Degrees of Freedom and Kinetic Theory of Gases

Using v = sqrt(gamma * P / rho) and P = (rho/M)RT, we find gamma. Since v = 330 m/s, rho = 1.3 kg/m^3, and P = 1.013e5 Pa, gamma = v^2 * rho / P = 1.71. However, for standard gases, gamma = 1 + 2/f. With gamma = 1.4 (diatomic), f = 5. The provided data yields gamma approx 1.4, corresponding to f = 5.

According to the Stefan-Boltzmann law, radiated energy E is proportional to T^4. If T increases by 10%, T_new = 1.1T. Then E_new = (1.1)^4 * E = 1.4641 * E. The increase is 46.41%.

The law is given by Claussius which states that for any dynamical system in a thermal equilibrium, the total energy is equally divided among the degrees of freedom.

Given that value of gamma is 1.66 i.e $\dfrac{5}{3}$ which implies that it is a monoatomic gas, and Neon (Ne) is the only monoatomic gas among the given options.

Degrees of freedom represent the number of independent coordinates required to specify the position of an object. An ant on a 2D plane requires two coordinates (x, y).

Gas molecules are in random motion having some momentum and while colliding with the walls they transfer their momentum to the walls and this collective transfer of momentum from all the molecules to the walls appears as pressure exerted by gas on the container wall.

There will be three degrees of freedom. Two are along x-direction and y-directions due to translation and the last degree of freedom due to angular rotation as the  man climbs up.

For a system of N particles having K independent relations among them, the degrees of freedom of the system is given by 3N-K. 3N  is due to three degrees of freedom associated with each particle if all the particles are independent of each other (i.e K=0) and due to K relation among them, degrees of freedom reduces to 3N-K

Units of R, N and K are $ Joule \times mole^{-1} \times K^{-1} $, $ mole^{-1} $ and $Joule \times  K^{-1}$
So $ R = N \times K $

Total kinetic energy = translational KE + rotational KE. For a rolling disc, K = (1/2)mv^2 + (1/2)I(omega^2). With I = (1/2)mr^2 and v = r*omega, K = (1/2)mv^2 + (1/2)(1/2)mr^2(v^2/r^2) = (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2.

A monoatomic gas has 3 degrees of freedom for translation motion.
Hence, option A is correct.

At a given temperature, the pressure of a container is determined by the number of times gas molecules strike the container walls. If the gas is compressed to a smaller volume, then the same number of molecules will strike against a smaller surface area; the number of collisions against the container will increase, and, by extension, the pressure will increase as well.
Increasing the kinetic energy of the particles will increase the pressure of the gas.
So, the internal energy of a gas Is the total kinetic energy of randomly moving molecules.
Hence, option C is correct.

Vibrational degree of freedom:
(a) For linear molecule = 3N - 5.
(b) For non-linear molecule = 3N - 6. where N is no. of atoms present in a molecule.

Diatomic, like your oxygen molecule, can move in three dimensions and spin in two directions.

 Monoatomic gas has only 3 degree of freedom (all translational)

$\because \gamma =1+\dfrac { 2 }{ f } $
$\Longrightarrow \dfrac { 2 }{ f } =\gamma -1\Longrightarrow f=\dfrac { 2 }{ \gamma -1 } $

For $n$ mole of any gas the total  kinetic energy is given as $E=\dfrac{3}{2}nRT$

The kinetic enrgy of one mole is given by:

For a process PT = constant, P * (PV/nR) = constant, so P^2 * V = constant. This is a polytropic process PV^x = constant with x = 1/2. Molar heat capacity C = R/(gamma-1) + R/(1-x). 4.5R = R/(2/f) + R/(1-0.5) = fR/2 + 2R. 4.5 = f/2 + 2, so f/2 = 2.5, f = 5.

The general epression for degree of freedom is $DOF=3N-n$

Let's assume we have $1$ mole of triatomic gas

Using PV = (m/M)RT, where m=2g, P=8.3e5 Pa, V=2e-3 m^3, T=400K, R=8.314. M = mRT/PV = (2 * 8.314 * 400) / (8.3e5 * 2e-3) = 6651.2 / 1660 = 4.006 g/mol.

Non-linear triatomic gas has f=6. If 50% dissociates into 3 atoms, the new mixture has 0.5 moles of triatomic (f=6) and 0.5 * 3 = 1.5 moles of monatomic (f=3). Average f = (0.56 + 1.53) / (0.5 + 1.5) = (3 + 4.5) / 2 = 7.5 / 2 = 3.75.

From First Law: dQ = dU + dW. This is always true for any process. For monatomic gas at constant pressure, dU = (3/2)nRdT and dW = PdV = nRdT, so dQ = (5/2)nRdT. Then dU/dQ = 3/5 and dW/dQ = 2/5, not 4/5. Option D is the fundamental First Law statement.

The relationship between the adiabatic index gamma and degrees of freedom f is gamma = 1 + 2/f. Solving for f gives f = 2 / (gamma - 1).

At normal room temperature, diatomic gas molecules have 3 translational and 2 rotational degrees of freedom, totaling 5.

An ant walking on a horizontal surface is constrained to move in a 2D plane (x and y coordinates). However, if the ant is considered a rigid body capable of rotation about its center of mass, it possesses 3 degrees of freedom (2 translational and 1 rotational).

$\begin{array}{l} w=nRT, , , \ln { \left( { \frac { { 4v } }{ v }  } \right)  }  \ =nRT, , \ln { 4 } , , , , , , & , , , \Delta u=0 \end{array}$

$\begin{array}{l} As, { y _{ mix } }=\dfrac { { { C _{ { p _{ mix } } } } } }{ { { C _{ { v _{ mix } } } } } } where, { C _{ { p _{ mix } } } }=\dfrac { { { n _{ 1 } }{ C _{ { p _{ 1 } } } }+{ n _{ 2 } }{ C _{ { p _{ 2 } } } } } }{ { { n _{ 1 } }+{ n _{ 2 } } } }  \ and, , { C _{ { v _{ mix } } } }=\dfrac { { { n _{ 1 } }{ C _{ { v _{ 1 } } } }+{ n _{ 2 } }{ C _{ { v _{ 2 } } } } } }{ { { n _{ 1 } }+{ n _{ 2 } } } }  \ So,, , { y _{ mix } }=\dfrac { { { n _{ 1 } }{ C _{ { p _{ 1 } } } }+{ n _{ 2 } }{ C _{ { p _{ 2 } } } } } }{ { { n _{ 1 } }+{ n _{ 2 } } } }  \ Given, ,, for, monoatomic, { C _{ p } },\dfrac { 5 }{ 2 } R, and, { C _{ { v _{ 1 } } } }=\dfrac { 3 }{ 2 } R \ For, diatomic, { C _{ { p _{ 2 } } } }=\dfrac { { 7R } }{ 2 } , and, { C _{ { v _{ 2 } } } }=\dfrac { 5 }{ 2 } R \ { y _{ mix } }=\dfrac { { { n _{ 1 } }\times \dfrac { 5 }{ 2 } R+{ n _{ 2 } }\times \dfrac { 7 }{ 2 } R } }{ { { n _{ 1 } }\times \dfrac { 3 }{ 2 } R+{ n _{ 2 } }\times \dfrac { 5 }{ 2 } R } } =\dfrac { { 5{ n _{ 1 } }+7{ n _{ 2 } } } }{ { 3{ n _{ 1 } }+5{ n _{ 2 } } } } =\dfrac { 3 }{ 2 }  \ 10{ n _{ 1 } }+14{ n _{ 2 } }=9{ n _{ 1 } }+15{ n _{ 2 } } \ { n _{ 1 } }={ n _{ 2 } } \ Hence, \ option, , A, , is, correct, , answer. \end{array}$

The final temperature of a mixture of non-reacting gases is the weighted average of their temperatures based on the number of moles (or molecules) of each gas, assuming equal degrees of freedom. The formula is T_final = (n1T1 + n2T2 + n3*T3) / (n1 + n2 + n3).

For a process PT = constant, the molar heat capacity is C = Cv + R / (1 - x), where PV^x = constant. Since PT = constant, P(PV/nR) = constant, so P^2V = constant, or PV^0.5 = constant. Thus x = 0.5. C = (f/2)R + R / (1 - 0.5) = (f/2)R + 2R. With C = 37.35 and R = 8.314, 37.35 = R(f/2 + 2) => 4.49 = f/2 + 2 => f/2 = 2.49 => f = 5.

Moles of He =$\displaystyle\ \frac{2}{4}$ = $\displaystyle\ \frac{1}{2}$
Molecules = $\displaystyle\ \frac{1}{2}\times6.02\times10^{23}$ = $3.01\times10^{23}$
As there are 3 degrees of freedom corresponding of 1 molecule of a monatomic gas.
$\therefore$ Total degrees of freedom = $3\times3.01\times10^{23}$
$= 9.03\times10^{23}$

Vibrational energy involves additional degrees of freedom. Thus the degrees of freedom for a diatomic gas increases at higher temperatures.

Average K.E. $= \displaystyle \frac{3}{2}$ RT
At 0$^o$C, average K.E. $= \displaystyle \frac{3}{2} \times R \times 273$
                                           $[T = (0 + 273) K]$
At 273$^o$C,    
average K.E. $= \displaystyle \frac{3}{2} \times R \times (273 + 273)$
$= \displaystyle \frac{3}{2} \times R \times 2 \times 273$
$\therefore $ Ratio = 2

$\displaystyle \frac{W}{Q}=\frac{P\Delta V}{nC _{P}\Delta T}=\frac{nR\Delta T}{nC _{P}\Delta T}=\frac{x/3}{x}$  (standard result)

$\displaystyle \Rightarrow C _{P}=3R=\left ( \frac{f}{2}+1 \right )R\Rightarrow f=4$

The average kinetic energy of gas molecules is directly proportional to absolute temperature only; this implies that all molecular motion ceases if the temperature is reduced to absolute zero.
Hence, option C is correct.

In three-dimensional space, three degrees of freedom are associated with the movement of a particle. A diatomic  gas molecule thus has 6 degrees of freedom. This set may be decomposed in terms of translations, rotations, and vibrations of the molecule. The center of mass motion of the entire molecule accounts for 3 degrees of freedom. In addition, the molecule has two rotational degrees of motion and one vibrational mode The rotations occur around the two axes perpendicular to the line between the two atoms. The rotation around the atom-atom bond is not a physical rotation. At normal temp,  vibration is not possible. Hence, the total no of degrees of freedom is $f= 3+ 2=5$

$Answer:-$ A

By the law of equipartition of energy, for one mole of polyatomic gas
$C _p=(4+f)R \ and \ C _v=(3+f)R$
$\therefore \dfrac{C _p}{C _v}=\dfrac{(4+f)R}{(3+f)R}=\dfrac{(4+f)}{(3+f)}$

For a gas, we know $\dfrac{R}{C _V}=\gamma -1$
or $0.67=\gamma -1$ or, $\gamma =1.67$
Hence the gas is monatomic.

$\Delta u=\dfrac{f}{2}RT=3RT$

At constant pressure, Q = nCpdeltaT and W = nRdeltaT. Given Q = 4Q (this seems to be a typo in the prompt, likely Q_heat = 4W). If Q_heat = 4W, then CpdeltaT = 4R*deltaT, so Cp = 4R. Since Cp = (f/2 + 1)R, then f/2 + 1 = 4, f/2 = 3, f = 6.

For monoatomic gas, degrees of freedom is 3.