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Kinetic Theory of Gases and Radiation — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsKinetic Theory of Gases and Radiation4 Marks
Question
Explain the degrees of freedom a polyatomic molecule can have. Hence write the expressions for
(i) the energy per molecule
(ii) the energy per mole
(iii) $\mathrm{C}_{\mathrm{V}}$
(iv) $\mathrm{C}_{\mathrm{p}}$
(v) the adiabatic constant, for a polyatomic gas.
✓

Answer


A polyatomic molecule has three degrees of translational freedom like any particle. Each molecule can also rotate about its centre of mass with an angular velocity components along each of the three axes. Therefore, each molecule has three degrees of rotational freedom. Additionally, if the molecule is soft at high enough temperatures, it can vibrate easily with many different frequencies, say, $f$, because there are many interatomic bonds. Hence, it has $2 f$ degrees of vibrational freedom. Then, according to the law of equipartition of energy, for each degree of freedom of translation and rotation, the molecule has the average energy $\frac{1}{2} \mathrm{k}_{\mathrm{B}} T$, but for each frequency of vibration the average energy is $\mathrm{k}_B T$, since each vibration involves potential energy and kinetic energy. $\mathrm{k}_{\mathrm{B}}$ is the Boltzmann constant and $\mathrm{T}$ is the thermodynamic temperature.
(i) The energy per molecule
$
=3\left(\frac{1}{2} k_{\mathrm{B}} T\right)+3\left(\frac{1}{2} k_{\mathrm{B}} T\right)+f\left(k_{\mathrm{B}} T\right)=(3+f) k_{\mathrm{B}} T
$
(ii) The energy per mole,
$
E=(3+f) k_{\mathrm{B}} T \times N_{\mathrm{A}}=(3+f) R T
$
(iii) $C_V=\frac{d E}{d T}=(3+f) R$
(iv) $C_P=C_V+R=(3+f) R+R=(4+f) R$
(v) The adiabatic constant, $\gamma=\frac{C_P}{C_V}=\frac{4+f}{3+f}$
[Note : As /increases, y decreases.]

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