Degree of Freedom and Law of Equipartition of Energy — NEET Physics PYQ MCQs with Solutions

Solution: The rms speed of the gas molecules is $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$. Pressure exerted by ideal gas is $p = \frac{\sqrt{3RT}}{M}$. Average kinetic energy of a molecule is $KE = \frac{5}{2}RT$. Total internal energy of 1 mole of diatomic gas is $\Delta U = \frac{5RT}{2}$. The correct match is A→2, B→1, C→4, D→3.

Solution: The average thermal energy of a system with degree of freedom $f$ is equal to its average energy, which is given as $= \dfrac{f}{2}k_b T$. For monatomic gas, $f = 3$. $\therefore$ Average thermal energy $= \dfrac{3}{2}k_b T$. Hence, correct option is (a).

Solution: The Poisson's ratio $\gamma = \frac{C_p}{C_v}$ where $C_p$ is molar heat capacity at constant pressure and $C_v$ is molar heat capacity at constant volume. Also, $C_v = \frac{f}{2}R$ (where $f$ is degree of freedom). For hydrogen and helium (monatomic), $f = 3$, so $\gamma = 1 + \frac{2}{3} = \frac{5}{3}$. For diatomic gas $X$ with vibrational mode, $f = 7$ (3 translational, 2 rotational, 2 vibrational), so $\gamma = 1 + \frac{2}{7} = \frac{9}{7}$. But hydrogen is diatomic so $\gamma = \frac{7}{5}$ for rigid molecules. Helium is monatomic so $\gamma = \frac{5}{3}$. The correct values are $\frac{5}{3}, \frac{5}{3}, \frac{9}{7}$ where helium and hydrogen are monatomic and the gas $X$ is diatomic with vibrational modes.

Solution: Total internal energy of system = Internal energy of oxygen molecules + Internal energy of argon molecules. For $O_2$ (diatomic): $U_{O_2} = n \times \dfrac{f}{2}RT = 2 \times \dfrac{5}{2}RT = \dfrac{5}{2} \times 2RT = 5RT$. For Ar (monatomic): $U_{Ar} = 4 \times \dfrac{3}{2}RT = \dfrac{3}{2} \times 4RT = 6RT$. Total = $5RT + 6RT = 11RT$.

Solution: As we know that for polytropic process of index $\alpha$ specific heat capacity $= C_v + \dfrac{R}{1 - \alpha}$. Process: $pV^{\gamma} = \text{constant} \Rightarrow \alpha = \gamma = 3$. $\therefore C = C_v + \dfrac{R}{fR} = \dfrac{R}{1 - 3}$. Where, $C_v = \dfrac{fR}{2} = \dfrac{3R}{2}$. For monatomic gas, $f = 3$. $\Rightarrow C = \dfrac{3R}{2} - \dfrac{R}{2} = R$.

Solution: $Q = \dfrac{nR \Delta T}{2}$. Amount of heat required, $Q = \dfrac{3}{2} \times N_A k_B (T_f - T_i) = \dfrac{3}{8} N_A k_B (T_f - T_i)$.

Solution: The specific heat of gas at constant volume in terms of degree of freedom $n$ is $C_v = \dfrac{n}{2}R$. Also, $C_p - C_v = R$. So, $C_p = \dfrac{n}{2}R + R = R\left(\dfrac{n}{2} + 1\right)$. Now, $\gamma = \dfrac{C_p}{C_v} = \dfrac{R\left(\dfrac{n}{2} + 1\right)}{\dfrac{nR}{2}} = \dfrac{n + 2}{n} = 1 + \dfrac{2}{n}$.

Solution: The gases carbon-monoxide (CO) and nitrogen (N$_2$) are diatomic, so both have equal kinetic energy $\dfrac{5}{2}kT$, i.e., $E_1 = E_2$.

Solution: The molecule of a triatomic gas has a tendency of rotating about any of three coordinate axes. So, it has 6 degrees of freedom. 3 translational and 3 rotational. At high enough temperature a triatomic molecule has 2 vibrational degree of freedom. But at temperature requirement is not given, so we answer simply by assuming triatomic gas molecule at room temperature.

Solution: Number of degree of freedom of a dynamical system is obtained by subtracting the number of independent relations from the total number of coordinates required to specify the positions of constituent particles of the system. If a number of particles in the system, $R$ = number of independent relations among the particles, $N$ = number of degree of freedom of the system, then $N = 3A - R$.

Solution: This is the case of monatomic gases. When $C_v = \dfrac{3}{2}R$. This is the case of monatomic gases, when $C_v = \dfrac{R}{0.67} = 1.5 R = \dfrac{3}{2}R$.

Solution: Concept: If there is sudden compression without any exchange of heat the process will be adiabatic. According to law of equipartition of energy for any dynamical system in thermal equilibrium, the total energy is distributed equally amongst all the degrees of freedom and the energy associated with each molecule per degree of freedom is $\dfrac{1}{2}kT$. For a polyatomic gas with $n$ degrees of freedom the mean energy per molecule $= \dfrac{n}{2}nkT$.