Kinetic molecular theory of gases and Molecular collisions Questions in English

(A) . In gases,molecular attraction is very weak and intermolecular spaces are large,hence the kinetic energy of gas molecules is the highest compared to solids and liquids.

(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ $(r \propto \frac{1}{\sqrt{M}})$.
For $NH_3$ $(M = 17 \ g/mol)$ and $HCl$ $(M = 36.5 \ g/mol)$,the rate of diffusion of $NH_3$ is higher than that of $HCl$ because $M_{NH_3} < M_{HCl}$.
Since $NH_3$ diffuses faster,it travels a longer distance in the same amount of time.
Therefore,the white $NH_4Cl$ ring will be formed closer to the $HCl$ bottle.

(D) For an ideal gas,the internal energy $(U)$ is defined as the sum of the kinetic energies of its molecules.
Since ideal gas molecules are considered point masses with no intermolecular forces,the internal energy depends solely on the translational kinetic energy.
According to the kinetic theory of gases,the average kinetic energy per mole is given by $U = \frac{3}{2} RT$ for a monoatomic ideal gas.
Since $R$ is the universal gas constant,it is clear that $U \propto T$.
Therefore,the internal energy of an ideal gas is a function of temperature only.

(C) According to Graham's law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$ of the gas,i.e.,$r \propto \frac{1}{\sqrt{M}}$.
Since the pressure is the same,the gas with the lowest molar mass will diffuse or fill the tube the fastest.
The molar masses are: $N_2 = 28 \ g/mol$,$O_2 = 32 \ g/mol$,$H_2 = 2 \ g/mol$,and $Ne = 20 \ g/mol$.
Since $H_2$ has the lowest molar mass,it will be filled first.

(B) The separation of a mixture of two gases based on their different rates of diffusion is governed by $Graham's \ law \ of \ diffusion$. According to this law,the rate of diffusion of a gas is inversely proportional to the square root of its molar mass $(Rate \ \propto \ \frac{1}{\sqrt{M}})$. By allowing the gas mixture to diffuse through a porous membrane,the lighter gas diffuses faster than the heavier gas,allowing for their separation.

(C) The postulates of the Kinetic Molecular Theory of Gases include:
$1)$ The molecules in a gas are small and very far apart. Most of the volume occupied by a gas is empty space.
$2)$ Gas molecules are in constant random motion.
$3)$ Molecules collide with each other and with the walls of the container.
$4)$ Collisions are perfectly elastic,meaning there is no loss of kinetic energy.
$5)$ The molecules exert no attractive or repulsive forces on one another except during collisions.
$6)$ The effect of gravity on the motion of gas molecules is considered negligible.
Therefore,option $C$ is a recognized postulate of the kinetic theory.

(C) The kinetic molecular theory of gases makes several assumptions:
$1)$ The volume occupied by the gas molecules is negligible compared to the total volume of the gas.
$2)$ There are no intermolecular forces of attraction or repulsion between the gas molecules.
$3)$ Gas molecules are in constant,random,and straight-line motion.
$4)$ Collisions between molecules and with the walls of the container are perfectly elastic,meaning there is no loss of kinetic energy.
$5)$ The average kinetic energy of the molecules is directly proportional to the absolute temperature.
Therefore,according to the kinetic theory of gases,there are no intermolecular attractions.

(D) The average kinetic energy of an ideal gas is given by the formula $KE = \frac{3}{2}RT$,where $R$ is the universal gas constant and $T$ is the absolute temperature.
Since $R$ is a constant,the kinetic energy depends only on the absolute temperature $(T)$ of the gas.
It does not depend on the molecular mass,atomic mass,or equivalent mass of the gas.
Therefore,the correct option is $(D)$.

(B) . For one mole of an ideal gas,the average kinetic energy is given by $K.E. = \frac{3}{2}RT$.
This equation shows that the kinetic energy depends only on the absolute temperature $T$.

(A) According to the kinetic theory of gases,the average kinetic energy of a gas molecule is given by $\frac{3}{2}kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature.
For one mole of gas,the total kinetic energy is obtained by multiplying the energy per molecule by the Avogadro number $(N_A)$:
$E = N_A \times \frac{3}{2}kT = \frac{3}{2}(N_A k)T$.
Since $N_A k = R$ (the universal gas constant),the energy per mole is $\frac{3}{2}RT$,which is equal to $1.5 \, RT$.

(A) For an ideal gas,the kinetic energy $(KE)$ is given by $KE = \frac{3}{2}PV$.
Since internal energy $(E)$ per unit volume is defined as $E = \frac{KE}{V}$,we have $E = \frac{3}{2}P$.
Rearranging this for pressure $(P)$,we get $P = \frac{2}{3}E$.

(C) The translational kinetic energy $(K.E.)$ of an ideal gas is given by the formula: $K.E. = \frac{3}{2}RT$.
This expression shows that the translational kinetic energy of an ideal gas depends only on the absolute temperature $(T)$ of the gas.

(B) The average kinetic energy $(KE)$ of an ideal gas is given by the formula $KE = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
Since the average kinetic energy depends only on the temperature $(T)$ and not on the mass of the gas particles,both helium $(He)$ and hydrogen $(H_2)$ will have the same average kinetic energy at the same temperature of $298 \ K$.
Therefore,the correct option is $B$.

(B) The translational kinetic energy for one mole of an ideal gas is given by $K.E. = \frac{3}{2} RT$.
At absolute zero $(T = 0 \ K)$,the kinetic energy becomes zero.
Since kinetic energy is directly related to the motion of particles,at $0 \ K$,the molecular motion also ceases and becomes zero.

(A) The average kinetic energy $(K.E.)$ of an ideal gas per mole is given by the formula: $K.E. = \frac{3}{2} RT$.
Given that the gas constant $R \approx 2 \ \text{cal} \ K^{-1} \ mol^{-1}$.
Substituting the value of $R$: $K.E. = \frac{3}{2} \times 2 \times T = 3T$.
Therefore,the average $K.E.$ is approximately $3$ times the absolute temperature $(T)$.

(D) According to the kinetic theory of gases,the mean translational kinetic energy of gas molecules is given by the expression $E_k = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
This shows that the mean translational kinetic energy is directly proportional to the absolute temperature $(E_k \propto T)$.
Therefore,option $(d)$ is the correct statement.

(A) The average molar kinetic energy of an ideal gas is given by the expression $\overline{K.E.} = \frac{3}{2}RT$.
Since the average molar kinetic energy depends only on the temperature $(T)$ and the universal gas constant $(R)$,it is independent of the nature of the gas or its molecular mass.
Therefore,at the same temperature,the average molar kinetic energy of $CO$ and $N_2$ molecules will be equal.
Thus,$\overline{KE}_{CO} = \overline{KE}_{N_2}$.

(A) The mean free path $(\lambda)$ is given by the formula $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$,where $P$ is the pressure.
As the pressure $(P)$ increases,the volume decreases and the density of gas molecules increases.
This causes the molecules to come closer to each other,resulting in more frequent collisions.
Therefore,the mean free path decreases as pressure increases.

(A) Absolute zero is the theoretical temperature of $0 \ K$ (or $-273.15 \ ^{\circ}C$) at which the enthalpy and entropy of a cooled ideal gas reach their minimum value,and all classical molecular motion ceases.

(C) According to the kinetic molecular theory of gases,molecules of an ideal gas are in constant random motion and possess a distribution of speeds (Maxwell-Boltzmann distribution).
Therefore,it is not true that all molecules move with the same speed.

(B) The average kinetic energy per molecule of an ideal gas is given by the formula: $E_k = \frac{3}{2}kT$.
Here,$k$ is the Boltzmann constant $(1.38 \times 10^{-23} \text{ J/K})$ and $T$ is the temperature in Kelvin.
Given $T = 25\,^oC = 25 + 273 = 298 \text{ K}$.
Substituting the values: $E_k = \frac{3}{2} \times 1.38 \times 10^{-23} \times 298 \approx 6.17 \times 10^{-21} \text{ J}$.

(A) According to the kinetic theory of gases,the average kinetic energy of gas molecules is directly proportional to the absolute temperature $(T)$.
As the temperature increases,the average speed of the molecules increases.
When these faster-moving molecules collide with the walls of the container,they transfer more momentum per collision.
Since pressure is defined as the force exerted per unit area,and force is the rate of change of momentum,the increased momentum transfer results in an increase in the pressure exerted by the gas on the container walls.
Therefore,the correct option is $(A)$.

(D) The correct answer is $D$. An ideal gas cannot be liquefied because there are no intermolecular forces of attraction between the molecules of an ideal gas. Liquefaction requires the presence of attractive forces to bring molecules closer together to form a liquid phase.

(D) The correct answer is $(D)$.
According to the kinetic theory of gases,an ideal gas is defined by the absence of inter-molecular forces of attraction.
Liquefaction of a gas requires the presence of inter-molecular forces to bring molecules together to form a liquid phase.
Since an ideal gas lacks these forces,it cannot be liquefied at any value of pressure $(P)$ and temperature $(T)$.

(A) The average kinetic energy $(K.E.)$ of an ideal gas is directly proportional to its absolute temperature $(T)$ in Kelvin: $K.E. \propto T$.
Given temperatures:
$T_1 = 20 + 273 = 293 \ K$
$T_2 = 40 + 273 = 313 \ K$
The factor by which the kinetic energy changes is the ratio of the final kinetic energy to the initial kinetic energy:
$\frac{K.E._2}{K.E._1} = \frac{T_2}{T_1} = \frac{313}{293}$.

(B) At $STP$,the molar volume of an ideal gas is $22.4 \ L$.
However,the question asks for the volume occupied by the molecules themselves (the actual volume of the particles).
The radius of a water molecule is approximately $1.5 \ \mathring{A} = 1.5 \times 10^{-10} \ m $ .
The volume of one water molecule is $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.14 \times (1.5 \times 10^{-10} \ m)^3 \approx 1.41 \times 10^{-29} \ m^3$.
For $1$ mole ($6.022 \times 10^{23}$ molecules),the total volume is $6.022 \times 10^{23} \times 1.41 \times 10^{-29} \ m^3 \approx 8.5 \times 10^{-6} \ m^3 = 8.5 \ mL$.
Since $22.4 \ L = 22400 \ mL$,the ratio is $\frac{8.5}{22400} \approx 0.00038$,which is approximately $0.038\%$.
This is significantly less than $1\%$ of $22.4 \ L$.

(D) The kinetic energy $(K.E.)$ of an ideal gas per mole is given by the formula: $K.E. = \frac{3}{2} RT$.
Here,$R$ is the universal gas constant and $T$ is the absolute temperature.
Since the temperature $(T)$ is the same for all gases,the kinetic energy per mole depends only on the temperature.
Therefore,all gases at the same temperature will have the same kinetic energy per mole,regardless of their molar mass.

(B) The total kinetic energy of an ideal gas is given by the formula $KE = 3/2 \ nRT$.
For $1 \ mole$ of gas,$n = 1$.
Therefore,the kinetic energy is $3/2 \ RT$.

(D) According to the Kinetic Molecular Theory of Gases,the pressure $P$ is given by the formula $P = \frac{1}{3} \rho u_{rms}^2$,where $\rho$ is the density and $u_{rms}$ is the root mean square velocity. This implies $P \propto u_{rms}^2$ (mean square velocity).
Furthermore,the average translational kinetic energy $E_k$ of gas molecules is given by $E_k = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
Thus,$E_k \propto T$.
Therefore,the statement that the average translational kinetic energy is directly proportional to the absolute temperature is correct.

(C) The total kinetic energy $(KE)$ of an ideal gas is given by the formula: $KE = \frac{3}{2} nRT$.
Given that the total kinetic energy of $Ar$ is equal to the total kinetic energy of $He$:
$\frac{3}{2} n_{Ar} R T_{Ar} = \frac{3}{2} n_{He} R T_{He}$.
Canceling the common terms $\frac{3}{2} R$ from both sides,we get:
$n_{Ar} \times T_{Ar} = n_{He} \times T_{He}$.
Substituting the given values:
$0.40 \ mol \times 400 \ K = 0.30 \ mol \times T_{He}$.
$160 = 0.30 \times T_{He}$.
$T_{He} = \frac{160}{0.30} = \frac{1600}{3} \approx 533.33 \ K$.
Thus,the temperature is approximately $533 \ K$.

(C) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ of a gas is inversely proportional to the square root of its molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
Comparing the molar masses: $M(H_2) = 2 \ g/mol$,$M(He) = 4 \ g/mol$,$M(N_2) = 28 \ g/mol$,and $M(O_2) = 32 \ g/mol$.
Since $H_2$ has the lowest molar mass,it will have the highest rate of diffusion.
Therefore,$H_2$ will leak out first.

(C) The kinetic gas equation is given by the expression $PV = \frac{1}{3}mNu^2$,where $m$ is the mass of one molecule,$N$ is the total number of molecules,and $u$ is the root mean square velocity.

(C) According to the kinetic molecular theory of gases,the following postulates are considered:
$1$. The volume occupied by the gas molecules is negligible compared to the total volume of the gas.
$2$. There are no intermolecular forces of attraction between the gas molecules.
$3$. Collisions between molecules are perfectly elastic,meaning the total kinetic energy remains constant.
Therefore,option $C$ is the correct statement.

(A) According to the kinetic theory of gases,the total kinetic energy $E$ of an ideal gas is given by $E = \frac{3}{2}PV$.
Rearranging this equation for pressure $P$,we get $P = \frac{2}{3} \times \frac{E}{V}$.
For unit volume $(V = 1)$,the relationship becomes $P = \frac{2}{3}E$.

(D) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ of a gas is inversely proportional to the square root of its molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
For an equimolar mixture,the ratio of the rates of diffusion of $H_2$ and $O_2$ is given by: $\frac{r_{H_2}}{r_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}}$.
The molar mass of $H_2$ $(M_{H_2})$ is $2 \ g/mol$ and the molar mass of $O_2$ $(M_{O_2})$ is $32 \ g/mol$.
Substituting these values: $\frac{r_{H_2}}{r_{O_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
Therefore,the ratio is $4 : 1$.

(A) The average kinetic energy of gas molecules is given by the formula $KE_{avg} = \frac{3}{2} RT$.
Since the temperature $(T)$ is the same for both gases at $STP$,the average kinetic energy of $H_2$ and $He$ molecules will be identical,as it depends only on the absolute temperature.

(A) The collision frequency $(Z)$ is given by the formula: $Z = \sqrt{2} \pi d^2 \bar{v} \frac{N}{V}$.
Here,$\bar{v}$ is the average speed of gas molecules,which is directly proportional to the square root of temperature $(\bar{v} \propto \sqrt{T})$.
According to the ideal gas law,$PV = nRT$,so $\frac{N}{V} = \frac{P}{kT}$.
Substituting these into the expression,$Z \propto \frac{\sqrt{T}}{T} = \frac{1}{\sqrt{T}}$.
However,for a fixed mass of gas at constant pressure,$V \propto T$. The number density $\frac{N}{V}$ decreases as $T$ increases.
Specifically,$Z \propto \frac{\sqrt{T}}{V} \propto \frac{\sqrt{T}}{T} = \frac{1}{\sqrt{T}}$.
Wait,let us re-evaluate: For a fixed amount of gas at constant pressure,$V \propto T$. The collision frequency $Z$ is proportional to $\frac{\bar{v}}{V}$. Since $\bar{v} \propto T^{1/2}$ and $V \propto T$,then $Z \propto T^{1/2} / T = T^{-1/2}$.
Thus,as temperature increases,the collision frequency decreases.

(A) According to the kinetic molecular theory of gases,the pressure exerted by a gas is due to the collisions of gas molecules with the walls of the container.
As the temperature increases,the average kinetic energy of the gas molecules increases,which leads to an increase in the average molecular speed.
Consequently,the molecules strike the walls of the container more frequently and with greater force,resulting in an increase in the pressure of the gas at a constant volume.

(A) The average kinetic energy $(K.E.)$ of a gas is directly proportional to its absolute temperature $(T)$ in Kelvin: $K.E. \propto T$.
Given,$T_1 = 20 + 273 = 293 \, K$ and $T_2 = 40 + 273 = 313 \, K$.
The ratio of the average kinetic energies is:
$\frac{K.E._2}{K.E._1} = \frac{T_2}{T_1} = \frac{313}{293}$.

(A) The kinetic energy of an ideal gas depends only on its absolute temperature $(T)$.
Since the temperature remains constant during the expansion,the average kinetic energy of the gas molecules,given by the formula $KE = \frac{3}{2}kT$,remains constant.
Therefore,the correct option is $A$.

(D) Diffusion is the movement of particles from a region of higher concentration to a region of lower concentration. In gases,the intermolecular forces are negligible,allowing particles to move freely and rapidly. In liquids,the molecules are held together by stronger intermolecular forces compared to gases,which restricts their movement and results in a slower rate of diffusion.

(A) According to the kinetic molecular theory of gases,gas molecules are in constant,random motion. Between two successive collisions,a gas molecule travels in a straight line path at a constant velocity. This distance traveled between two successive collisions is known as the mean free path.

(B) According to the kinetic molecular theory of gases,the average kinetic energy of gas molecules is directly proportional to the absolute temperature $(T)$ of the gas.
Mathematically,$KE_{avg} = \frac{3}{2} RT$.
Since the total kinetic energy is the sum of the kinetic energies of all individual molecules,it also depends solely on the temperature of the gas.

(C) The kinetic energy $(K.E.)$ of an ideal gas depends only on its absolute temperature $(T)$.
According to the kinetic theory of gases,$K.E. = \frac{3}{2} nRT$.
Since the temperature $(T)$ is constant during the expansion of the gas,the average kinetic energy of the gas molecules remains constant.

(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ of the gas: $r \propto \frac{1}{\sqrt{M}}$.
For two gases to diffuse at the same rate,they must have the same molar mass.
Calculate the molar masses of the given compounds:
$CO_2 = 12 + (2 \times 16) = 44 \ g/mol$.
$N_2O$ (Nitrous oxide) $= (2 \times 14) + 16 = 44 \ g/mol$.
Since both $CO_2$ and $N_2O$ have a molar mass of $44 \ g/mol$,they will diffuse at the same rate.

(A) According to the kinetic molecular theory of gases,the average kinetic energy of $1 \ mol$ of an ideal gas is given by the formula $KE = \frac{3}{2} RT$,where $R$ is the universal gas constant and $T$ is the absolute temperature in Kelvin.

(C) The average kinetic energy $(KE_{avg})$ of an ideal gas is directly proportional to its absolute temperature $(T)$ in Kelvin.
$KE_{avg} = \frac{3}{2} k_B T$
Initial temperature $T_1 = 20 + 273 = 293\,K$.
Final temperature $T_2 = 40 + 273 = 313\,K$.
The ratio of average kinetic energy is $\frac{KE_2}{KE_1} = \frac{T_2}{T_1} = \frac{313}{293}$.
Therefore,the factor by which the average kinetic energy changes is $313/293$.

(D) The kinetic energy per mole 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 of the gas.
Therefore,at the same temperature,all gases will have the same kinetic energy per mole.

(A) The average kinetic energy of an ideal gas is given by the formula $K.E. = \frac{3}{2}RT$.
Since the average kinetic energy depends only on the absolute temperature $(T)$ and not on the nature of the gas,it remains the same for all gases at the same temperature.
Therefore,the average kinetic energy of $N_2$ at $27\,^oC$ will also be $E$.

(A) The mean free path $(\lambda)$ is inversely proportional to the collision cross-section $(\sigma)$,which depends on the size of the gas molecules: $\lambda \propto \frac{1}{d^2}$.
Since $H_2$ has the smallest molecular size among the given options,it will have the largest mean free path.