Correct answer is (D)

The average kinetic energy per molecule of a gas is given by the expression $\frac{3}{2}kT$, where $k$ is the Boltzmann constant and $T$ is the temperature of the gas in kelvins. Since the temperature of the mixture is given in Celsius, we need to convert it to kelvins by adding 273.15 to get $303.15$ K.

Let us assume that the total mass of the mixture is $3x$ (where $x$ is a constant), then the mass of hydrogen and argon in the mixture will be $2x$ and $x$ respectively, according to the given ratio.

The number of moles of hydrogen and argon can be calculated using their respective masses and molar masses, which are 1 g/mol and 39.9 g/mol respectively. Therefore:

Number of moles of hydrogen = $\frac{2x}{1\mathrm{g}/\mathrm{mol}}=2x$ mol

Number of moles of argon = $\frac{x}{39.9\mathrm{g}/\mathrm{mol}}=\frac{x}{39.9}$ mol

The total number of moles of gas in the mixture is the sum of the number of moles of hydrogen and argon:

Total number of moles of gas = $2x+\frac{x}{39.9}=\frac{79.9x}{39.9}$ mol

The average kinetic energy per molecule of hydrogen is:

$\frac{3}{2}k{T}_{{\mathrm{H}}_2}=\frac{3}{2}\times 1.38\times {10}^{-23}\times 303.15\mathrm{K}=6.12\times {10}^{-21}\mathrm{J}$

The average kinetic energy per molecule of argon is:

$\frac{3}{2}k{T}_{\mathrm{Ar}}=\frac{3}{2}\times 1.38\times {10}^{-23}\times 303.15\mathrm{K}=6.12\times {10}^{-21}\mathrm{J}$

Therefore, the ratio of the average kinetic energy per molecule of argon to hydrogen is:

$\frac{K_{\mathrm{Ar}}}{K_{{\mathrm{H}}_2}}=\frac{\frac{3}{2}k{T}_{\mathrm{Ar}}}{\frac{3}{2}k{T}_{{\mathrm{H}}_2}}=\frac{6.12\times {10}^{-21}\mathrm{J}}{6.12\times {10}^{-21}\mathrm{J}}=1$

Hence, the ratio of average kinetic energy per molecule of argon to hydrogen is $1$.

Correct answer is (D)

The internal energy (U) of a gas depends on its degrees of freedom (f).

For a monatomic gas like neon, the degrees of freedom are f = 3 (translational). For a diatomic gas like oxygen, the degrees of freedom are f = 5 (3 translational + 2 rotational).

The internal energy for each component of the gas mixture can be calculated using the formula:

$U=\frac{f}{2}nRT$

where n is the number of moles, R is the universal gas constant, and T is the temperature.

For the 4 moles of neon (monatomic):

$U_{Ne}=\frac{3}{2}\cdot 4RT=6RT$

For the 2 moles of oxygen (diatomic):

$U_{O_2}=\frac{5}{2}\cdot 2RT=5RT$

Now, to find the total internal energy, we sum the internal energies of the individual components:

$U_{total}=U_{Ne}+U_{O_2}=6RT+5RT=11RT$

Thus, the total internal energy of the system is 11RT.

Correct answer is (C)

The kinetic energy of an ideal gas is given by the equation:

$KE=\frac{3}{2}kT$

where (k) is Boltzmann's constant and (T) is the absolute temperature in kelvins. Therefore, the kinetic energy of a gas is directly proportional to its temperature.

If the kinetic energy doubles, the temperature must also double. The original temperature is given as (${27}^{\circ }C$), which is equal to (300 K) in absolute terms. Therefore, the final temperature ($T_f$) in kelvins is:

$T_f=2\cdot 300K=600K$

Converting this back to degrees Celsius gives:

$T_f=600K-273={327}^{\circ }C$

Correct answer is (C)

$\mathrm{\Delta }Q=n{C}_P\mathrm{\Delta }T=$

$n\left(\frac{f}{2}+1\right)R\mathrm{\Delta }T$

= $n\left(\frac{5}{2}+1\right)R\mathrm{\Delta }T$

= $n\left(\frac{7}{2}\right)R\mathrm{\Delta }T$

Given, $\mathrm{\Delta }Q$ = 735

$\Rightarrow$ $n\left(\frac{7}{2}\right)R\mathrm{\Delta }T$ = 735

$\Rightarrow$ $nR\mathrm{\Delta }T=735\times \frac{2}{7}$

Also, $\mathrm{\Delta }$U = $\frac{f}{2}nR\mathrm{\Delta }T$

= $\frac{5}{2}nR\mathrm{\Delta }T$

= $\frac{5}{2}\times 735\times \frac{2}{7}$

= 525 J

Correct answer is (A)

K.E. per molecule $=\left(\frac{f}{2}KT\right)$

$\frac{\mathrm{average}{(\mathrm{K}.\mathrm{E})}_{\mathrm{hydrogen}}}{\mathrm{average}{(\mathrm{K}.\mathrm{E})}_{\mathrm{oxygen}}}=\frac{f_{\mathrm{hydrogen}}}{f_{\mathrm{oxygen}}}=1$