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At a given temperature internal energy o...

At a given temperature internal energy of a monoatomic, diatomic and non - linear gas is `(U_(0)` each. Find their translational and rotiational kinetic energies separately.

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The correct Answer is:
A
Monoatomic gas Translational degree of freedom of monoatomic gas is (3) and rotational degree of freedom is zero. Therefore, whole internal energy is in the form of translational kinetic energy.
:. `K_T =U_0 and K_R =0`
Diatomic gas Translational degree of freedom of diatomic gas is 3 and rotational degree of freedom is (2). Hence, ratio of translational and rotational kinetic energy will be `3: 2`.
:. `K_T =3/5U_0 and K_R= 2/5U_0`
Non- linear polyatomic gas In non - linear polyatomic gas translational and rotational both degrees of freedom are (3) each. So, translational and rotational kinetic energy are equal. Hence,
`K_T = K_R = (U_0)/2`.
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