)k_(B)`T, where `k_(B)` is the Boltzmann constant. (i) Monatomic gas : Sincce an atom or a monatomic molecule can have only translational motion, it has only three degrees of freedom. Hence, the energy per molecule is `3((1)/(2)k_(B)T)=(3)/(2)k_(B)T`. Therefore, the energy per mole is `E=(3)/(2)k_(B)TxxN_(A)=(3)/(2)RT" "`......(1) where `N_(A)` is the Avogadro constant. (ii) Diatomic gas : A rigid diatomic molecule has three translational degress of freedom and two rotational degress of freedom, each contributing quadratic term to the energy. Hence, the energy per molecule of a gas of rigid diatomic molecules is `3((1)/(2)k_(B)T)+2((1)/(2)k_(B)T)=5(5)/(2)k_(B)T`. Therefore, the energy per mole is `E=(5)/(2)k_(B)TxxN_(A)=(5)/(2)RT" "`...(2) A soft (or non-rigid) diatomic molecules has, in addition, one frequency of vibration which contributes two quadratic terms to the energy. Hence, the energy per molecule of a soft diatomic molecule is `E=3((1)/(2)k_(B)T)+2((1)/(2)k_(B)T)+2((1)/(2)k_(B)T)=(7)/(2)k_(B)T` Therefore, the energy per mole of a gas soft diatomic molecules is `E=(7)/(2)k_(B)TxxN_(A)=(7)/(2)RT" "`......(3) |
|