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Calculate the degree of freedom of diato...

Calculate the degree of freedom of diatomic molecule?

A

Non-steady flow

B

Compressible liquids

C

Streamlined flow

D

Turbulent flow

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To calculate the degree of freedom of a diatomic molecule, we can follow these steps: ### Step 1: Understand the Concept of Degree of Freedom The degree of freedom (f) of a molecule refers to the number of independent ways in which the molecule can move. For a diatomic molecule, we need to consider both translational and rotational motions. ### Step 2: Identify the Formula The degree of freedom can be calculated using the formula: \[ f = 3n - c \] where: - \( n \) is the number of molecules, - \( c \) is the number of constraints. ### Step 3: Determine the Values of n and c For a diatomic molecule: - At high temperatures, the number of degrees of freedom due to translational and rotational motion is \( n = 2 \) (since it consists of two atoms). - At high temperature, there are no constraints, so \( c = 0 \). ### Step 4: Calculate Degree of Freedom at High Temperature Substituting the values into the formula: \[ f = 3(2) - 0 = 6 \] Thus, the degree of freedom at high temperature is 6. ### Step 5: Consider Low Temperature Conditions At low temperatures, the number of degrees of freedom remains \( n = 2 \) (still a diatomic molecule), but there is a constraint due to the vibrational motion being frozen out. Here, \( c = 1 \). ### Step 6: Calculate Degree of Freedom at Low Temperature Substituting the values into the formula: \[ f = 3(2) - 1 = 5 \] Thus, the degree of freedom at low temperature is 5. ### Conclusion - Degree of freedom at high temperature: 6 - Degree of freedom at low temperature: 5
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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 ?

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