Thermodynamics and Gas Laws

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

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

The heat supplied at constant pressure is given by Q = n Cp delta T. Using Cp = (f + 2)R / 2, substitute the given values: Q = 70 cal, n = 2 moles, R = 2 cal/mol K, and delta T = 5 K. Solving for f gives 5.

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 = (n1*T1 + n2*T2 + n3*T3) / (n1 + n2 + n3).

According to the equipartition theorem of energy, each active degree of freedom of a gas molecule contributes an average kinetic energy of (1/2)kT per molecule, where k is the Boltzmann constant and T is the absolute temperature.

Statement 1 is true because the total translational kinetic energy of an ideal gas is given by (3/2)nRT, which equals 1.5 PV since PV = nRT. Statement 2 is also a true physical fact about gas molecules colliding, but intermolecular collisions are not the explanatory cause of why the total kinetic energy equals 1.5 PV.

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.

The heat absorbed at constant pressure is Q = n Cp delta T = 150 J, and the increase in internal energy is delta U = n Cv delta T. The ratio Cv / Cp is 1 / gamma, where gamma = 1 + (2 / f). With f = 3, gamma = 5/3, so delta U = Q / gamma = 150 * (3/5) = 90 J.

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 $f= 3+ 2=5$

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.