Pressure and Energy Questions in English

(B) The average translational kinetic energy of a gas molecule is given by the formula $K = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature in Kelvin.
From the relation $K \propto T$,we can see that the kinetic energy is directly proportional to the absolute temperature.
Given initial temperature $T_i = -78^{\circ} C$. Converting this to Kelvin: $T_i = 273 + (-78) = 195 \ K$.
We want the new kinetic energy $K' = 2K$. Since $K \propto T$,if the kinetic energy doubles,the absolute temperature must also double.
Therefore,the new absolute temperature $T_f = 2 \times T_i = 2 \times 195 \ K = 390 \ K$.
Converting the final temperature back to Celsius: $T_f(^{\circ} C) = 390 - 273 = 117^{\circ} C$.

(B) The translational kinetic energy of $N$ molecules is given by $K = \frac{3}{2} N k_B T = \frac{3}{2} n R T$,where $n$ is the number of moles.
Given mass $m = 50 \text{ g}$,molar mass of $\text{CO}_2$ $M = 44 \text{ g/mol}$.
Number of moles $n = \frac{m}{M} = \frac{50}{44} \approx 1.136 \text{ mol}$.
Temperature $T = 17 + 273.15 = 290.15 \text{ K}$.
Using $R = 8.314 \text{ J/(mol K)}$:
$K = \frac{3}{2} \times \left( \frac{50}{44} \right) \times 8.314 \times 290.15$.
$K = 1.5 \times 1.13636 \times 8.314 \times 290.15 \approx 4108.6 \text{ J}$.
Comparing with the given options,the closest value is $4102.8 \text{ J}$.

(B) The internal energy $U$ of an ideal gas is given by $U = n C_v T$.
For a diatomic gas,the degrees of freedom $f = 5$,so $C_v = \frac{fR}{2} = \frac{5R}{2}$.
Substituting this into the internal energy equation: $U = n \left( \frac{5R}{2} \right) T = \frac{5}{2} nRT$.
Using the ideal gas law $PV = nRT$,we can substitute $nRT$ with $PV$: $U = \frac{5}{2} PV$.
Given: Pressure $P = 1 \text{ atm} \approx 10^5 \text{ Pa}$,Volume $V = 4 \times 4 \times 3 = 48 \text{ m}^3$.
Calculating $U$: $U = \frac{5}{2} \times 10^5 \times 48 = 5 \times 10^5 \times 24 = 120 \times 10^5 \text{ J} = 12 \times 10^6 \text{ J}$.

(D) The average kinetic energy (per molecule) of an ideal gas 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.
Since both helium $(He)$ and argon $(Ar)$ are monatomic gases,they have the same number of degrees of freedom $(f = 3)$.
Given that both gases are at the same temperature $(T = 300 \ K)$,the average kinetic energy per molecule depends only on the temperature.
Therefore,the ratio of the average kinetic energies is: $\frac{K.E._{He}}{K.E._{Ar}} = \frac{\frac{3}{2} k_B T}{\frac{3}{2} k_B T} = 1: 1$.

(D) The translational kinetic energy of a gas molecule depends only on its translational degrees of freedom,which is $3$ for any gas molecule (monoatomic,diatomic,or polyatomic).
From the kinetic theory of gases,the pressure $P$ of an ideal gas is given by $P = \frac{1}{3} \frac{N}{V} m v_{rms}^2$.
This can be rewritten as $PV = \frac{1}{3} N m v_{rms}^2 = \frac{2}{3} (\frac{1}{2} N m v_{rms}^2)$.
Since the total translational kinetic energy $K_{trans} = \frac{1}{2} N m v_{rms}^2$,we have $PV = \frac{2}{3} K_{trans}$.
Therefore,$K_{trans} = \frac{3}{2} PV$.
This relationship is independent of the atomicity of the gas because it only considers the translational motion.

(B) In an ideal gas,the molecules are considered as point masses with no intermolecular forces of attraction or repulsion.
Since there are no intermolecular forces,the potential energy associated with the interaction between molecules is zero.
Therefore,the internal energy of an ideal gas consists entirely of the kinetic energy of its molecules due to their random translational motion.
For a monoatomic ideal gas,the internal energy $U$ is given by $U = \frac{3}{2} nRT$,which represents the total translational kinetic energy.

(A) According to the kinetic theory of gases,the absolute temperature $T$ of an ideal gas is directly proportional to the average translational kinetic energy of its molecules.
Mathematically,the average kinetic energy per molecule is given by $K_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant.
Therefore,temperature is a measure of the average kinetic energy of the gas molecules.

(A) The average kinetic energy $(E_{avg})$ of a gas molecule is given by the formula $E_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
Since $k_B$ is a constant,the average kinetic energy is directly proportional to the absolute temperature $(E_{avg} \propto T)$.
Given temperatures are $T_A = 350 \ K$ and $T_B = 420 \ K$.
The ratio of the average kinetic energy of gas $B$ to that of gas $A$ is $\frac{E_B}{E_A} = \frac{T_B}{T_A}$.
Substituting the values,we get $\frac{E_B}{E_A} = \frac{420}{350} = \frac{42}{35} = \frac{6}{5}$.
Thus,the ratio is $6: 5$.

(A) The pressure $P$ exerted by an ideal gas on the walls of a container is given by the kinetic theory of gases as $P = \frac{1}{3} \rho v_{rms}^{2}$.
From the ideal gas equation,$PV = nRT$,we have $P = \frac{nRT}{V}$.
Since $n$,$R$,and $V$ are constant for a closed container,the pressure $P$ is directly proportional to the temperature $T$,i.e.,$P \propto T$.
The force $F$ exerted on the walls is given by $F = P \times A$,where $A$ is the surface area of the wall.
Since $A$ is constant for a closed container,$F \propto P$.
Therefore,$F \propto T$,which implies $F \propto T^{1}$.
Comparing this with $T^{x}$,we get $x = 1$.

(D) The mean kinetic energy $(K)$ of an ideal gas molecule is given by the formula $K = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature in Kelvin.
From this relation,we can see that $K \propto T$.
Given,at $T_1 = 399^{\circ} C = (399 + 273) K = 672 K$,the kinetic energy is $E_1 = E$.
We need to find the temperature $T_2$ at which the kinetic energy $E_2 = E/2$.
Using the proportionality $E_1 / E_2 = T_1 / T_2$,we get:
$E / (E/2) = 672 / T_2$
$2 = 672 / T_2$
$T_2 = 672 / 2 = 336 K$.
To convert this back to Celsius: $T_2(^{\circ} C) = 336 - 273 = 63^{\circ} C$.
Therefore,the correct option is $D$.

(C) According to the kinetic theory of gases,the average kinetic energy of a gas molecule is directly proportional to the absolute temperature of the gas.
Specifically,the average kinetic energy 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.
Since the temperature $T$ is constant for all molecules of a gas in thermal equilibrium,the average kinetic energy remains the same for all molecules.
Other quantities like velocity,momentum,and angular momentum vary from molecule to molecule due to collisions and the Maxwell-Boltzmann distribution of speeds.
Therefore,the correct option is $C$.

(B) The average kinetic energy $(K_{avg})$ of an ideal gas molecule is directly proportional to its absolute temperature $(T)$ in Kelvin,given by the formula: $K_{avg} = \frac{3}{2} k_B T$.
Given:
Initial temperature $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$.
Initial kinetic energy = $E$.
Final temperature $T_2 = 327^{\circ} C = 327 + 273 = 600 \ K$.
Since $K_{avg} \propto T$,we have:
$\frac{K_2}{K_1} = \frac{T_2}{T_1}$
$\frac{K_2}{E} = \frac{600 \ K}{300 \ K}$
$\frac{K_2}{E} = 2$
$K_2 = 2 E$.
Therefore,the final average kinetic energy will be $2 E$.

(D) The average translational 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.
From this relation,it is clear that the translational kinetic energy depends only on the temperature $T$ and is independent of the molar mass of the gas.
Given for nitrogen $(N_2)$: $E_1 = 0.042 \ eV$ at temperature $T_1 = T$.
For oxygen $(O_2)$: The temperature is doubled,so $T_2 = 2T$.
Since $E \propto T$,we have:
$\frac{E_2}{E_1} = \frac{T_2}{T_1} = \frac{2T}{T} = 2$
Therefore,$E_2 = 2 \times E_1 = 2 \times 0.042 \ eV = 0.084 \ eV$.

(A) The pressure $P$ of an ideal gas is related to its total translational kinetic energy $E$ and volume $V$ by the formula: $P = \frac{2}{3} \frac{E}{V}$.
Given,$E = 7.5 \times 10^3 \text{ J}$ and $V = 10 \text{ litre} = 10 \times 10^{-3} \text{ m}^3 = 10^{-2} \text{ m}^3$.
Substituting the values into the formula:
$P = \frac{2}{3} \times \frac{7.5 \times 10^3}{10^{-2}}$
$P = \frac{2}{3} \times 7.5 \times 10^5$
$P = 5 \times 10^5 \text{ N/m}^2$.

(D) The pressure $P$ exerted by an ideal gas is given by the ideal gas equation $PV = N K_B T$,where $N$ is the number of molecules,$K_B$ is the Boltzmann constant,and $V$ is the volume.
Since pressure $P$ is defined as force $F$ per unit area $A$,we have $P = F/A$.
Substituting this into the ideal gas equation:
$(F/A) V = N K_B T$
$F = (N K_B T A) / V$
Since $N$,$K_B$,$A$,and $V$ are constants for a given container and gas sample,the force $F$ is directly proportional to the temperature $T$.
$F \propto T^1$
Comparing this with $F \propto T^x$,we get $x = 1$.

(B) 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 $\rho = \frac{M}{V}$ is the density of the gas and $v_{rms}$ is the root-mean-square speed.
Substituting $\rho$ in the equation:
$P = \frac{1}{3} \left( \frac{M}{V} \right) v_{rms}^2$
Multiply and divide the right side by $2$:
$P = \frac{2}{3} \left( \frac{1}{2} \frac{M}{V} v_{rms}^2 \right)$
Since the total kinetic energy $K.E. = \frac{1}{2} M v_{rms}^2$,the kinetic energy per unit volume is $u = \frac{K.E.}{V} = \frac{1}{2} \rho v_{rms}^2$.
Therefore,$P = \frac{2}{3} u$.

(D) The pressure $P$ and volume $V$ of an ideal gas are related to the number of molecules $N$ and the average kinetic energy per molecule $\langle K \rangle$ by the relation:
$PV = \frac{2}{3} N \langle K \rangle$
Given:
$V = 500 \text{ c.c.} = 500 \times 10^{-6} \text{ m}^3 = 5 \times 10^{-4} \text{ m}^3$
$P = 2 \times 10^5 \text{ N/m}^2$
$\langle K \rangle = 6 \times 10^{-21} \text{ J}$
Rearranging the formula to solve for $N$:
$N = \frac{3PV}{2\langle K \rangle}$
Substituting the values:
$N = \frac{3 \times (2 \times 10^5) \times (5 \times 10^{-4})}{2 \times (6 \times 10^{-21})}$
$N = \frac{3 \times 10^2}{12 \times 10^{-21}} = \frac{300}{12} \times 10^{21} = 25 \times 10^{21} = 2.5 \times 10^{22}$
Thus,the number of gas molecules is $2.5 \times 10^{22}$.

(B) The pressure $P$ of an ideal gas is given by the kinetic theory formula:
$P = \frac{1}{3} \frac{N m}{V} v_{rms}^2$
where $N$ is the number of molecules,$m$ is the mass of each molecule,$V$ is the volume,and $v_{rms}$ is the root-mean-square speed.
From this relation,we can see that $P \propto m \cdot v_{rms}^2$.
Let the initial mass be $m_1 = m$ and initial speed be $v_1 = v$. The initial pressure is $P_1 = P$.
Given the new conditions:
$m_2 = \frac{m}{2}$
$v_2 = 2v$
Now,calculating the new pressure $P_2$:
$\frac{P_2}{P_1} = \frac{m_2}{m_1} \times \left( \frac{v_2}{v_1} \right)^2$
$\frac{P_2}{P} = \left( \frac{m/2}{m} \right) \times \left( \frac{2v}{v} \right)^2$
$\frac{P_2}{P} = \left( \frac{1}{2} \right) \times (2)^2$
$\frac{P_2}{P} = \frac{1}{2} \times 4 = 2$
Therefore,$P_2 = 2P$.

(B) The pressure of an ideal gas is given by the kinetic theory formula: $P = \frac{1}{3} \frac{N m}{V} v_{rms}^2$.
From this expression,we can see that the pressure $P$ is directly proportional to the mass of the molecules $m$ and the square of their root-mean-square velocity $v_{rms}^2$,assuming the number of molecules $N$ and volume $V$ remain constant: $P \propto m v_{rms}^2$.
Let the initial state be $P_1 = P_0$,mass $m_1 = m$,and velocity $v_1 = v$.
In the final state,the mass is halved: $m_2 = \frac{m}{2}$,and the velocity is doubled: $v_2 = 2v$.
The ratio of the final pressure $P_2$ to the initial pressure $P_1$ is given by:
$\frac{P_2}{P_1} = \frac{m_2 v_2^2}{m_1 v_1^2} = \frac{(\frac{m}{2}) (2v)^2}{m v^2}$.
Simplifying the expression:
$\frac{P_2}{P_1} = \frac{(\frac{m}{2}) (4v^2)}{m v^2} = \frac{2 m v^2}{m v^2} = 2$.
Therefore,the final pressure is $P_2 = 2 P_1 = 2 P_0$.

(C) The average kinetic energy $K$ of a gas molecule is given by the formula $K = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
Since $K \propto T$,the ratio of the average kinetic energy of gas $B$ to gas $A$ is given by $\frac{K_B}{K_A} = \frac{T_B}{T_A}$.
Given $T_A = 360 \ K$ and $T_B = 420 \ K$.
Substituting these values,we get $\frac{K_B}{K_A} = \frac{420}{360} = \frac{7}{6}$.
Thus,the ratio is $7: 6$.

(B) The kinetic energy per 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.
Rearranging the formula to solve for $R$:
$R = \frac{2 E}{3 T}$

(D) The average translational kinetic energy $K$ of a gas molecule is given by the formula $K = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
From this relation,it is clear that $K \propto T$.
If we want to double the kinetic energy $(K' = 2K)$,we must have $T' = 2T$.
Therefore,the temperature must be increased to $2T$.

(B) The translational kinetic energy of a gas molecule is given by the formula $E = \frac{3}{2}kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature.
From this relation,it is clear that the kinetic energy $E$ is directly proportional to the absolute temperature $T$ $(E \propto T)$.
To double the kinetic energy $(E' = 2E)$,we must double the absolute temperature $(T' = 2T)$.
Therefore,the temperature must be increased to $2T$.

(A) The average translational kinetic energy of a gas molecule is given by $K.E. = \frac{3}{2} kT$.
Since $k = \frac{R}{N}$,we have $K.E. = \frac{3}{2} \frac{RT}{N}$.
The kinetic energy of an electron accelerated through a potential difference $V$ is $K.E. = eV$.
Equating the two energies:
$\frac{3}{2} \frac{RT}{N} = eV$
Solving for $T$:
$T = \frac{2 eVN}{3 R}$.

(B) 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 and $v_{rms}$ is the root mean square velocity.
We know that the kinetic energy per unit volume $(u)$ is given by $u = \frac{1}{2} \rho v_{rms}^2$.
Therefore,$\rho v_{rms}^2 = 2u$.
Substituting this into the pressure equation: $P = \frac{1}{3} (2u) = \frac{2}{3} u$.
Thus,the pressure exerted by an ideal gas is $2/3$ of the kinetic energy per unit volume of the gas.

(B) The pressure $p$ exerted by an ideal gas is given by the kinetic theory of gases as:
$p = \frac{1}{3} \rho \bar{v}^2$
where $\rho$ is the density and $\bar{v}^2$ is the mean square speed.
Since density $\rho = \frac{M}{V}$,we can write:
$p = \frac{1}{3} \frac{M}{V} \bar{v}^2$
Multiplying and dividing by $2$,we get:
$p = \frac{2}{3} \left( \frac{1}{2} \frac{M}{V} \bar{v}^2 \right)$
Here,$\frac{1}{2} M \bar{v}^2$ is the total kinetic energy of the gas molecules.
Thus,$\frac{1}{2} \frac{M}{V} \bar{v}^2$ represents the kinetic energy per unit volume,which is given as $E$.
Therefore,$p = \frac{2}{3} E$.

(D) The change in momentum of one molecule striking the wall and rebounding is given by $\Delta p = 2mv \cos \theta$, where $m$ is the mass, $v$ is the velocity, and $\theta$ is the angle with the normal.
Given: $m = 3 \times 10^{-27} \,kg$, $v = 500 \,m/s$, $\theta = 60^{\circ}$, $N = 10^{23}$ molecules/sec, and $A = 1 \,cm^2 = 10^{-4} \,m^2$.
Pressure $P = \frac{F}{A} = \frac{1}{A} \times \frac{\Delta p}{\Delta t} = \frac{N \times (2mv \cos 60^{\circ})}{A}$.
Substituting the values: $P = \frac{10^{23} \times 2 \times (3 \times 10^{-27}) \times 500 \times \cos 60^{\circ}}{10^{-4}}$.
Since $\cos 60^{\circ} = 0.5$, we have $P = \frac{10^{23} \times 6 \times 10^{-27} \times 500 \times 0.5}{10^{-4}} = \frac{3000 \times 10^{-4} \times 0.5}{10^{-4}} = 1500 \,N/m^2$.

(B) According to the kinetic theory of gases,the pressure $P$ exerted by an ideal gas is given by the formula: $P = \frac{1}{3} \rho v_{rms}^2$,where $\rho$ is the density and $v_{rms}$ is the root mean square speed.
Since $\rho = \frac{M}{V}$,we can write $P = \frac{1}{3} \frac{N m}{V} v_{rms}^2$.
Also,the mean square speed is defined as $\langle v^2 \rangle = \frac{1}{N} \sum v_i^2 = v_{rms}^2$.
Therefore,the pressure $P$ is directly proportional to the mean of the square of the speeds of the gas molecules,which is $\langle v^2 \rangle$.

(D) According to the kinetic theory of gases,the pressure $P$ of an ideal gas is given by the relation $P = \frac{2}{3} \frac{N}{V} \langle K \rangle$,where $\langle K \rangle$ is the average kinetic energy of the molecules.
Since the volume $V$ and the number of molecules $N$ are constant,the pressure $P$ is directly proportional to the average kinetic energy of the molecules.
Mathematically,$P \propto \langle K \rangle$.
Therefore,the correct option is $D$.

(C) The mean kinetic energy $(E)$ of an ideal gas is directly proportional to its absolute temperature ($T$ in Kelvin),given by the formula $E = \frac{3}{2} k_B T$.
Therefore,$E \propto T$.
Given initial temperature $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$.
Given final temperature $T_2 = 327^{\circ} C = 327 + 273 = 600 \ K$.
Using the proportionality ratio: $\frac{E_2}{E_1} = \frac{T_2}{T_1}$.
Substituting the values: $E_2 = E_1 \times \frac{600}{300}$.
$E_2 = 2 E_1$.

(C) The average kinetic energy $(K)$ of molecules of an ideal gas is directly proportional to its absolute temperature $(T)$,given by the relation $K = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant.
Since the temperature $(T)$ is kept constant,the kinetic energy of the molecules remains unchanged,regardless of the change in pressure.

(B) According to the kinetic theory of gases,the average kinetic energy of a molecule of an ideal gas is given by the equipartition theorem.
For a monatomic ideal gas,the molecule has $3$ degrees of freedom.
The mean energy per molecule is given by $E = \frac{f}{2} k_{B} T$,where $f$ is the number of degrees of freedom.
For a monatomic gas,$f = 3$,so the mean energy is $E = \frac{3}{2} k_{B} T$.
Therefore,the correct option is $B$.

(C) The average translational kinetic energy $(K_{avg})$ of a gas molecule is given by the formula: $K_{avg} = \frac{3}{2} k_B T$.
Given:
Temperature $T = 127^{\circ} C = 127 + 273 = 400 \,K$.
Boltzmann constant $k_B = 1.38 \times 10^{-23} \,J \,K^{-1}$.
Substituting the values into the formula:
$K_{avg} = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 400$.
$K_{avg} = 1.5 \times 1.38 \times 400 \times 10^{-23}$.
$K_{avg} = 1.5 \times 552 \times 10^{-23}$.
$K_{avg} = 828 \times 10^{-23} \,J$.
$K_{avg} = 8.28 \times 10^{-21} \,J$.

(C) The internal energy $U$ of a monoatomic gas is given by the formula $U = \frac{3}{2} nRT$,where $n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the temperature in Kelvin.
Given: $n = 4$ moles,$T = 77^{\circ} C = 77 + 273 = 350 \ K$.
Substituting the values into the formula:
$U = \frac{3}{2} \times 4 \times R \times 350$
$U = 6 \times R \times 350$
$U = 2100 R$.

(C) For an ideal gas, the internal energy $U$ is given by the formula $U = \frac{3}{2} PV$ for a monatomic gas.
Given values are $V = 3 \,m^3$ and $P = 3 \times 10^5 \,Pa$.
Substituting these values into the formula:
$U = \frac{3}{2} \times (3 \times 10^5 \,Pa) \times (3 \,m^3)$
$U = \frac{3}{2} \times 9 \times 10^5 \,J$
$U = 1.5 \times 9 \times 10^5 \,J$
$U = 13.5 \times 10^5 \,J$.

(B) The relationship between pressure $P$,volume $V$,and total translational kinetic energy $E$ is given by the formula: $P = \frac{2}{3} \frac{E}{V}$.
Given:
Volume $V = 10 \text{ liters} = 10 \times 10^{-3} \text{ m}^3 = 10^{-2} \text{ m}^3$.
Total kinetic energy $E = 4.5 \times 10^5 \text{ J}$.
Substituting these values into the formula:
$P = \frac{2}{3} \times \frac{4.5 \times 10^5}{10^{-2}}$
$P = \frac{2}{3} \times 4.5 \times 10^7$
$P = 3 \times 10^7 \text{ Nm}^{-2} = 30 \times 10^6 \text{ Nm}^{-2}$.
Thus,the correct option is $B$.

(D) The average kinetic energy $(E)$ of an ideal gas is given by the formula:
$E = \frac{3}{2} nRT$
For $n = 1$ mole of gas,the equation becomes:
$E = \frac{3}{2} RT$
From the ideal gas equation,we know that $PV = nRT$. For $n = 1$,this simplifies to:
$PV = RT$
Substituting $RT = PV$ into the kinetic energy equation:
$E = \frac{3}{2} PV$
Rearranging the terms to solve for pressure $(P)$:
$P = \frac{2}{3} \frac{E}{V}$

(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 $\frac{E_2}{E_1} = \frac{T_2}{T_1}$.
Given: $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$, $E_1 = 3.3 \times 10^{-20} \ J$, and $T_2 = 127^{\circ} C = 127 + 273 = 400 \ K$.
Substituting the values: $E_2 = E_1 \times \frac{T_2}{T_1} = 3.3 \times 10^{-20} \times \frac{400}{300}$.
$E_2 = 3.3 \times 10^{-20} \times \frac{4}{3} = 1.1 \times 4 \times 10^{-20} = 4.4 \times 10^{-20} \ J$.

(C) Temperature,$T = 77^{\circ} C = 273 + 77 = 350 \,K$.
Boltzmann constant,$K_{B} = 1.38 \times 10^{-23} \,J/K$.
The average kinetic energy of a monoatomic gas atom is given by $E = \frac{3}{2} K_{B} T$.
Substituting the values:
$E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 350 \,J$.
To convert the energy from Joules to electron-volts $(eV)$,divide by the charge of an electron $(1.6 \times 10^{-19} \,C)$:
$E(eV) = \frac{1.5 \times 1.38 \times 350 \times 10^{-23}}{1.6 \times 10^{-19}} \,eV$.
$E(eV) = \frac{724.5 \times 10^{-23}}{1.6 \times 10^{-19}} \,eV$.
$E(eV) \approx 4.52 \times 10^{-2} \,eV$.

(C) According to the kinetic theory of gases,the pressure $P$ exerted by an ideal gas is given by the relation $P = \frac{2}{3} \frac{K}{V}$,where $K$ is the total kinetic energy of the gas molecules and $V$ is the volume.
Since the volume $V$ is constant for a given amount of gas,the pressure $P$ is directly proportional to the total kinetic energy $K$ of the gas molecules.
This is because pressure arises from the collisions of gas molecules with the walls of the container,and the force of these collisions depends on the kinetic energy of the molecules.
Therefore,the correct option is $C$.

(A) 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.
From this formula,it is clear that the average kinetic energy depends only on the absolute temperature $T$ and is independent of the mass or nature of the gas molecule.
Given that the average kinetic energy of $H_2$ molecules at $300 \ K$ is $E$,and since the temperature for $O_2$ molecules is also $300 \ K$,the average kinetic energy of $O_2$ molecules will also be $E$.

(A) The mean kinetic energy per molecule of an ideal gas is given by the formula $\bar{E} = \frac{3}{2} K_B T$,where $K_B$ is the Boltzmann constant and $T$ is the absolute temperature.
Since the gases are in the same vessel,they are in thermal equilibrium,meaning they are at the same temperature $T$.
Because the mean kinetic energy $\bar{E}$ depends only on the temperature $T$ and is independent of the mass or nature of the gas molecules,the ratio of the mean kinetic energies of hydrogen and oxygen is $1: 1$.

(D) The average translational kinetic energy of a gas molecule is given by the formula $E = \frac{3}{2} k T$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature in Kelvin.
Given,$E = 0.69 \ eV = 0.69 \times 1.6 \times 10^{-19} \ J$.
Substituting the values into the formula:
$0.69 \times 1.6 \times 10^{-19} = \frac{3}{2} \times 1.38 \times 10^{-23} \times T$
$T = \frac{0.69 \times 1.6 \times 10^{-19} \times 2}{3 \times 1.38 \times 10^{-23}}$
$T = \frac{2.208 \times 10^{-19}}{4.14 \times 10^{-23}} \approx 5333 \ K$
To convert the temperature to Celsius,we use $T(^{\circ}C) = T(K) - 273.15$.
$T(^{\circ}C) = 5333 - 273 = 5060^{\circ} C$.
Therefore,the correct option is $D$.

(C) The average translational kinetic energy 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.
Since the average translational kinetic energy depends only on the temperature $T$ and not on the nature or mass of the gas molecules,it is the same for all ideal gases at the same temperature.
Given that the average translational kinetic energy of $O_2$ molecules is $0.048 \ eV$ at temperature $T$,the average translational kinetic energy of $N_2$ molecules at the same temperature $T$ will also be $0.048 \ eV$.

(C) The temperature of an ideal gas is directly proportional to the average translational kinetic energy of its molecules,given by the relation $K.E. = \frac{3}{2} k_B T$. Thus,Assertion $(A)$ is true.
However,the reason provided is incorrect. Thermal energy is not produced by collisions; rather,the kinetic energy of the molecules *is* the thermal energy of the gas. Collisions between molecules are elastic in an ideal gas,meaning they do not result in a loss or gain of kinetic energy that would 'produce' thermal energy; they simply redistribute the existing kinetic energy. Therefore,Reason $(R)$ is false.

(B) Helium $(He)$ is a monoatomic gas. The degrees of freedom $(f)$ for a monoatomic gas is $3$.
The total random kinetic energy $(U)$ of an ideal gas is given by the formula: $U = \frac{f}{2} nRT$,where $n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the absolute temperature.
Given: Mass $(m)$ = $1 \,g$,Molar mass of Helium $(M)$ = $4 \,g/mol$,Temperature $(T)$ = $100 \,K$,$R = 8.3 \,J \,mol^{-1} \,K^{-1}$.
Number of moles $(n)$ = $\frac{m}{M} = \frac{1}{4} = 0.25 \,mol$.
Substituting the values into the formula: $U = \frac{3}{2} \times 0.25 \times 8.3 \times 100$.
$U = 1.5 \times 0.25 \times 830$.
$U = 0.375 \times 830 = 311.25 \,J$.
Therefore,the total random kinetic energy is $311.25 \,J$.

(D) The average translational kinetic energy of a gas molecule is given by $KE = \frac{3}{2} k_B T$.
The kinetic energy gained by an electron accelerated through a potential difference $V$ is $KE = eV$.
Given that these two energies are equal:
$\frac{3}{2} k_B T = eV$
Substituting the values $V = 10 \,V$, $e = 1.6 \times 10^{-19} \,C$, and $k_B = 1.38 \times 10^{-23} \,JK^{-1}$:
$T = \frac{2eV}{3k_B} = \frac{2 \times 1.6 \times 10^{-19} \times 10}{3 \times 1.38 \times 10^{-23}}$
$T = \frac{32 \times 10^{-19}}{4.14 \times 10^{-23}} \approx 7.73 \times 10^4 \,K = 77.3 \times 10^3 \,K$.

(A) The average kinetic energy of a gas molecule is given by the formula $KE_{av} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature of the gas.
Since the argon and chlorine gases are in the same flask and in thermal equilibrium,they are at the same temperature $T = 27^{\circ} C = 300 \ K$.
Because the average kinetic energy per molecule depends only on the temperature $T$ and is independent of the mass or the nature of the gas molecules,the ratio of the average kinetic energies of the two gases is $1:1$.