Thermodynamics | Translation, Rotational, Vibrational Degree of Freedom | Class 11th | Jay Daiya Sir

Statement-1: A reas gas nearly behaves like an ideal gas at low pressure and high temperature. Statement-2: If the ratio of translational and rotational degree of freedom is 1.5 the gas must be diatomic Statement-3: Most probable speed of a gas is proportional to absolute temperature of the gas.

The number of translational degree of freedom for a diatomic gas is

A gas have 3 translation and 2 Rotational degree of freedom. Find (C_p)/(C_v) ?

For a diatomic gas having 3 translational and 2 rotational degree of freedom ,the energy is given by ?

Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom . The gas is maintained at a temperature of T. The total internal energy , U of a mole of this gas, and the value of gamma(= (C_P)/(C_(v))) are given, respectively , by :

According to the tranistion state theory, for the formation of on activation complex, one of the vibrational degree of freedom is converted into the tranistion degree of freedom. Reason (R ): The energy of the activated complex is higher than the energy of the reactant molecules.

Assertion Prssure of a gas is 2/3 times translational kinetic energy of gas molecules. ReasonTranslational degree of freedom of any type of gas is three, whether the gas is monoatomic, diatomic or polyatomic.

Calculate gamma (ratio of C_(p) and C_(v) ) for triatomic linear gas at high temperature. Assume that the contribution of vibrational degree of freedom is 75% :

The total energy of molecules is divided equally amongst the various degrees of freedom of a molecule. The distribution of kinetic energy along x, y, z axis are E_(K_(x)), E_(K_(y)), E_(K_(z)) Total K.e =E_(K_(x)) + E_(K_(y)) + E_(K_(z)) Since the motion of molecule is equally probable in all the three directions, therefore E_(K_(x)) = E_(K_(y)) = E_(K_(z)) =1/3 E_(K) =1/3 xx 3/2 kT = 1/2 kT , where k =R/N_(A) = Botzman constant. K.E. = 1/2 kT per molecule or =1/2 RT per mole. In vibration motion, molecules possess both kinetic energy as well as potential energy. This means energy of vibration involves two degrees of fiuedom. Vibration energy =2 xx 1/2kT =2 xx 1/2RT [ therefore two degrees of freedom per mole] If the gas molecules have n_(1) translational degrees of freedom, n_2 rotational degrees of freedom and n_(3) vibrational degrees of freedom, that total energy = n_(1)[(kT)/2] + n_(2) [(kT)/2] + n_(3) [(kT)/2] xx 2 Where 'n' is atomicity of gas. How many total degrees of freedom are present in H_(2) molecules in all types of motions ?