Pressure and Energy Questions in English

(A) According to the kinetic theory of gases,the mean square velocity $\overline{v^2}$ of gas molecules is directly proportional to the absolute temperature $T$ of the gas.
Mathematically,$\frac{1}{2} m \overline{v^2} = \frac{3}{2} k_B T$,which implies $\overline{v^2} \propto T$.
Since the mean square velocity is a direct measure of the average kinetic energy of the molecules,which in turn depends on the absolute temperature,option $(a)$ is the correct way to relate or calculate the temperature.

(C) Absolute zero is defined as $0 \,K$ or $-273.15 \,^\circ C$.
According to the kinetic theory of gases, the root mean square velocity $(v_{rms})$ of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
At $T = 0 \,K$, the velocity $v_{rms}$ becomes zero.
This implies that the random thermal motion of molecules ceases at absolute zero.
Therefore, option $(C)$ is correct.

(A) Molecular motion is directly related to the average kinetic energy of the molecules in a system.
According to the kinetic theory of gases,the absolute temperature $T$ of a gas is proportional to the average translational kinetic energy of its molecules,given by the relation $K_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant.
Therefore,molecular motion manifests itself as temperature.
Internal energy is the sum of both the kinetic energy and the potential energy of the molecules,so it is not solely a measure of molecular motion.

(B) According to the kinetic theory of gases,the average kinetic energy of gas molecules is directly proportional to the absolute temperature $(KE_{avg} \propto T)$.
As the temperature increases,the average velocity $(v_{rms} \propto \sqrt{T})$ of the gas molecules increases.
Since the molecules move faster,they strike the walls of the container more frequently per unit time.
Therefore,the number of collisions per unit time increases.

(A) The average kinetic energy due to translational motion of molecules of an ideal gas at absolute temperature $T$ is given by the formula:
$K.E. = \frac{3}{2} k_B T$
where $k_B$ is the Boltzmann constant.
Since $\frac{3}{2}$ and $k_B$ are constants,the kinetic energy is directly proportional to the absolute temperature $T$.
Therefore,$K.E. \propto T$.
Given the options provided,the expression $kT$ represents the dependency on temperature $T$ where $k$ is the Boltzmann constant.

(C) According to the kinetic theory of gases,gas molecules are in constant random motion.
When these molecules collide with the walls of the container,they undergo a change in velocity,which results in a change in momentum.
According to Newton's second law of motion,the rate of change of momentum is equal to the force exerted.
Since force is exerted on the walls over a specific area,this results in the pressure exerted by the gas.
Therefore,the correct answer is $C$.

(B) The pressure exerted by an ideal gas is given by the formula $P = \frac{1}{3} \rho v_{rms}^2$,where $\rho = \frac{M}{V}$ is the density and $v_{rms}$ is the root-mean-square speed.
Since $\rho = \frac{N \cdot m}{V}$ (where $N$ is the number of molecules and $m$ is the mass of one molecule),we have $P = \frac{1}{3} \frac{N \cdot m}{V} v_{rms}^2$.
This implies $P \propto m \cdot v_{rms}^2$.
Let the initial mass be $m_1 = m$ and initial speed be $v_1 = v$. The initial pressure is $P_0 \propto m \cdot v^2$.
After the changes,the new mass is $m_2 = \frac{m}{2}$ and the new speed is $v_2 = 2v$.
The new pressure $P$ is proportional to $m_2 \cdot v_2^2 = (\frac{m}{2}) \cdot (2v)^2 = (\frac{m}{2}) \cdot (4v^2) = 2mv^2$.
Comparing the new pressure $P$ with the initial pressure $P_0$,we get $P = 2P_0$.

(D) The pressure $P$ exerted by an ideal gas is given by the kinetic theory of gases as $P = \frac{1}{3} \rho v_{rms}^2$,where $\rho$ is the density of the gas and $v_{rms}$ is the root mean square velocity.
We know that the average kinetic energy per unit volume $E$ is given by $E = \frac{1}{2} \rho v_{rms}^2$.
From this,we can write $\rho v_{rms}^2 = 2E$.
Substituting this into the pressure equation: $P = \frac{1}{3} (2E) = \frac{2}{3}E$.
Therefore,the correct relation is $P = \frac{2}{3}E$.

(D) The total average kinetic energy of translatory motion for an ideal gas is given by the formula $E = \frac{3}{2}PV$,where $P$ is the pressure and $V$ is the volume.
Given: $E = 1.5 \times 10^5 \text{ J}$ and $V = 20 \text{ litres} = 20 \times 10^{-3} \text{ m}^3$.
Rearranging the formula for pressure: $P = \frac{2E}{3V}$.
Substituting the values: $P = \frac{2 \times (1.5 \times 10^5)}{3 \times (20 \times 10^{-3})}$.
$P = \frac{3 \times 10^5}{60 \times 10^{-3}} = \frac{3 \times 10^5}{6 \times 10^{-2}} = 0.5 \times 10^7 = 5 \times 10^6 \text{ N/m}^2$.

(A) The pressure exerted by a gas is a scalar quantity that depends on the random motion of molecules relative to the center of mass of the gas.
The root mean square velocity $\bar{c}$ is defined relative to the center of mass of the gas.
When the entire system moves with a velocity $v$,the random motion of the molecules (which causes pressure) remains unchanged because the relative velocities of the molecules with respect to the container remain the same.
The pressure $P$ is given by the kinetic theory formula: $P = \frac{1}{3}\rho\bar{c}^2$.
Since the pressure is independent of the bulk motion of the gas,the pressure remains $\frac{1}{3}\rho\bar{c}^2$.

(A) When a molecule of mass $m$ collides elastically with a wall of a vessel with velocity $V$,it rebounds with the same speed in the opposite direction.
Initial momentum of the molecule,$p_i = mV$.
Final momentum of the molecule,$p_f = -mV$.
The change in linear momentum,$\Delta p = p_f - p_i = -mV - (mV) = -2mV$.
Since the question asks for the magnitude of the change in linear momentum,the change is $2mV$.

(D) According to the kinetic theory of gases,the pressure $P$ exerted by an ideal gas is given by $P = \frac{2}{3} \times \frac{E}{V}$,where $E$ is the total internal kinetic energy and $V$ is the volume of the gas.
Rearranging this equation,we get $PV = \frac{2}{3} E$.
Therefore,the correct relationship is $PV = \frac{2}{3} E$.

(C) The molar mass of oxygen $(O_2)$ is $32\, g/mol$. Therefore,$32\, g$ of oxygen corresponds to $n = 1\, mole$.
The temperature in Kelvin is $T = -73 + 273 = 200\, K$.
The mean kinetic energy $(E)$ of an ideal gas is given by the formula $E = \frac{3}{2}nRT$.
Substituting the values: $E = \frac{3}{2} \times 1 \times 8.3 \times 200$.
$E = 3 \times 8.3 \times 100 = 2490\, J$.

(C) The average kinetic energy of a gas molecule is given by the formula $E = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
The total kinetic energy of $N$ molecules is given by $K.E. = N \times \frac{3}{2} k_B T$.
For hydrogen $(H_2)$: $E_{H_2} = N_{H_2} \times \frac{3}{2} k_B T$.
For oxygen $(O_2)$: $E_{O_2} = N_{O_2} \times \frac{3}{2} k_B T$.
Given that the number of molecules of $H_2$ is double that of $O_2$,we have $N_{H_2} = 2 N_{O_2}$.
Therefore,the ratio of the total kinetic energy is:
$\frac{E_{H_2}}{E_{O_2}} = \frac{N_{H_2} \times \frac{3}{2} k_B T}{N_{O_2} \times \frac{3}{2} k_B T} = \frac{N_{H_2}}{N_{O_2}} = \frac{2 N_{O_2}}{N_{O_2}} = \frac{2}{1}$.
Thus,the ratio is $2:1$.

(D) The average kinetic energy $(E)$ of an ideal gas molecule is given by the formula: $E = \frac{3}{2}kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature in Kelvin.
Note that the average kinetic energy depends only on the temperature $T$ and is independent of the pressure and the nature of the gas.
Given:
$T_1 = -23^{\circ}C = 273 - 23 = 250 \, K$
$E_1 = 5 \times 10^{-14} \, erg$
$T_2 = 227^{\circ}C = 273 + 227 = 500 \, K$
Using the ratio:
$\frac{E_1}{E_2} = \frac{T_1}{T_2}$
$\frac{5 \times 10^{-14}}{E_2} = \frac{250}{500}$
$\frac{5 \times 10^{-14}}{E_2} = \frac{1}{2}$
$E_2 = 2 \times 5 \times 10^{-14} = 10 \times 10^{-14} \, erg$.

(D) The mean kinetic energy $(E)$ of a gas molecule is given by the formula $E = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
Since the mean kinetic energy depends only on the temperature $(T)$ and not on the nature or mass of the gas molecules,for any two gases at the same temperature,their mean kinetic energies will be equal.
Therefore,the ratio of the mean kinetic energy of hydrogen to that of oxygen is $1:1$.

(B) The mean kinetic energy $(E)$ of a gas molecule is given by the formula $E = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
Since $k_B$ is a constant,the mean kinetic energy is directly proportional to the absolute temperature $(E \propto T)$.
Given temperatures are $T_1 = 300 \, K$ and $T_2 = 450 \, K$.
The ratio of the mean kinetic energies is $\frac{E_1}{E_2} = \frac{T_1}{T_2}$.
Substituting the values,we get $\frac{E_1}{E_2} = \frac{300}{450} = \frac{2}{3}$.
Therefore,the ratio is $2:3$.

(B) According to the kinetic theory of gases,an ideal gas consists of molecules that are point masses and do not exert any intermolecular forces of attraction on each other.
Since there are no intermolecular forces,the potential energy $(PE)$ of the gas is zero.
Therefore,the total internal energy of an ideal gas is entirely due to the kinetic energy $(KE)$ of its molecules.
Thus,the total energy is equal to the kinetic energy.

(C) The average kinetic energy $(K_{avg})$ of a gas molecule is given by the formula: $K_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature of the gas.
Since $k_B$ is a universal constant,the average kinetic energy depends solely on the temperature $T$ of the gas.
Therefore,knowing the temperature of the gas is sufficient to determine the average kinetic energy of its molecules.

(A) For a monoatomic gas,the degrees of freedom $(f)$ is $3$.
According to the equipartition theorem,the average kinetic energy per molecule is $\frac{f}{2}kT = \frac{3}{2}kT$,where $k$ is the Boltzmann constant.
For one gram mole of gas,the number of molecules is equal to the Avogadro number $(N_A)$.
Therefore,the average kinetic energy per mole is $E = N_A \times \frac{3}{2}kT = \frac{3}{2}(N_A k)T$.
Since $R = N_A k$,the expression becomes $E = \frac{3}{2}RT$.

(C) The average kinetic energy $(K.E.)$ of an ideal gas atom is given by the formula: $K.E. = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature in Kelvin.
From this relation,it is clear that $K.E. \propto T$.
Given that the initial temperature $T_1 = 300\, K$ and the final temperature $T_2 = 600\, K$.
Since the temperature is doubled $(T_2 = 2 T_1)$,the average kinetic energy will also be doubled.

(B) According to the kinetic theory of gases,the average kinetic energy $(E)$ of a single molecule of an ideal gas is given by the formula: $E = \frac{3}{2}kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature of the gas.
From this relation,it is evident that the average kinetic energy is directly proportional to the absolute temperature of the gas $(E \propto T)$.
Therefore,the correct option is $B$.

(C) The average kinetic energy $(E)$ of gas molecules is directly proportional to the absolute temperature $(T)$ in Kelvin: $E \propto T$.
Given the initial temperature $T_1 = 20^{\circ}C = 20 + 273 = 293 \, K$.
We want the final kinetic energy $E_2$ to be twice the initial kinetic energy $E_1$,so $E_2 = 2E_1$.
Using the ratio: $\frac{E_1}{E_2} = \frac{T_1}{T_2}$.
Substituting the values: $\frac{E_1}{2E_1} = \frac{293}{T_2}$.
$T_2 = 2 \times 293 = 586 \, K$.
Converting back to Celsius: $T_2(^{\circ}C) = 586 - 273 = 313^{\circ}C$.

(A) Given data:
Mass of oxygen,$m = 20 \, g$
Molar mass of oxygen,$M = 32 \, g/mol$
Number of moles,$n = \frac{m}{M} = \frac{20}{32} = 0.625 \, mol$
Temperature,$T = 47^{\circ}C = 47 + 273 = 320 \, K$
Gas constant,$R = 8.3 \, J/mol \cdot K$
The kinetic energy of translation for an ideal gas is given by $E = \frac{3}{2} nRT$.
Substituting the values:
$E = \frac{3}{2} \times 0.625 \times 8.3 \times 320$
$E = 1.5 \times 0.625 \times 8.3 \times 320$
$E = 2490 \, J$
Thus,the kinetic energy of translation is $2490 \, J$.

(B) The translatory kinetic energy of $n$ moles of an ideal gas is given by $E = \frac{3}{2}nRT$.
For $1 \ gm$ of gas,the number of moles $n = \frac{1}{M}$,where $M$ is the molar mass of the gas.
Substituting this into the energy equation,we get the kinetic energy per gram as $E_{gm} = \frac{3}{2} \left( \frac{1}{M} \right) RT = \frac{3}{2} \frac{RT}{M}$.

(D) The kinetic energy of one mole (one gram molecule) of an ideal gas is given by the formula:
$K = \frac{3}{2} RT$
Given values are:
$R = 8.31\,J/mol\cdot K$
$T = 273\,K$ (Standard temperature at $STP$)
Substituting these values into the formula:
$K = \frac{3}{2} \times 8.31 \times 273$
$K = 1.5 \times 8.31 \times 273$
$K = 12.465 \times 273$
$K \approx 3402.945\,J$
Rounding to the appropriate significant figures,we get:
$K \approx 3.4 \times 10^3\,J$

(C) The average kinetic energy $(E)$ of a gas molecule is directly proportional to its absolute temperature $(T)$,given by the relation $E = \frac{3}{2} k_B T$.
Therefore,the ratio of kinetic energies at two different temperatures is given by $\frac{E_1}{E_2} = \frac{T_1}{T_2}$.
Given:
$T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$.
$E_2 = 2E_1$.
Substituting these values into the ratio formula:
$\frac{E_1}{2E_1} = \frac{300 \ K}{T_2}$
$\frac{1}{2} = \frac{300 \ K}{T_2}$
$T_2 = 600 \ K$.
Converting back to Celsius:
$T_2 = 600 - 273 = 327^{\circ}C$.

(A) The translational kinetic energy per molecule of an ideal gas is given by $K.E._{molecule} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
For one mole of gas,the number of molecules is equal to Avogadro's number,$N_A$.
Therefore,the total translational kinetic energy for one mole is $K.E._{molar} = N_A \times (\frac{3}{2} k_B T)$.
Since the universal gas constant $R = N_A k_B$,we substitute this into the equation.
Thus,$K.E._{molar} = \frac{3}{2} RT$.

(A) The average kinetic energy $(KE)$ of an ideal gas is directly proportional to its absolute temperature $(T)$: $KE \propto T$.
Given:
Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$.
Final temperature $T_2 = 327^{\circ}C = 327 + 273 = 600 \ K$.
Initial kinetic energy = $E_1$.
Using the proportionality relation:
$\frac{E_1}{E_2} = \frac{T_1}{T_2}$
Substituting the values:
$\frac{E_1}{E_2} = \frac{300 \ K}{600 \ K} = \frac{1}{2}$
Therefore,$E_2 = 2E_1$.

(C) The average kinetic energy $(E)$ of a gas molecule is directly proportional to its absolute temperature $(T)$: $E \propto T$.
Therefore,the ratio is given by: $\frac{E_1}{E_2} = \frac{T_1}{T_2}$.
Given:
$E_1 = 6.21 \times 10^{-21} \, J$
$T_1 = 27^oC = 27 + 273 = 300 \, K$
$T_2 = 227^oC = 227 + 273 = 500 \, K$
Substituting the values:
$\frac{6.21 \times 10^{-21}}{E_2} = \frac{300}{500} = \frac{3}{5}$.
Solving for $E_2$:
$E_2 = \frac{6.21 \times 10^{-21} \times 5}{3} = 2.07 \times 5 \times 10^{-21} \, J = 10.35 \times 10^{-21} \, J$.

(C) The average translational kinetic energy of an ideal gas molecule is given by the formula:
$E = \frac{3}{2} k_B T$
where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
From this formula,it is clear that the translational kinetic energy depends only on the temperature $T$ of the gas.
It is independent of the nature of the gas (i.e.,independent of molar mass).
Since the temperature $T$ is the same for both ${O_2}$ and ${N_2}$ molecules,their average translational kinetic energies will be equal.
Therefore,$E_{N_2} = E_{O_2} = 0.048 \; eV$.

(A) The average kinetic energy of a gas molecule is given by the formula $K.E. = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature in Kelvin.
At $T = 0 \; K$,the kinetic energy becomes $K.E. = \frac{3}{2} k_B (0) = 0$.
Since the root mean square velocity $v_{rms} = \sqrt{\frac{3RT}{M}}$,at $T = 0 \; K$,$v_{rms} = 0$.
Therefore,the kinetic energy of the gas molecules becomes zero at absolute zero temperature.

(D) The average kinetic energy $(K.E.)$ of an ideal gas is directly proportional to its absolute temperature $(T)$ in Kelvin.
$K.E. \propto T$
Therefore,the ratio of kinetic energies is given by:
$\frac{E_1}{E_2} = \frac{T_1}{T_2}$
Convert the temperatures from Celsius to Kelvin:
$T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$
$T_2 = 927^{\circ}C = 927 + 273 = 1200 \ K$
Substitute the values into the ratio:
$\frac{E_1}{E_2} = \frac{300}{1200} = \frac{1}{4}$
This implies:
$E_2 = 4E_1$
Thus,the kinetic energy becomes four times the initial value.

(D) The mean kinetic energy of a gas molecule depends only on the absolute temperature $T$ and is given by the formula $KE_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant.
Since both gases are at the same temperature $T$,the mean kinetic energy per molecule for both hydrogen and oxygen will be identical.
Therefore,the ratio of the mean kinetic energies is $\frac{KE_{H_2}}{KE_{O_2}} = \frac{\frac{3}{2} k_B T}{\frac{3}{2} k_B T} = 1:1$.
The volume and mass of the gases do not affect the mean kinetic energy per molecule.

(B) The number of translational degrees of freedom for any gas molecule (monoatomic,diatomic,or polyatomic) is always $3$.
The translational kinetic energy $(KE_{trans})$ of a gas is given by the formula $KE_{trans} = \frac{3}{2}nRT$.
From the ideal gas equation,we know that $PV = nRT$.
Substituting this into the kinetic energy formula,we get $KE_{trans} = \frac{3}{2}PV$.
Since the volume $V$ and pressure $P$ are the same for both the monoatomic and diatomic gases,the total translational kinetic energy remains $\frac{3}{2}PV$ for the diatomic gas as well.

(A) The average kinetic energy $(KE)$ of gas molecules is directly proportional to the absolute temperature $(T)$ of the gas.
This is given by the relation: $KE = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant.
At absolute zero temperature,$T = 0 \ K$.
Substituting this value into the equation,we get $KE = \frac{3}{2} k_B (0) = 0$.
Therefore,at absolute zero temperature,the kinetic energy of the molecules becomes zero.

(A) The average kinetic energy $(E)$ of gas molecules is directly proportional to the absolute temperature $(T)$ in Kelvin: $E \propto T$.
Given the initial temperature $T_1 = -68^\circ C$. Converting this to Kelvin: $T_1 = 273 + (-68) = 205 \ K$.
We want the new kinetic energy $E_2$ to be twice the initial kinetic energy $E_1$,so $E_2 = 2E_1$.
Using the relation $\frac{E_2}{E_1} = \frac{T_2}{T_1}$,we get:
$2 = \frac{T_2}{205 \ K}$.
Solving for $T_2$: $T_2 = 2 \times 205 = 410 \ K$.
Converting the final temperature back to Celsius: $T_2(^\circ C) = 410 - 273 = 137^\circ C$.

(A) The average kinetic energy $E$ of a monoatomic gas molecule is given by the formula $E = \frac{3}{2}kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature.
Given temperature $T = 30^oC = 30 + 273 = 303\,K$.
Boltzmann constant $k = 1.38 \times 10^{-23}\,J/K$.
Substituting the values: $E = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 303 = 6.27 \times 10^{-21}\,J$.
To convert this energy into electron-volts $(eV)$,divide by $1.6 \times 10^{-19}\,J/eV$:
$E = \frac{6.27 \times 10^{-21}}{1.6 \times 10^{-19}} \approx 0.039\,eV$.
Since $0.039\,eV < 1\,eV$,the correct option is $A$.

(B) The average kinetic energy $(E)$ of a gas molecule is directly proportional to its absolute temperature $(T)$,given by the relation $E = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant.
Since $E \propto T$,the ratio of the average kinetic energies of the molecules of two gases at temperatures $T_1$ and $T_2$ is given by $\frac{E_1}{E_2} = \frac{T_1}{T_2}$.
Given $T_1 = 300 \ K$ and $T_2 = 350 \ K$.
Therefore,$\frac{E_1}{E_2} = \frac{300}{350} = \frac{6}{7}$.

(A) According to the Kinetic Theory of Gases,gas molecules are in constant random motion.
When these molecules collide with the walls of a closed container,they undergo a change in velocity (specifically,a change in the component of velocity perpendicular to the wall).
Since momentum is defined as $p = mv$,a change in velocity implies a change in momentum.
By Newton's Second Law,the rate of change of momentum is equal to the force exerted.
Therefore,the molecules exert a force on the walls by transferring momentum to them,which manifests as gas pressure.

(D) The mean kinetic energy $(E)$ of an ideal gas is directly proportional to its absolute temperature $(T)$,given by the relation $E \propto T$ or $\frac{E_1}{E_2} = \frac{T_1}{T_2}$.
Given:
$T_1 = 300\, K$
$E_1 = 100\, J$
$T_2 = 450\, K$
Substituting the values into the formula:
$\frac{100}{E_2} = \frac{300}{450}$
$E_2 = 100 \times \frac{450}{300}$
$E_2 = 100 \times 1.5 = 150\, J$.
Therefore,the mean energy at $450\, K$ is $150\, J$.

(D) The total momentum of the gas molecules is the vector sum of the individual momenta of all molecules.
$\vec{P}_{total} = \sum m_i \vec{v}_i = N \cdot m \cdot \vec{v}_{avg}$
In a container at rest,the gas molecules move in random directions.
The average velocity $\vec{v}_{avg}$ of the gas molecules at any temperature is always zero because for every molecule moving with velocity $\vec{v}$,there is a statistically equal probability of another molecule moving with velocity $-\vec{v}$.
Therefore,the total momentum $\vec{P}_{total} = M \vec{v}_{avg} = 0$.

(D) The average translational kinetic energy of a gas molecule is given by $E = \frac{3}{2}kT$,where $k$ is the Boltzmann constant $(1.38 \times 10^{-23}\, J/K)$ and $T$ is the absolute temperature.
The energy gained by an electron accelerated through a potential difference of $V = 1\, V$ is $E = eV$,where $e = 1.6 \times 10^{-19}\, C$.
Equating the two energies:
$\frac{3}{2}kT = eV$
Solving for $T$:
$T = \frac{2eV}{3k}$
Substituting the values:
$T = \frac{2 \times 1.6 \times 10^{-19}}{3 \times 1.38 \times 10^{-23}}$
$T = \frac{3.2 \times 10^{-19}}{4.14 \times 10^{-23}}$
$T \approx 0.7729 \times 10^4\, K = 7.7 \times 10^3\, K$.

(D) The kinetic energy $(E)$ of one mole of an ideal gas is given by the formula $E = \frac{3}{2}RT$.
At normal temperature and pressure $(NTP)$,the standard temperature is $T = 273 \text{ K}$.
Given the gas constant $R = 8.31 \text{ J/mol-K}$.
Substituting these values into the formula:
$E = \frac{3}{2} \times 8.31 \times 273$
$E = 1.5 \times 8.31 \times 273$
$E = 12.465 \times 273$
$E \approx 3402.945 \text{ J}$
Rounding to the appropriate significant figures,we get $E \approx 3.4 \times 10^3 \text{ J}$.

(C) The average translational kinetic energy of a gas molecule is given by the formula $E_{avg} = \frac{3}{2} k_B T$.
At $NTP$ (Normal Temperature and Pressure),the temperature $T = 273 \text{ K}$.
Given $k_B = 1.38 \times 10^{-23} \text{ J/K}$.
Substituting the values: $E_{avg} = \frac{3}{2} \times 1.38 \times 10^{-23} \times 273$.
$E_{avg} = 1.5 \times 1.38 \times 273 \times 10^{-23} \text{ J}$.
$E_{avg} = 565.11 \times 10^{-23} \text{ J} = 0.565 \times 10^{-20} \text{ J}$.
Rounding to the nearest option,we get $0.56 \times 10^{-20} \text{ J}$.

(C) The formula for $r.m.s.$ velocity is given by $V_{rms} = \sqrt{\frac{3P}{\rho}}$.
Rearranging the formula to solve for pressure $P$,we get $P = \frac{\rho V_{rms}^2}{3}$.
Substituting the given values: $\rho = 8.99 \times 10^{-2} \ kg/m^3$ and $V_{rms} = 1840 \ m/s$.
$P = \frac{8.99 \times 10^{-2} \times (1840)^2}{3}$.
$P = \frac{8.99 \times 10^{-2} \times 3385600}{3}$.
$P = \frac{304365.44}{3} \approx 101455.14 \ N/m^2$.
Rounding to the appropriate significant figures,$P \approx 1.01 \times 10^5 \ N/m^2$.

(A) The average kinetic energy of one mole of an ideal gas is given by the formula $E = \frac{3}{2}RT,$ where $R$ is the universal gas constant and $T$ is the absolute temperature in Kelvin.
Since $R$ is a constant,the kinetic energy $E$ is directly proportional to the temperature $T$ $(E \propto T)$.
Therefore,the ratio of kinetic energies at two different temperatures $T$ and $T'$ is given by $\frac{E'}{E} = \frac{T'}{T}.$
Given $T = 300 \ K$ and $T' = 400 \ K,$
$\frac{E'}{E} = \frac{400}{300} = \frac{4}{3} = 1.33.$

(A) Thermal neutrons are neutrons that are in thermal equilibrium with their surroundings,typically at room temperature $(T \approx 300 \text{ K})$.
The average kinetic energy $(E_{avg})$ of a particle in thermal equilibrium is given by the formula:
$E_{avg} = \frac{3}{2} K_B T$
Given:
$K_B = 8 \times 10^{-5} \text{ eV/K}$
$T = 300 \text{ K}$
Substituting the values:
$E_{avg} = \frac{3}{2} \times (8 \times 10^{-5} \text{ eV/K}) \times 300 \text{ K}$
$E_{avg} = 1.5 \times 2400 \times 10^{-5} \text{ eV}$
$E_{avg} = 3600 \times 10^{-5} \text{ eV}$
$E_{avg} = 0.036 \text{ eV}$
This value is of the order of $0.03 \text{ eV}$.
Therefore,the correct option is $A$.

(C) According to the kinetic theory of gases,gas molecules are in constant random motion. When these molecules collide with the walls of the container,they undergo a change in momentum. According to Newton's second law of motion,the rate of change of momentum is equal to the force exerted. Since the molecules exert a force on the walls over a specific area,this results in the pressure exerted by the gas. Therefore,the pressure is due to the transfer of momentum from the molecules to the walls of the container.

(D) According to the kinetic theory of gases,the average kinetic energy $(K_{avg})$ of an ideal gas is given by the formula: $K_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature in Kelvin.
At absolute zero temperature $(T = 0 \ K)$,the average kinetic energy of the molecules becomes $K_{avg} = \frac{3}{2} k_B (0) = 0$.
Therefore,at absolute zero,the motion of atoms or molecules ceases,and their kinetic energy becomes zero.