Degree of Freedom Questions in English

(A) monoatomic gas molecule consists of a single atom.
Since it can only move in three-dimensional space ($x, y, z$ axes),it possesses only translational degrees of freedom.
It does not have rotational degrees of freedom because the moment of inertia about its center of mass is negligible.
Therefore,a monoatomic gas molecule has $3$ degrees of freedom.

(C) The number of degrees of freedom $f$ for a molecule is given by the formula $f = 3N - k$,where $N$ is the number of atoms and $k$ is the number of constraints.
For a rigid diatomic molecule,there are $N = 2$ atoms and $k = 1$ constraint (the fixed distance between the atoms).
Substituting these values,we get $f = 3(2) - 1 = 5$.
Thus,a diatomic molecule has $5$ degrees of freedom ($3$ translational and $2$ rotational).

(C) For a non-linear triatomic gas molecule,the degrees of freedom $(f)$ is calculated as follows:
$1$. Translational degrees of freedom: $3$ (along $x, y, z$ axes).
$2$. Rotational degrees of freedom: $3$ (about $x, y, z$ axes).
Therefore,the total degrees of freedom $f = 3 + 3 = 6$.

(A) The molar specific heat at constant volume is given by the formula $C_V = \frac{f}{2}R$,where $f$ is the number of degrees of freedom.
For a diatomic gas,the atoms are assumed to be at a constant distance (rigid rotator model),which means the molecule has $3$ translational degrees of freedom and $2$ rotational degrees of freedom.
Thus,the total degrees of freedom $f = 3 + 2 = 5$.
Substituting this value into the formula,we get $C_V = \frac{5}{2}R$.

(D) Argon $(Ar)$ is a monoatomic gas.
For a monoatomic gas,the degrees of freedom $(f)$ is $3$,which corresponds entirely to translational motion.
Monoatomic gases do not possess rotational degrees of freedom because they are point-like particles.
Therefore,the total heat supplied to the gas at constant volume is used entirely for increasing the translational kinetic energy.
The share of translational energy is $100\%$ and the share of rotational energy is $0\%$.

(B) For a diatomic gas molecule,the motion can be described in three-dimensional space.
The number of translational degrees of freedom is always $3$ for any gas molecule,corresponding to motion along the $x, y,$ and $z$ axes.
The number of rotational degrees of freedom for a rigid diatomic molecule is $2$.
Therefore,the number of translational degrees of freedom is $3$.

(B) The adiabatic index $\gamma$ is defined as the ratio of the molar heat capacity at constant pressure $(C_p)$ to the molar heat capacity at constant volume $(C_v)$,given by $\gamma = C_p/C_v = 1 + 2/f$,where $f$ is the degrees of freedom.
For a monatomic gas,$f = 3$,so $\gamma = 1 + 2/3 = 5/3 \approx 1.67$.
For a diatomic gas,$f = 5$ (at room temperature),so $\gamma = 1 + 2/5 = 7/5 = 1.4$.
Since the given value is $\gamma = 7/5$,the gas must be diatomic.
Among the given options,Helium $(He)$,Argon $(Ar)$,and Neon $(Ne)$ are monatomic noble gases,while Hydrogen $(H_2)$ is a diatomic gas.
Therefore,the gas is Hydrogen.

(A) rigid body in space has $6$ degrees of freedom ($3$ translational and $3$ rotational).
When a rigid body is constrained to rotate about a fixed axis,its translational motion is restricted,and its rotational motion is restricted to only one axis.
However,in the context of statistical mechanics and kinetic theory,a rigid body rotating about a fixed axis is often considered to have $1$ degree of freedom associated with its rotational kinetic energy $(K = \frac{1}{2} I \omega^2)$.
Given the standard options provided for this specific physics problem,the intended answer is $1$.

(B) The kinetic energy $E$ per gram mole of an ideal gas is given by the formula $E = \frac{f}{2} RT$,where $f$ is the number of degrees of freedom.
For a diatomic gas at room temperature,the molecule has $3$ translational and $2$ rotational degrees of freedom,so $f = 3 + 2 = 5$.
Substituting $f = 5$ into the formula,we get $E = \frac{5}{2} RT$.

(C) According to the law of equipartition of energy,each degree of freedom contributes $\frac{1}{2}kT$ to the average kinetic energy of a molecule.
For a gas molecule with $n$ degrees of freedom,the total mean energy per molecule is the sum of the energies associated with each degree of freedom.
Therefore,the mean energy per molecule $E = n \times (\frac{1}{2}kT) = \frac{nkT}{2}$.
Thus,the correct option is $C$.

(C) According to the law of equipartition of energy,the total average kinetic energy of a gas molecule is given by $KE = \frac{f}{2} k_B T$,where $f$ is the number of degrees of freedom and $k_B$ is the Boltzmann constant.
The mean kinetic energy per degree of freedom is obtained by dividing the total kinetic energy by the number of degrees of freedom $f$.
Therefore,$\text{Mean kinetic energy per degree of freedom} = \frac{KE}{f} = \frac{\frac{f}{2} k_B T}{f} = \frac{1}{2} k_B T$.

(C) The speed of sound in a gas is given by the formula $v = \sqrt{\frac{\gamma P}{\rho}}$,where $\gamma$ is the adiabatic index,$P$ is the pressure,and $\rho$ is the density.
Rearranging for $\gamma$,we get $\gamma = \frac{v^2 \rho}{P}$.
At $STP$,$P = 1.013 \times 10^5 \ Pa$.
Substituting the given values: $\gamma = \frac{(330)^2 \times 1.3}{1.013 \times 10^5} = \frac{108900 \times 1.3}{101300} \approx 1.4$.
For an ideal gas,$\gamma = 1 + \frac{2}{f}$,where $f$ is the degree of freedom.
Thus,$1.4 = 1 + \frac{2}{f} \Rightarrow 0.4 = \frac{2}{f} \Rightarrow f = \frac{2}{0.4} = 5$.

(B) The degrees of freedom $(f)$ of a gas molecule are defined as the number of independent ways in which a molecule can possess energy.
For any gas molecule,the translational degrees of freedom are always $3$,corresponding to motion along the $x$,$y$,and $z$ axes.
Since a diatomic gas molecule can move in three-dimensional space,it possesses $3$ translational degrees of freedom.
Therefore,the correct option is $B$.

$(A)$ According to the equipartition theorem, the average kinetic energy associated with each degree of freedom per molecule is $\frac{1}{2} kT$.
Both oxygen $(O_2)$ and nitrogen $(N_2)$ are diatomic molecules.
For a diatomic molecule, there are $2$ rotational degrees of freedom.
Therefore, the average rotational kinetic energy per molecule for both gases is $2 \times \frac{1}{2} kT = kT$.
Since the temperature $T$ is the same for both gases, the average rotational kinetic energy per molecule is identical for both $O_2$ and $N_2$.
The ratio of the average rotational kinetic energy of $O_2$ to $N_2$ is $1 : 1$.

(A) According to the equipartition theorem,each rotational degree of freedom contributes $\frac{1}{2}kT$ to the average energy of a molecule.
For a rigid diatomic molecule,there are $2$ rotational degrees of freedom.
Therefore,the average rotational kinetic energy is given by: $K.E._{rot} = 2 \times (\frac{1}{2}kT) = kT$.
The rotational kinetic energy is also expressed as: $K.E._{rot} = \frac{1}{2}I\omega_{rms}^2$.
Equating the two expressions: $\frac{1}{2}I\omega_{rms}^2 = kT$.
Solving for $\omega_{rms}$: $\omega_{rms}^2 = \frac{2kT}{I}$.
Thus,$\omega_{rms} = \sqrt{\frac{2kT}{I}}$.

(A) For a non-linear triatomic gas molecule,the degrees of freedom are as follows:
Translation degrees of freedom $(f_t)$ = $3$.
Rotation degrees of freedom $(f_r)$ = $3$.
Vibration degrees of freedom $(f_v)$ = $2$ (since each vibrational mode contributes $2$ degrees of freedom).
Total degrees of freedom $(f)$ = $f_t + f_r + f_v = 3 + 3 + 2 = 8$.
However,in standard physics problems where the total degrees of freedom is given as $7$ (implying $f_t=3, f_r=2, f_v=2$ for a linear molecule or specific conditions),we use the formula $\gamma = 1 + \frac{2}{f}$.
Given $f = 7$,$\gamma = 1 + \frac{2}{7} = \frac{9}{7} \approx 1.28$.
Thus,$C_P/C_V = 1.28$.

(A) The ratio of specific heats $\gamma$ is defined as the ratio of the molar specific heat at constant pressure $(C_p)$ to the molar specific heat at constant volume $(C_v)$.
For an ideal gas,$C_v = \frac{f}{2}R$ and $C_p = (\frac{f}{2} + 1)R$,where $f$ is the number of degrees of freedom.
Thus,$\gamma = \frac{C_p}{C_v} = \frac{(\frac{f}{2} + 1)R}{(\frac{f}{2})R} = \frac{f + 2}{f}$.
Rearranging the equation to solve for $f$:
$\gamma f = f + 2$
$\gamma f - f = 2$
$f(\gamma - 1) = 2$
$f = \frac{2}{\gamma - 1}$.

(C) Oxygen $(O_2)$ is a diatomic molecule. As a rigid rotator,it has $5$ degrees of freedom ($3$ translational and $2$ rotational).
The number of moles $n$ in $200 \, g$ of $O_2$ is given by $n = \frac{\text{mass}}{\text{molar mass}} = \frac{200}{32} = 6.25 \, \text{mol}$.
The total thermal energy $U$ is given by $U = n \cdot \frac{f}{2} \cdot R \cdot T$,where $f = 5$,$R = 8.314 \, J \cdot mol^{-1} \cdot K^{-1}$,and $T = 27 + 273 = 300 \, K$.
$U = 6.25 \times \frac{5}{2} \times 8.314 \times 300$
$U = 6.25 \times 2.5 \times 8.314 \times 300 = 38953.125 \, J \approx 3.9 \times 10^4 \, J$.
Given the options provided,the closest value is $3.8 \times 10^4 \, J$.

(D) At $NTP$,$22400 \ cm^3$ of any ideal gas contains $6.02 \times 10^{23}$ molecules.
Since $H_2$ is a diatomic molecule,the number of atoms per molecule is $2$. However,the question asks for the degrees of freedom of the molecules in $1 \ cm^3$.
The number of molecules in $1 \ cm^3$ is $n = \frac{6.02 \times 10^{23}}{22400} \approx 0.26875 \times 10^{20}$ molecules.
$A$ diatomic molecule $(H_2)$ has $5$ degrees of freedom ($3$ translational + $2$ rotational).
Total degrees of freedom = (Number of molecules) $\times$ (Degrees of freedom per molecule)
Total degrees of freedom = $0.26875 \times 10^{20} \times 5 = 1.34375 \times 10^{20}$.

(C) The total internal energy $E$ of an ideal gas is given by $E = \frac{f}{2}RT = \frac{f}{2}PV$,where $f$ is the degrees of freedom.
For a monoatomic gas,the degrees of freedom $f = 3$. Therefore,$E = \frac{3}{2}PV$.
For a diatomic gas at moderate temperatures,the degrees of freedom $f = 5$ ($3$ translational + $2$ rotational).
Therefore,the kinetic energy of a diatomic gas molecule is $E = \frac{5}{2}PV$.

(D) Argon $(Ar)$ is a monoatomic gas.
For a monoatomic gas,the degrees of freedom $(f)$ is $3$,which corresponds entirely to translational motion.
Monoatomic gases do not possess rotational degrees of freedom because they are point-like particles.
Therefore,the total internal energy of a monoatomic gas is purely translational kinetic energy.
Translational kinetic energy = $100\%$ of the total energy.
Rotational kinetic energy = $0\%$ of the total energy.
Thus,the correct option is $D$.

(D) The total kinetic energy $E$ of a gas is given by $E = \mu \frac{f}{2} RT$,where $\mu$ is the number of moles,$f$ is the degrees of freedom,$R$ is the universal gas constant,and $T$ is the absolute temperature.
Since $R = N_A k$,we can write $E = \mu \frac{f}{2} N_A k T = n \frac{f}{2} kT$,where $n$ is the total number of molecules.
For $CO_2$,which is a linear triatomic molecule,the degrees of freedom $f = 7$ ($3$ translational + $2$ rotational + $2$ vibrational).
The molar mass $M$ of $CO_2$ is $12 + 2(16) = 44 \, g/mol$.
The number of moles $\mu = \frac{m}{M} = \frac{1}{44} \, mol$.
Substituting these values: $E = \frac{1}{44} \times \frac{7}{2} NkT = \frac{7}{88} NkT$.

(C) Given: Speed of sound $v_s = 330 \ m/s$,density $\rho = 1.3 \ kg/m^3$,and pressure at $STP$ $P = 1.01 \times 10^5 \ N/m^2$.
The formula for the speed of sound in a gas is $v_s = \sqrt{\frac{\gamma P}{\rho}}$.
Rearranging for $\gamma$: $\gamma = \frac{v_s^2 \rho}{P}$.
Substituting the values: $\gamma = \frac{(330)^2 \times 1.3}{1.01 \times 10^5} = \frac{108900 \times 1.3}{101000} \approx \frac{141570}{101000} \approx 1.4$.
The relation between the adiabatic index $\gamma$ and degrees of freedom $f$ is $\gamma = 1 + \frac{2}{f}$.
Solving for $f$: $f = \frac{2}{\gamma - 1} = \frac{2}{1.4 - 1} = \frac{2}{0.4} = 5$.

(D) For a diatomic molecule,the translational degrees of freedom are $f_t = 3$ and the rotational degrees of freedom are $f_r = 2$.
According to the equipartition theorem,the average energy associated with each degree of freedom is $\frac{1}{2} k_B T$.
Average translational kinetic energy $E_t = f_t \times \frac{1}{2} k_B T = 3 \times \frac{1}{2} k_B T = \frac{3}{2} k_B T$.
Average rotational kinetic energy $E_r = f_r \times \frac{1}{2} k_B T = 2 \times \frac{1}{2} k_B T = 1 k_B T$.
The ratio of average translational kinetic energy to rotational kinetic energy is $\frac{E_t}{E_r} = \frac{\frac{3}{2} k_B T}{k_B T} = \frac{3}{2}$.

(B) The ratio of specific heats is given by $\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}$,where $f$ is the total number of degrees of freedom.
Given $\gamma = \frac{9}{7}$,we have $\frac{9}{7} = 1 + \frac{2}{f}$.
Subtracting $1$ from both sides,we get $\frac{2}{7} = \frac{2}{f}$,which implies $f = 7$.
However,the question specifically asks for the degrees of freedom for translational motion.
For any gas molecule,the number of degrees of freedom for translational motion is always $3$ (along the $x, y,$ and $z$ axes),regardless of the total degrees of freedom.

(B) For an adiabatic process, the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = \text{constant}.$
Taking the natural logarithm on both sides, we get $\ln P + \gamma \ln V = \ln(\text{constant}).$
Rearranging this, we get $\ln P = -\gamma \ln V + \ln(\text{constant}).$
This is the equation of a straight line $y = mx + c,$ where the slope $m = -\gamma.$
The magnitude of the slope is equal to the adiabatic index $\gamma.$
From the given graph, the slope of line $X$ is greater than the slope of line $Y,$ so $\gamma_x > \gamma_y.$
Since $\gamma = 1 + \frac{2}{f},$ a larger $\gamma$ corresponds to a smaller degree of freedom $f.$
Therefore, $\gamma_x > \gamma_y \implies f_1 < f_2,$ which means $f_2 > f_1.$

(B) diatomic molecule is treated as a rigid rotator with $2$ degrees of rotational freedom. According to the law of equipartition of energy,each degree of freedom contributes $1/2 kT$ to the average kinetic energy. Therefore,the mean rotational kinetic energy is $2 \times (1/2 kT) = kT$.

(B) For an ideal gas at constant pressure,the heat supplied is $Q = n C_P \Delta T$ and the work done is $W = n R \Delta T$.
Given $Q = x$ and $W = x/3$,we have the ratio $\frac{W}{Q} = \frac{n R \Delta T}{n C_P \Delta T} = \frac{R}{C_P} = \frac{x/3}{x} = \frac{1}{3}$.
We know that $C_P = C_V + R = \frac{f}{2}R + R = R(\frac{f}{2} + 1) = R(\frac{f+2}{2})$.
Substituting this into the ratio: $\frac{R}{R(\frac{f+2}{2})} = \frac{2}{f+2} = \frac{1}{3}$.
Solving for $f$: $f+2 = 6$,which gives $f = 4$.

(A) The ratio of specific heats at constant pressure $(C_p)$ and constant volume $(C_v)$ is given by the adiabatic index $\gamma$.
The formula for the ratio is $\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}$,where $f$ is the number of degrees of freedom.
Given that the gas has $f = 5$ degrees of freedom:
$\gamma = 1 + \frac{2}{5} = 1 + 0.4 = 1.4$.
Therefore,the ratio of specific heats is $1.4$.

(D) For gas $A$: The molar heat capacity at constant volume is $C_V = 22 \, J\, mol^{-1}\, K^{-1}$.
Using the relation $C_V = \frac{f}{2}R$,where $R \approx 8.314 \, J\, mol^{-1}\, K^{-1}$,we get $f = \frac{2 C_V}{R} = \frac{2 \times 22}{8.314} \approx 5.29$.
Since $f$ is slightly greater than $5$ (the degrees of freedom for a rigid diatomic molecule),it indicates the presence of $1$ vibrational mode.
For gas $B$: The molar heat capacity at constant volume is $C_V = 21 \, J\, mol^{-1}\, K^{-1}$.
Using $f = \frac{2 C_V}{R} = \frac{2 \times 21}{8.314} \approx 5.05$.
This value is approximately $5$,which corresponds to a rigid diatomic molecule with no vibrational modes.
Thus,$A$ has a vibrational mode but $B$ has none.

(C) The internal energy $U$ of an ideal gas is given by $U = \frac{f}{2} nRT$,where $f$ is the degrees of freedom,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the absolute temperature.
For $He$ (a monoatomic gas),the degrees of freedom $f_1 = 3$. Given $n_1$ moles at $10T$,the internal energy is $U_1 = \frac{3}{2} n_1 R (10T) = 15 n_1 RT$.
For $H_2$ (a diatomic gas),the degrees of freedom $f_2 = 5$. Given $n_2$ moles at $6T$,the internal energy is $U_2 = \frac{5}{2} n_2 R (6T) = 15 n_2 RT$.
Equating the internal energies,$U_1 = U_2$,we get $15 n_1 RT = 15 n_2 RT$.
Therefore,$n_1 = n_2$,which implies $\frac{n_1}{n_2} = 1$.

(D) The total internal energy $U$ of $\mu$ moles of an ideal gas is given by $U = \frac{f}{2} \mu RT$,where $f$ is the degree of freedom,$\mu$ is the number of moles,$R$ is the universal gas constant,and $T$ is the absolute temperature.
For $O_2$ (a diatomic gas),the degree of freedom $f_1 = 5$ and $\mu_1 = 1$.
So,$U_{O_2} = \frac{5}{2} (1) RT = 2.5 RT$.
For $He$ (a monoatomic gas),the degree of freedom $f_2 = 3$ and $\mu_2 = 2$.
So,$U_{He} = \frac{3}{2} (2) RT = 3 RT$.
The ratio is $\frac{U_{O_2}}{U_{He}} = \frac{2.5 RT}{3 RT} = \frac{2.5}{3} = \frac{5}{6}$.

(D) According to the equipartition theorem,the average kinetic energy associated with each degree of freedom for a single molecule is $\frac{1}{2} kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature.
For $1 \, mole$ of gas,the number of molecules is equal to the Avogadro number,$N_A$.
Therefore,the mean kinetic energy for $1 \, mole$ per degree of freedom is $\frac{1}{2} kT \times N_A$.
Since $R = k \times N_A$ (where $R$ is the universal gas constant),the expression becomes $\frac{1}{2} RT$.

(B) monoatomic gas atom has $3$ degrees of freedom because it can only undergo translational motion along the $X-$,$Y-$,and $Z-$ axes. Rotational and vibrational degrees of freedom are not present for a single point-like atom.
Therefore,the Assertion is correct.
The expression $\frac{C_P}{C_V} = \gamma$ is a standard thermodynamic relation for the ratio of molar specific heats,which is also correct.
However,the value of the degrees of freedom is determined by the structure of the molecule (monoatomic,diatomic,etc.),not by the ratio of specific heats $\gamma$. Thus,the Reason is not the correct explanation of the Assertion.

(A) The adiabatic index is given by $\gamma = 1 + \frac{2}{f}$,where $f$ is the number of degrees of freedom.
$1$. Hydrogen $(H_2)$ is a diatomic gas at room temperature. Its degrees of freedom $f = 5$ ($3$ translational + $2$ rotational). Thus,$\gamma = 1 + \frac{2}{5} = \frac{7}{5}$.
$2$. Helium $(He)$ is a monoatomic gas. Its degrees of freedom $f = 3$ ($3$ translational). Thus,$\gamma = 1 + \frac{2}{3} = \frac{5}{3}$.
$3$. Gas $X$ is a diatomic gas with an additional vibrational mode. $A$ vibrational mode contributes $2$ degrees of freedom ($1$ kinetic + $1$ potential). So,$f = 5 + 2 = 7$. Thus,$\gamma = 1 + \frac{2}{7} = \frac{9}{7}$.
Therefore,the values are $\frac{7}{5}, \frac{5}{3}, \text{and } \frac{9}{7}$.

(B) For a rigid diatomic molecule,the degrees of freedom $(f_A)$ are $5$ ($3$ translational + $2$ rotational).
For a diatomic molecule with an additional vibrational mode,the degrees of freedom $(f_B)$ are $7$ ($3$ translational + $2$ rotational + $2$ vibrational).
The molar specific heat at constant volume is given by $C_v = \frac{f}{2}R$.
Therefore,$C_v^A = \frac{5}{2}R$ and $C_v^B = \frac{7}{2}R$.
The ratio $\frac{C_v^A}{C_v^B} = \frac{\frac{5}{2}R}{\frac{7}{2}R} = \frac{5}{7}$.

(N/A) The degree of freedom is defined as the number of independent ways in which a molecule of a gas can possess energy or undergo motion.
The degree of freedom depends on the following factors:
$(1)$ The number of atoms in the molecule (atomicity).
$(2)$ The temperature of the gas (which determines the activation of vibrational modes).
Mathematically,the total number of independent quadratic terms in the expression for the total energy of a molecule is called its degree of freedom $(f)$.

(N/A) Law of Equipartition of Energy: In thermal equilibrium,the total energy of a system is equally distributed among all its active degrees of freedom,and each degree of freedom is associated with an average energy equal to $\frac{1}{2} k_{B} T$,where $k_{B}$ is the Boltzmann constant and $T$ is the absolute temperature.
Consider a monoatomic gas molecule in thermal equilibrium at temperature $T$. The average kinetic energy of the molecule is given by:
$\langle E_{t} \rangle = \langle \frac{1}{2} m v_{x}^{2} \rangle + \langle \frac{1}{2} m v_{y}^{2} \rangle + \langle \frac{1}{2} m v_{z}^{2} \rangle$
From the kinetic theory of gases,we know that the average kinetic energy of a molecule is $\frac{3}{2} k_{B} T$:
$\langle E_{t} \rangle = \frac{3}{2} k_{B} T$
Since the gas is isotropic,the average kinetic energy along each axis is equal:
$\langle \frac{1}{2} m v_{x}^{2} \rangle = \langle \frac{1}{2} m v_{y}^{2} \rangle = \langle \frac{1}{2} m v_{z}^{2} \rangle$
Substituting this into the total energy equation:
$\frac{3}{2} k_{B} T = 3 \langle \frac{1}{2} m v_{x}^{2} \rangle$
Therefore,the energy associated with each degree of freedom is:
$\langle \frac{1}{2} m v_{x}^{2} \rangle = \frac{1}{2} k_{B} T$
This confirms that for every degree of freedom,the associated average energy is $\frac{1}{2} k_{B} T$. This is known as the Law of Equipartition of Energy.

(N/A) diatomic gas molecule like $O_{2}$,$N_{2}$,$H_{2}$,or $CO$ has three translational degrees of freedom corresponding to motion along the $x$,$y$,and $z$ axes.
$E_{t} = \langle \frac{1}{2} m v_{x}^{2} \rangle + \langle \frac{1}{2} m v_{y}^{2} \rangle + \langle \frac{1}{2} m v_{z}^{2} \rangle$
Additionally,it possesses two $(2)$ rotational degrees of freedom,as shown in the diagram. For a diatomic molecule,there are two independent rotational motions about axes perpendicular to the interatomic axis.
Let $\omega_{1}$ and $\omega_{2}$ be the angular speeds about axes $1$ and $2$,and $I_{1}$ and $I_{2}$ be the moments of inertia about these axes.
The total energy of the molecule is the sum of translational and rotational kinetic energy $(KE)$:
$E = E_{t} + E_{r}$
$E = \langle \frac{1}{2} m v_{x}^{2} \rangle + \langle \frac{1}{2} m v_{y}^{2} \rangle + \langle \frac{1}{2} m v_{z}^{2} \rangle + \langle \frac{1}{2} I_{1} \omega_{1}^{2} \rangle + \langle \frac{1}{2} I_{2} \omega_{2}^{2} \rangle$
Here,there are $3$ terms for translational $KE$ and $2$ terms for rotational $KE$,resulting in a total of $5$ degrees of freedom at room temperature.
At high temperatures,these molecules also exhibit a vibrational mode. The atoms oscillate along the interatomic axis like a one-dimensional harmonic oscillator,contributing two additional degrees of freedom (one for kinetic energy and one for potential energy),making the total $7$ degrees of freedom.

Let $N$ be the number of molecules of a given gas. If each molecule has $f$ degrees of freedom,then according to the law of equipartition of energy,the internal energy associated with each degree of freedom is $\frac{1}{2} k_{B} T$.
The total internal energy $U$ of the gas is given by:
$U = [\text{Total number of molecules}] \times [\text{Degrees of freedom per molecule}] \times [\text{Internal energy per degree of freedom}]$
$\therefore U = N \times f \times \frac{1}{2} k_{B} T$
Since $N = \mu N_{A}$,where $\mu$ is the number of moles and $N_{A}$ is Avogadro's number:
$U = \mu N_{A} \times f \times \frac{1}{2} k_{B} T$
Using the relation $R = N_{A} k_{B}$:
$U = \mu f \left( N_{A} k_{B} \right) \frac{1}{2} T$
$\therefore U = \frac{1}{2} \mu f RT$

(N/A) The degree of freedom $(f)$ of a dynamical system is defined as the total number of independent coordinates or ways in which a system can possess energy or move without violating any constraints imposed on it.
Mathematically,for a system of $N$ particles with $k$ constraints,the degree of freedom is given by $f = 3N - k$.

(N/A) The degree of freedom $(f)$ is defined as the number of independent ways in which a molecule can possess energy.
$(i)$ For a monoatomic gas (e.g.,$He, Ne, Ar$),the molecule can only have translational motion along the $x, y,$ and $z$ axes. Thus,it has $3$ translational degrees of freedom and $0$ rotational degrees of freedom. Total degrees of freedom $f = 3$.
$(ii)$ For a diatomic gas (e.g.,$H_2, O_2, N_2$) at room temperature,the molecule has $3$ translational degrees of freedom and $2$ rotational degrees of freedom (about the axes perpendicular to the bond axis). Total degrees of freedom $f = 3 + 2 = 5$.

(B) diatomic molecule,modeled as a rigid rotator,consists of two atoms separated by a fixed distance.
It has $3$ translational degrees of freedom corresponding to motion along the $x, y,$ and $z$ axes.
Additionally,it has $2$ rotational degrees of freedom because it can rotate about the two axes perpendicular to the line connecting the atoms.
Therefore,the total degree of freedom $f$ is the sum of translational and rotational degrees of freedom: $f = 3 + 2 = 5$.

(N/A) The law of equipartition of energy states that for a dynamic system in thermal equilibrium,the total energy of the system is distributed equally amongst all the degrees of freedom. The energy associated with each degree of freedom per molecule is $\frac{1}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature of the system.

(C) The degree of freedom $(f)$ of a gas molecule is defined as the number of independent ways in which a molecule can possess energy.
For a non-linear polyatomic gas molecule,there are $3$ translational degrees of freedom and $3$ rotational degrees of freedom.
Therefore,the total degree of freedom for a non-linear polyatomic gas is $f = 3 + 3 = 6$.
If the polyatomic gas is linear,it has $3$ translational and $2$ rotational degrees of freedom,resulting in $f = 5$ (at room temperature).
In general,for a non-linear polyatomic molecule,the degree of freedom is $6$.

(C) The degree of freedom of a particle or an object is defined as the number of independent coordinates required to specify its position in space.
Since a room is a three-dimensional space,a honey bee flying within it can move independently along the $x$,$y$,and $z$ axes.
Therefore,the number of degrees of freedom for the honey bee is $3$.

(A) Yes,the degree of freedom changes with an increase in temperature. At low temperatures,only translational and rotational degrees of freedom are active. As the temperature increases,the vibrational mode of motion becomes active,which adds to the total degrees of freedom of the gas molecules.

(C) The degree of freedom $(f)$ of a gas molecule is determined by its atomicity (monatomic,diatomic,or polyatomic),which defines the number of independent ways a molecule can store energy (translational,rotational,and vibrational).
At low temperatures,only translational and rotational degrees of freedom are active.
As the temperature increases,vibrational degrees of freedom may also become active,thus making the degree of freedom dependent on both atomicity and temperature.

(N/A) The total number of degrees of freedom in a thermodynamic system is given by the product of the degrees of freedom per molecule and the total number of molecules.
At $NTP$,the volume occupied by $1$ mole of an ideal gas is $22400$ $cc$.
Using Avogadro's number $(N_A = 6.023 \times 10^{23})$,the number of molecules $(n)$ in $1$ $cc$ is:
$n = \frac{6.023 \times 10^{23}}{22400} \approx 2.688 \times 10^{19}$ molecules.
Hydrogen $(H_2)$ is a diatomic molecule. At room temperature,its degrees of freedom $(f)$ is $5$ ($3$ translational + $2$ rotational).
Total degrees of freedom $= f \times n = 5 \times 2.688 \times 10^{19} = 1.344 \times 10^{20}$.

(D) For a non-linear (triangular) rigid molecule,the degrees of freedom $(f)$ are calculated as follows:
$1$. Translational degrees of freedom: $3$ (along $x, y, z$ axes).
$2$. Rotational degrees of freedom: $3$ (about the three principal axes of rotation for a non-linear molecule).
Total degrees of freedom $(f)$ = $3 + 3 = 6$.
The internal energy $(U)$ of $n$ moles of an ideal gas is given by $U = \frac{f}{2} nRT$.
For $n = 1$ mole and $f = 6$:
$U = \frac{6}{2} \times 1 \times RT = 3RT$.
Thus,the internal energy is $3RT$.