Kinetic molecular theory of gases and Molecular collisions Questions in English

(A) Gases do not settle to the bottom of a container because their particles possess high kinetic energy,which keeps them in constant,random motion.
Due to this high kinetic energy and their very small mass,the effect of gravity on individual gas molecules is negligible,preventing them from settling at the bottom.

(D) The Assertion is incorrect because molecules in a gas possess a distribution of speeds (Maxwell-Boltzmann distribution) at a given temperature,not the same speed.
The Reason is also incorrect because,for a pure gas,all molecules are identical in size and shape.
Therefore,both Assertion and Reason are incorrect.

(N/A) The kinetic molecular theory provides a microscopic model for the behavior of gases based on the following postulates:
$1$. $Gases$ consist of a large number of identical particles (atoms or molecules) that are so small and so far apart that the actual volume of the molecules is negligible compared to the empty space between them. They are considered as 'point masses',which explains the high compressibility of gases.
$2$. There is no force of attraction between the particles of a gas at ordinary temperature and pressure. This explains why gases expand to occupy all available space.
$3$. Particles of a gas are in constant,random motion in straight lines. If they were at rest,the gas would have a fixed shape,which is not observed.
$4$. Gas pressure is exerted due to the collision of particles with the walls of the container.
$5$. Collisions of gas molecules are perfectly elastic. This means the total kinetic energy of the molecules remains constant before and after the collision,even if individual energies change.
$6$. At any given temperature,although individual particles have different speeds and kinetic energies that change constantly due to collisions,the overall distribution of speeds remains constant.

(N/A) The microscopic model of gases is based on the postulates of the Kinetic Molecular Theory of Gases:
$1$. $Particles$ $as$ $point$ $masses$: Gases consist of a large number of identical particles (atoms or molecules) that are so small and so far apart that the actual volume of the molecules is negligible compared to the empty space between them. This explains the high compressibility of gases.
$2$. $No$ $intermolecular$ $forces$: There is no force of attraction between gas particles at ordinary temperature and pressure. This explains why gases expand to occupy all available space.
$3$. $Constant$ $random$ $motion$: Gas particles are in constant,random,straight-line motion. If they were at rest,gases would have a fixed shape,which is not observed.
$4$. $Pressure$ $exertion$: Gas particles collide with each other and with the walls of the container. The pressure exerted by a gas is the result of these collisions with the container walls.
$5$. $Elastic$ $collisions$: Collisions between gas molecules are perfectly elastic. While individual energies may change during collisions,the total kinetic energy of the system remains constant at a given temperature.
$6$. $Distribution$ $of$ $speeds$: At any given temperature,although individual particles have different and constantly changing speeds due to collisions,the overall distribution of speeds remains constant.

(N/A) $1$. Kinetic Energy: The total kinetic energy of a gas is the sum of the kinetic energies of all its individual molecules. For an ideal gas,it is given by $E_k = \frac{3}{2} nRT$.
$2$. Average Translational Kinetic Energy: This is the average kinetic energy per mole of gas molecules,given by $E_{avg} = \frac{3}{2} RT$. For a single molecule,it is $\frac{3}{2} kT$,where $k$ is the Boltzmann constant.
$3$. Root Mean Square Speed $(u_{rms})$: It is defined as the square root of the mean of the squares of the speeds of different molecules of a gas. It is given by the formula $u_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.

(N/A) The two main unfair assumptions of the Kinetic Molecular Theory of gases are:
$1$. The volume occupied by gas molecules is negligible compared to the total volume of the gas.
Justification: At high pressures,the gas molecules are pushed close together,and their actual volume becomes significant relative to the total volume of the container.
$2$. There are no forces of attraction or repulsion between gas molecules.
Justification: At low temperatures and high pressures,gas molecules move slowly and come closer to each other,leading to significant intermolecular forces of attraction,which causes the gas to deviate from ideal behavior and eventually liquefy.

(C) In an ideal gas,it is assumed that there are no intermolecular forces of attraction or repulsion between the molecules.
This is a hypothetical concept.

(C) The postulates of the kinetic molecular theory are related to the behavior of gas particles. It provides a microscopic model for the behavior of gases.

(B) Gas molecules are in constant,random,linear motion in all possible directions. $ \newline $ When these molecules collide with the walls of the container,they exert a force per unit area,which is defined as pressure.

(A) When a gas is heated,the kinetic energy of the gas molecules increases.
As a result,the molecules move faster and collide more frequently and with greater force against the walls of the container,leading to an increase in pressure if the volume is constant.

(N/A) An elastic collision is a collision in which there is no net loss of total kinetic energy in the system as a result of the collision. The kinetic energy is only transferred between the colliding molecules.

(B) The average translational kinetic energy of a gas molecule is given by the expression $ \frac{3}{2} k_B T $.
Here,$ k_B $ is the Boltzmann constant and $ T $ is the absolute temperature in Kelvin.
This is derived from the kinetic theory of gases where the total kinetic energy for $ 1 \text{ mole} $ of gas is $ \frac{3}{2} RT $.

(C) In the kinetic molecular theory of gases,collisions are assumed to be perfectly elastic,meaning there is no loss of kinetic energy. If collisions were not elastic,the molecules would lose kinetic energy with each collision. Eventually,the molecules would lose all their kinetic energy,causing them to stop moving and hit the container walls with zero force,resulting in the pressure becoming $0$.

(C) The kinetic energy of an ideal gas depends only on its absolute temperature,given by the formula $KE = \frac{3}{2} nRT$ for $n$ moles or $KE = \frac{3}{2} kT$ per molecule.
Since both $N_2$ and $O_2$ are at the same temperature of $300 \ K$,their average kinetic energy per mole (or per molecule) will be the same.
Therefore,the kinetic energy is equal for both.

(B) The average kinetic energy $(KE)$ of an ideal gas is directly proportional to its absolute temperature $(T)$.
The relationship is given by the equation: $KE = \frac{3}{2} kT$,where $k$ is the Boltzmann constant.

(A) Since $He$ and $Ne$ are monoatomic noble gases,they do not possess rotational or vibrational degrees of freedom. They only exhibit translational kinetic energy.

(D) The kinetic energy $(KE)$ of an ideal gas is given by the formula $KE = \frac{3}{2} n R T$ (for translational kinetic energy of any ideal gas).
Number of moles of $H_2$ $(n_{H_2})$ = $\frac{3 \ g}{2 \ g/mol} = 1.5 \ mol$.
Number of moles of $O_2$ $(n_{O_2})$ = $\frac{4 \ g}{32 \ g/mol} = 0.125 \ mol = \frac{1}{8} \ mol$.
Since the temperature $T$ is the same for both,the ratio of kinetic energies is:
$\frac{KE_{H_2}}{KE_{O_2}} = \frac{\frac{3}{2} n_{H_2} R T}{\frac{3}{2} n_{O_2} R T} = \frac{n_{H_2}}{n_{O_2}} = \frac{1.5}{0.125} = \frac{1.5 \times 8}{1} = \frac{12}{1}$.
Thus,the ratio is $12: 1$.

(D) The expressions for most probable velocity $(V_{mp})$,average velocity $(\bar{V})$,and root mean square velocity $(V_{rms})$ are as follows:
$V_{mp} = \sqrt{\frac{2RT}{M}}$
$\bar{V} = \sqrt{\frac{8RT}{\pi M}} \approx \sqrt{\frac{2.55RT}{M}}$
$V_{rms} = \sqrt{\frac{3RT}{M}}$
Comparing these values,we get the order: $V_{mp} < \bar{V} < V_{rms}$.
In the Maxwell-Boltzmann distribution curve,the peak corresponds to $V_{mp}$,followed by $\bar{V}$,and then $V_{rms}$ on the right side of the peak.
Thus,the correct plot is the one where the order from left to right is $V_{mp}$,$\bar{V}$,$V_{rms}$.

(D) The correct option is $D$.
Average kinetic energy depends only upon the temperature and is independent of the mass or type of gas molecules.
For any ideal gas,the average kinetic energy per molecule is given by the formula $(K.E.)_{avg} = \frac{3}{2}kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature.
Since both deuterium $(D_2)$ and hydrogen $(H_2)$ are at the same temperature $(298 \, K)$,their average kinetic energies are equal.

(A) The diffusion coefficient $D$ is proportional to the mean free path $\lambda$ and the mean speed $v$,so $D \propto \lambda \cdot v$.
We know that the mean free path $\lambda \propto \frac{T}{P}$ and the mean speed $v \propto \sqrt{T}$.
Therefore,$D \propto \frac{T}{P} \cdot \sqrt{T} = \frac{T^{3/2}}{P}$.
Given that the temperature $T_f = 4T_i$ and the pressure $P_f = 2P_i$,we can write the ratio:
$\frac{D_f}{D_i} = \left( \frac{T_f}{T_i} \right)^{3/2} \cdot \left( \frac{P_i}{P_f} \right)$
$\frac{D_f}{D_i} = (4)^{3/2} \cdot \left( \frac{1}{2} \right)$
$\frac{D_f}{D_i} = 8 \cdot \frac{1}{2} = 4$.
Thus,$D_f = 4D_i$,which means $x = 4$.

(A) . According to the postulates of the kinetic theory of gases,collisions between molecules and with the walls of the container are perfectly elastic.
$B$. The momentum transferred to the wall is $\Delta p = 2mu$. Since momentum is proportional to mass $(m)$,heavier molecules transfer more momentum upon collision.
$C$. According to the Maxwell-Boltzmann distribution,the fraction of molecules with very high or very low velocities is extremely small.
$D$. Between two successive collisions,molecules move in straight lines with constant velocity due to the absence of intermolecular forces.

(A) $1.$ According to Graham's law,if all conditions are identical,$r \propto \frac{1}{\sqrt{M}}$.
Since all conditions are identical for $X$ and $Y$,we have $\frac{r_x}{r_y} = \sqrt{\frac{M_y}{M_x}}$.
$\frac{d}{24-d} = \sqrt{\frac{40}{10}} = 2$.
$d = 48 - 2d \implies 3d = 48 \implies d = 16 \ cm$.
$2.$ The mean free path $\lambda$ is given by $\lambda = \frac{RT}{\sqrt{2} \pi d^2 N_A P}$.
Since $d$ and $P$ are the same for both gases,their $\lambda$ values are identical. However,$X$ has a lower molar mass,so it has a higher average speed $(v_{avg} \propto \frac{1}{\sqrt{M}})$. Due to this higher drift speed,$X$ experiences a higher collision frequency with the inert gas molecules compared to $Y$. Consequently,$X$ encounters more resistance,causing it to cover a shorter distance than predicted by Graham's law.

(A) $Argon$ is a monatomic gas,thus its atoms can move in any direction in space.
Since it is a single atom,it does not have bonds to vibrate or an axis to rotate about.
Hence,it can have only three independent degrees of freedom,all of which are translational.
Therefore,$Argon$ possesses only translational motion.

(B) According to Graham's Law of diffusion,the rate of diffusion $r \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass.
The molar masses are: $H_{2} = 2 \text{ g/mol}$,$CH_{4} = 16 \text{ g/mol}$,$SO_{2} = 64 \text{ g/mol}$.
Since $M(H_{2}) < M(CH_{4}) < M(SO_{2})$,the rate of diffusion is $r(H_{2}) > r(CH_{4}) > r(SO_{2})$.
Therefore,the amount of gas remaining in the container after time $t$ will be in the reverse order of their diffusion rates.
The amount left is in the order: $SO_{2} > CH_{4} > H_{2}$.
Since partial pressure $p$ is directly proportional to the number of moles $n$ $(p = \frac{nRT}{V})$,the order of partial pressures is $pSO_{2} > pCH_{4} > pH_{2}$.

(B) The average kinetic energy $(KE)$ of one mole of an ideal gas is given by the formula: $KE = \frac{3}{2} RT$.
Since $R$ is a constant,$KE \propto T$.
For oxygen $(O_{2})$: $T_{O_{2}} = 273 \ K$.
For sulphur dioxide $(SO_{2})$: $T_{SO_{2}} = 546 \ K$.
Comparing the kinetic energies: $\frac{KE_{O_{2}}}{KE_{SO_{2}}} = \frac{T_{O_{2}}}{T_{SO_{2}}} = \frac{273}{546} = \frac{1}{2}$.
Therefore,$KE_{SO_{2}} = 2 \times KE_{O_{2}}$,which implies $KE_{SO_{2}} > KE_{O_{2}}$ or $KE_{O_{2}} < KE_{SO_{2}}$.

(D) The kinetic energy $(KE)$ of an ideal gas is given by the formula $KE = \frac{3}{2} nRT$.
For $H_2$:
Molar mass $M(H_2) = 2 \ g/mol$.
Number of moles $n_1 = \frac{4 \ g}{2 \ g/mol} = 2 \ mol$.
Temperature $T_1 = 27 + 273 = 300 \ K$.
$x = \frac{3}{2} \times 2 \times R \times 300 = 900R$.
For $O_2$:
Molar mass $M(O_2) = 32 \ g/mol$.
Number of moles $n_2 = \frac{6.4 \ g}{32 \ g/mol} = 0.2 \ mol$.
Temperature $T_2 = 127 + 273 = 400 \ K$.
$KE_2 = \frac{3}{2} \times 0.2 \times R \times 400 = 120R$.
Now,find the ratio: $\frac{KE_2}{x} = \frac{120R}{900R} = \frac{12}{90} = \frac{2}{15}$.
Therefore,$KE_2 = \frac{2x}{15} \ J$.

(A) The root mean square velocity $(u_{rms})$ is given by the formula $u_{rms} = \sqrt{\frac{3RT}{M}}$.
Given $u_{rms} = 412 \ ms^{-1}$ and molar mass $M = 44 \times 10^{-3} \ kg \ mol^{-1}$.
Squaring both sides: $u_{rms}^2 = \frac{3RT}{M} \implies \frac{3}{2}RT = \frac{1}{2} M u_{rms}^2$.
The kinetic energy $(KE)$ of $1 \ mol$ of an ideal gas is given by $KE = \frac{3}{2}RT = \frac{1}{2} M u_{rms}^2$.
Substituting the values: $KE = \frac{1}{2} \times (44 \times 10^{-3} \ kg \ mol^{-1}) \times (412 \ ms^{-1})^2$.
$KE = 0.5 \times 0.044 \times 169744 \ J \ mol^{-1} = 3734.368 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$: $KE = \frac{3734.368}{1000} \approx 3.7344 \ kJ \ mol^{-1}$.
Thus,the correct option is $A$.

(A) The average kinetic energy of an ideal gas is given by $K.E. = \frac{3}{2} nRT$ for $n$ moles of gas.
For $1 \ mol$ of any ideal gas,$K.E. = \frac{3}{2} RT$.
This shows that the kinetic energy of $1 \ mol$ of an ideal gas depends only on the temperature $T$ and is independent of the nature of the gas (mass of the molecule).
Therefore,the statement that kinetic energy depends on the mass of the gas molecule is false.

(D) The kinetic energy $(KE)$ of an ideal gas is directly proportional to its absolute temperature $(T)$.
For one mole of an ideal gas,the kinetic energy is given by the formula:
$KE = \frac{3}{2} RT$
Since $R$ is the universal gas constant,$KE$ depends only on the absolute temperature $T$.

(D) The total kinetic energy $(KE)$ of an ideal gas is given by the formula: $KE = \frac{3}{2} n R T$.
For the kinetic energies to be equal,we set the expressions for $He$ and $Ar$ equal to each other:
$\frac{3}{2} n_{He} R T_{He} = \frac{3}{2} n_{Ar} R T_{Ar}$.
Canceling the common terms $\frac{3}{2}$ and $R$,we get: $n_{He} \times T_{He} = n_{Ar} \times T_{Ar}$.
Given: $n_{He} = 0.30 \ mol$,$n_{Ar} = 0.40 \ mol$,and $T_{Ar} = 400 \ K$.
Substituting the values: $0.30 \times T_{He} = 0.40 \times 400$.
$T_{He} = \frac{0.40 \times 400}{0.30} = \frac{160}{0.30} = 533.33 \ K \approx 533 \ K$.

(B) Gas molecules are in constant random motion with high velocities and change direction upon collision with other molecules. Therefore,there is a complete disorder of molecules in gases.

(B) The average kinetic energy per molecule of an ideal gas is given by the formula: $KE_{avg} = \frac{3}{2} k T$
Where $k$ is the Boltzmann constant $(k = \frac{R}{N_A})$ and $T$ is the temperature in Kelvin.
Given $T = 27^{\circ} C = 27 + 273 = 300 \ K$.
Substituting the values:
$KE_{avg} = \frac{3}{2} \times \frac{8.314 \ J \ K^{-1} \ mol^{-1}}{6.022 \times 10^{23} \ mol^{-1}} \times 300 \ K$
$KE_{avg} = 1.5 \times 1.38 \times 10^{-23} \times 300 \ J$
$KE_{avg} \approx 6.21 \times 10^{-21} \ J \ molecule^{-1}$

(C) The kinetic energy $(KE)$ of an ideal gas is given by the formula $KE = nRT$,where $n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the temperature in Kelvin.
For helium: $n_1 = 0.3 \ mol$,$T_1 = T$.
$KE_{He} = 0.3 \times R \times T$.
For argon: $n_2 = 0.4 \ mol$,$T_2 = 400 \ K$.
$KE_{Ar} = 0.4 \times R \times 400$.
According to the problem,$KE_{He} = KE_{Ar}$.
$0.3 \times R \times T = 0.4 \times R \times 400$.
Dividing both sides by $R$:
$0.3 \times T = 160$.
$T = \frac{160}{0.3} = 533.33 \ K \approx 533 \ K$.

(D) According to Graham's Law of Diffusion,the rate of diffusion is proportional to the distance traveled in the same time interval.
$\frac{r_A}{r_B} = \frac{d_A}{d_B} = \sqrt{\frac{M_B}{M_A}}$
Given: $d_A = 40 \ m$,$d_B = 100 - 40 = 60 \ m$,$M_B = 18$.
Substituting the values:
$\frac{40}{60} = \sqrt{\frac{18}{M_A}}$
$\frac{2}{3} = \sqrt{\frac{18}{M_A}}$
Squaring both sides:
$\frac{4}{9} = \frac{18}{M_A}$
$M_A = \frac{18 \times 9}{4} = 40.5 \ g/mol$.
Thus,the molecular weight of $A$ is $40.5 \ g/mol$.

(C) According to the kinetic molecular theory of gases:
$a$) The actual volume of the molecules is negligible in comparison to the empty space between them. This means the particle size is considered a point mass.
$b$) Collisions of gas molecules are perfectly elastic,not inelastic. Thus,statement $b$ is incorrect.
$c$) At any particular time,different particles in the gas have different speeds and different kinetic energies. Only the average kinetic energy is constant at a given temperature. Thus,statement $c$ is incorrect.
$d$) Pressure is exerted by the gas as a result of the collision of the particles with the walls of the container. This is a correct statement.
Therefore,statements $a$ and $d$ are correct.

(D) The correct answer is $(i)$,$(ii)$,and $(iii)$.
According to the kinetic molecular theory of gases:
$(i)$ The actual volume occupied by gas molecules is negligible compared to the total volume of the gas.
$(ii)$ The collisions between gas molecules and with the walls of the container are perfectly elastic,meaning there is no loss of kinetic energy.
$(iii)$ The particles are considered point masses,and their size is extremely small compared to the average distance between them.

(C) The explanation of the given statements is as follows:
$(a)$ Gas particles are considered as point masses or rigid particles.
$(b)$ Kinetic energy of gas molecules is directly proportional to absolute temperature. $\text{K.E.} \propto T$. Therefore,$\text{K.E.}$ increases with increasing temperature.
$(c)$ Rigid elastic gas particles undergo collisions and follow the law of conservation of energy. Therefore,the total energy of molecules before and after the collisions remains equal.
$(d)$ The distribution of molecular speed of a gas remains constant at a particular temperature.
Hence,option $(C)$ is incorrect.

(D) The kinetic energy of $n$ moles of an ideal gas is given by the formula: $KE = \frac{3}{2} nRT$
Given values are:
$n = 1 \ mol$
$T = 27 + 273 = 300 \ K$
$R = 8.314 \ J \ mol^{-1} \ K^{-1}$
Substituting these values into the formula:
$KE = \frac{3}{2} \times 1 \times 8.314 \times 300$
$KE = 1.5 \times 8.314 \times 300$
$KE = 3741.3 \ J$
Rounding to the nearest integer,we get $3741 \ J$.

(C) The average kinetic energy of gas molecules depends only on the absolute temperature $(T)$ and is independent of the nature of the gas.
The formula for average kinetic energy is given by $\text{Average Kinetic Energy} = \frac{3}{2} KT$,where $K$ is the Boltzmann constant.
Since both hydrogen and helium are at the same temperature $(298 \ K)$,their average kinetic energies will be equal.

(A) According to the Maxwell-Boltzmann distribution law,the $x$-axis represents the speed of the molecules and the $y$-axis represents the fraction or number of molecules.
At lower temperatures,the molecules have lower average kinetic energy,so the peak of the distribution is shifted towards lower speeds and the curve is narrower and taller.
As the temperature increases,the average kinetic energy of the molecules increases,causing the peak of the distribution to shift towards higher speeds,and the curve becomes broader and flatter.
Therefore,for temperatures $T_1 < T_2 < T_3$,the correct plot shows the peak shifting to the right as temperature increases,which corresponds to the graph in option $A$.

(B) Number of moles of $N_2$ $(n)$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{280 \ g}{28 \ g \ mol^{-1}} = 10 \ mol$.
Temperature $(T)$ = $27^{\circ} C = (27 + 273) \ K = 300 \ K$.
The formula for kinetic energy of an ideal gas is $K.E. = \frac{3}{2} nRT$.
Substituting the values: $K.E. = \frac{3}{2} \times 10 \ mol \times 8.314 \ J \ mol^{-1} \ K^{-1} \times 300 \ K$.
$K.E. = 1.5 \times 10 \times 8.314 \times 300 = 37413 \ J$.
Converting to $kJ$: $37413 \ J = 37.413 \ kJ \approx 37.4 \ kJ$.

(A) The average kinetic energy of an ideal gas is given by the formula $KE = \frac{3}{2} RT$.
Since the kinetic energy depends only on the temperature $T$ and the gas constant $R$,it is independent of the nature or molar mass of the gas.
Therefore,at the same temperature $T$,the kinetic energy of $CH_4$ and $O_2$ will be equal.
Thus,$KE_{(CH_4)} = KE_{(O_2)} = X$.

(D) The kinetic energy $(KE)$ of $n$ moles of an ideal gas is given by $KE = n \times \frac{3}{2} RT$.
For $4$ $g$ of $H_2$,the number of moles $n_{H_2} = \frac{4 \ g}{2 \ g/mol} = 2$ $mol$.
So,$KE_{H_2} = 2 \times \frac{3}{2} RT = 3RT$.
For $8$ $g$ of $O_2$,the number of moles $n_{O_2} = \frac{8 \ g}{32 \ g/mol} = 0.25$ $mol$ or $\frac{1}{4}$ $mol$.
So,$KE_{O_2} = \frac{1}{4} \times \frac{3}{2} RT = \frac{3}{8} RT$.
The ratio $KE_{H_2} : KE_{O_2} = 3RT : \frac{3}{8} RT = 1 : \frac{1}{8} = 8 : 1$.

(B) The average kinetic energy per molecule of an ideal gas is given by the formula: $KE_{avg} = \frac{3}{2} k T$.
Here,$k$ is the Boltzmann constant $(k = \frac{R}{N_A})$,$T$ is the temperature in Kelvin,and $N_A$ is Avogadro's number.
Given: $T = 27^{\circ} C = 27 + 273 = 300 \ K$.
Substituting the values:
$KE_{avg} = \frac{3}{2} \times \left( \frac{8.314 \ J \ K^{-1} \ mol^{-1}}{6.022 \times 10^{23} \ mol^{-1}} \right) \times 300 \ K$.
$KE_{avg} = 1.5 \times 1.38 \times 10^{-23} \ J \ K^{-1} \times 300 \ K$.
$KE_{avg} \approx 6.21 \times 10^{-21} \ J \ \text{molecule}^{-1}$.

(A) The kinetic energy $(KE)$ of $n$ moles of an ideal gas is given by $KE = \frac{3}{2} nRT$.
For $16 \ g$ of methane ($CH_4$,molar mass = $16 \ g/mol$),$n = \frac{16}{16} = 1 \ mol$.
For $32 \ g$ of oxygen ($O_2$,molar mass = $32 \ g/mol$),$n = \frac{32}{32} = 1 \ mol$.
Since both samples contain $1 \ mol$ of gas at the same temperature $(300 \ K)$,their kinetic energies are equal. Thus,$(A)$ is true.
The kinetic energy of $1 \ mol$ of any ideal gas is $\frac{3}{2} RT$,which depends only on temperature. Thus,$(R)$ is true and correctly explains $(A)$.