Home
Class 12
PHYSICS
For a diatomic gas having 3 translationa...

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

Text Solution

AI Generated Solution

To find the energy associated with a diatomic gas having 3 translational and 2 rotational degrees of freedom, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Degrees of Freedom**: - For a diatomic gas, there are 3 translational degrees of freedom (movement in x, y, and z directions). - There are also 2 rotational degrees of freedom (rotation about two axes perpendicular to the molecular axis). ...
Promotional Banner

Topper's Solved these Questions

  • OSCILLATIONS

    AAKASH INSTITUTE ENGLISH|Exercise EXAMPLE|21 Videos

  • OSCILLATIONS

    AAKASH INSTITUTE ENGLISH|Exercise Exercise|18 Videos

  • OSCILLATIONS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT ( SECTION-I(SUBJECTIVE TYPE QUESTIONS) )|7 Videos

  • NUCLEI

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT (SECTION-D)|10 Videos

  • PHYSICAL WORLD

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT (Section-B)|4 Videos

Similar Questions

Explore conceptually related problems

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

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.

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 :

At ordinary temperatures, the molecules of an ideal gas have only translational and rotational kinetic energies. At high temperatures they may also have vibrational energy. As a result of this, at higher temperature

At ordinary temperatures, the molecules of an ideal gas have only translational and rotational kinetic energies. At high temperatures they may also have vibrational energy. As a result of this, at higher temperature

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 ?

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. The rotational kinetic energy of H20 molecule is equal to

At ordinary temperatures, the molecules of an ideal gas have only translational and rotational kinetic energies. At high temperatures, they may also have vibrational energy. As a result of this, at higher temperatures, molar specific heat capacity at constant volume, C_(V) is

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. The vibrational kinetic energy of CO_2 molecule is

A polyatomic gas with (n) degress of freedom has a mean energy per molecule given by.