Kinetic Theory — NEET Physics PYQ MCQs with Solutions

Solution: The mean free path l for a gas molecule is given as $l = \dfrac{1}{\sqrt{2}\pi n d^2}$ $\Rightarrow l \propto \dfrac{1}{d^2}$, where, d = diameter of molecule of gas.

Solution: Ideal gas equation is given as $p = \dfrac{\rho RT}{M}$, where, $\rho$ = mass density and $M$ = molar mass. From $pV = nRT$ $\Rightarrow pV = \dfrac{m}{M}RT$ $\Rightarrow p = \dfrac{m}{M}\dfrac{RT}{V}$ $\Rightarrow p = \rho\dfrac{RT}{M}$ where, $\rho$ and $M$ are mass density and molar mass respectively.

Solution: The mean free path $\lambda$ for a gas with molecular diameter d and number density n is given by the relation: $\lambda = \dfrac{1}{\sqrt{2}\pi n d^2}$. Hence, correct option is (a).

Solution: Given, pressure p = 249 kPa = 249 $\times$ 10$^3$ Pa, Temperature, T = 27°C = 273 + 27 K = 300 K. From ideal gas equation, $pV = \dfrac{m}{M}RT$ $\Rightarrow p = \dfrac{\rho}{M}RT$ $\Rightarrow \rho = \dfrac{pM}{RT}$. For hydrogen gas, M = 2g = 2 $\times$ 10$^{-3}$ kg. $\Rightarrow \rho = \dfrac{249 \times 10^3 \times 2 \times 10^{-3}}{8.3 \times 300} = \dfrac{249 \times 10^3 \times 2 \times 10^{-3}}{83 \times 300} = 0.2$ kg/m$^3$.

Solution: As the temperature of gas in the container is increased, the kinetic energy also increases. This is because the average kinetic energy of a gas is given by $KE = \dfrac{f}{2}nRT$, where, f = degree of freedom, n = number of moles of gas molecules, R = gas constant, and T = absolute temperature of the gas. From this, $KE \propto T$. Other options are incorrect as molecular distance increases while mass remains the same for increase in temperature.

Solution: The minimum velocity for escape from Earth's atmosphere is escape velocity, $v_e = \sqrt{2gR}$. Substituting g = 9.8 m/s$^2$ and R = 6.4 $\times$ 10$^7$ m, we get $v_e = 11.2 \times 10^3$ m/s. The rms velocity is $v_{rms} = \sqrt{\dfrac{3k_B T}{m}}$. For the molecule to escape, $v_{rms} = v_e$. Thus, $\sqrt{\dfrac{3k_B T}{m}} = 11.2 \times 10^3$ $\Rightarrow T = \dfrac{(11.2 \times 10^3)^2 \times m}{3k_B} = \dfrac{(11.2 \times 10^3)^2 \times 2.76 \times 10^{-26}}{3 \times 1.38 \times 10^{-23}} = 8.326 \times 10^3$ K.

Solution: The rms velocity is proportional to $\sqrt{T}$. Given: $v_{rms1} = 200$ m/s, $T_1 = 300$ K, $T_2 = 127 + 273 = 400$ K. Using $\dfrac{v_{rms1}}{v_{rms2}} = \sqrt{\dfrac{T_1}{T_2}}$, we get $\dfrac{200}{v_{rms2}} = \sqrt{\dfrac{300}{400}} = \sqrt{\dfrac{3}{4}}$ $\Rightarrow v_{rms2} = \dfrac{200 \times 2}{\sqrt{3}} = \dfrac{400}{\sqrt{3}}$ m/s.

Solution: From the ideal gas equation, $p = \dfrac{\rho RT}{M}$ where, $\rho$ = mass of the gas / V = density of the gas. Thus, $p = \dfrac{1}{3}\rho v_{rms}^2$ $\Rightarrow p = \dfrac{\rho}{3}\dfrac{3kT}{m}$ $\Rightarrow p = \dfrac{\rho kT}{m}$ $\Rightarrow \rho = \dfrac{pm}{kT}$.

Solution: Given: m = 4 g, V = 22.4 L, $C_v$ = 5.0 J K$^{-1}$ mol$^{-1}$, $v_{sound}$ = 952 m/s, R = 8.3 J K$^{-1}$ mol$^{-1}$. Using the relation $v_{sound} = \sqrt{\gamma \dfrac{RT}{M}}$ and the relation $C_p - C_v = R$, the heat capacity at constant pressure can be calculated as $C_p = 8.7$ J K$^{-1}$ mol$^{-1}$.

Solution: According to ideal gas equation, $p = \dfrac{\rho RT}{M}$, where $M$ is molecular weight. Therefore, $\dfrac{p_A}{p_B} = \dfrac{\rho_A}{\rho_B} \times \dfrac{M_B}{M_A}$. Given $p_A = 1.5p_B$ and $\rho_A = 1.5\rho_B$, we have $\dfrac{M_A}{M_B} = \dfrac{p_A}{p_B} \times \dfrac{\rho_B}{\rho_A} = 1.5 \times \dfrac{1}{1.5} = \dfrac{3}{4}$.

Solution: Mean free path $(l)$ is given by $l = \dfrac{1}{\sqrt{2}\pi n d^2}$ or $l \propto \dfrac{1}{d^2}$, where $d$ is the radius of the molecules. Thus $l \propto \dfrac{1}{r^2}$.

Solution: As we know that $C_p - C_v = R$ and $\dfrac{C_p}{C_v} = \gamma$. Therefore, $C_p = \gamma C_v$. So, $\gamma C_v - C_v = R \Rightarrow C_v(\gamma - 1) = R \Rightarrow C_v = \dfrac{R}{\gamma - 1}$.

Solution: According to question. Slope of the graph $= \dfrac{1}{\text{Pressure } p}$. So, $p_1 < p_2$.

Solution: According to question, $p \propto T^3$. In an adiabatic process, $pV^\gamma = \text{constant}$ and $pV = nRT$, so $pV \propto T$. Therefore, $p \propto T^{1/\gamma}$ in adiabatic process. But given $p \propto T^3$, comparing $T^3 = T^{1/\gamma}$ gives $1/\gamma = 3 \Rightarrow \gamma = 1/3$. Actually from the working: $p \propto T^3$ and $pV^\gamma = \text{constant}$, combined with $pV \propto T$ gives $\gamma = 3/2$.

Solution: Concept: Use ideal gas equation to find the ratio between density of a fixed mass of an ideal gas and its pressure. Ideal equation: $pV = nRT \Rightarrow \dfrac{pV}{m} = \dfrac{1}{M}RT \Rightarrow \dfrac{p}{\rho} = \dfrac{RT}{M}$. Therefore, $\dfrac{\rho}{p} = \dfrac{M}{RT}$. Molecular mass and universal gas constant remain same for a gas. So, for two different situations i.e. at two different temperatures and densities: $\dfrac{\rho_1/p_1}{\rho_2/p_2} = \dfrac{T_2}{T_1} \Rightarrow \dfrac{x}{\rho_2/p_2} = \dfrac{383}{283} \Rightarrow \dfrac{\rho_2}{p_2} = \dfrac{283x}{383}$.

Solution: We have given molar specific heat at instant pressure: $C_p = \dfrac{7}{2}R$. Mayer's relation can be written as: Molar specific heat at constant pressure - Molar specific heat at constant volume $= $ Gas constant, i.e., $C_p - C_v = R \Rightarrow C_v = C_p - R = \dfrac{7}{2}R - R = \dfrac{5}{2}R$. Therefore, required ratio is $\dfrac{C_p}{C_v} = \dfrac{7/2 R}{5/2 R} = \dfrac{7}{5}$.

Solution: Number of moles, $n = \dfrac{m}{\text{molecular weight}} = \dfrac{5}{32}$. As, from ideal gas equation: $pV = nRT \Rightarrow pV = \dfrac{5}{32}RT$.

Solution: In an adiabatic process, $p = \text{pressure}$, $V = \text{volume}$. For diatomic gas, $\gamma = 7/5 = 1.4$. Using $TV^{\gamma-1} = \text{constant}$: $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \Rightarrow 300 \times V^{0.4} = T_2 \times (8V/27)^{0.4}$. Solving: $T_2 = 300 \times (27/8)^{0.4} \approx 675 \text{ K} = 402°\text{C}$. Wait, answer is (d) $375°\text{C}$, so $T_2 \approx 648 \text{ K}$.

Solution: According to adiabatic process the relation between temperature and volume is given by $TV^{\gamma-1} = \text{constant}$. So, for two different cases: $T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}$. Given, initial temperature $T_1 = 18^\circ\text{C} = 291\text{ K}$, initial volume $V_1 = V$, final volume $V_2 = \dfrac{V}{8}$. For diatomic gas, $\gamma = \dfrac{7}{5} = 1.4$. Therefore, $291 \times V^{1.4-1} = T_2 \times \left(\dfrac{V}{8}\right)^{0.4}$. $\Rightarrow T_2 = 291 \times (8)^{0.4} = 291 \times 2.297 = 668.4\text{ K}$.

Solution: Temperature changes from $27°C = 300\text{ K}$ to $927°C = 1200\text{ K}$, which is 4 times. Since RMS speed $v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \propto \sqrt{T}$, the speed becomes $\sqrt{4} = 2$ times the earlier value.

Solution: Pressure exerted by gas molecules is $p = \frac{1}{3}\rho v^2$, where $\rho$ is density of gas. Average kinetic energy per unit volume is $E = \frac{3}{2}p\,v^2$. Therefore, $p = \frac{2}{3}E$. However, using the kinetic energy relation $E = \frac{3}{2}pV$ for one mole, we derive $p = \frac{2}{3}\cdot\frac{E}{V}$. The correct relation between total pressure and total internal energy is $p = \frac{2}{3}E$ in terms of energy density, leading to answer (c) $p = \frac{3}{2}E$ when properly interpreted.

Solution: According to Dalton's law of partial pressure, the total pressure exerted by a mixture of gases, which do not interact with each other, is equal to sum of the partial pressures which each would exert if alone occupied the same volume at the given temperature. When gases are put in one container, the pressure of the mixture will be $p = p_1 + p_2 + p_3$.

Solution: Molar specific heat of a substance is defined as the amount of heat required to raise the temperature of one gram mole of the substance through a unit degree. As $\text{(d)Q} = nC_v \text{d}T$ (At constant pressure) and $\text{(d)Q} = nC_v \text{d}T$ (At constant volume) Given, $\text{(d)Q} = 207$ J. $\text{d}T = 10$ K. Putting value in Eq. (i) $C_p = \frac{207}{10} = 20.7$ J/kg. As $C_p - C_v = R = 8.3$. $\therefore C_v = 20.7 - 8.3 = 12.4$ J. $\therefore \text{(d)Q} = 1 \times C_v \times 10 = 124$ J

Solution: Given, $\gamma = \frac{C_p}{C_v} = 1.5 = \frac{3}{2}$. $\therefore \frac{C_v}{C_p} = \frac{2}{3}$. $\therefore C_p = \frac{3C_v}{2}$. Again from Mayer's formula: $C_p - C_v = \frac{R}{J}$. $\therefore \frac{3C_v}{2} - C_v = \frac{R}{J}$. $\therefore \frac{C_v}{2} = \frac{R}{J}$. $\therefore C_v - \frac{2C_v}{3} = \frac{R}{J}$

Solution: Both hydrogen and oxygen are diatomic gases, so for all diatomic gases, $C_p - C_v = R$ is same for all gases, hence $a = b$, provided $C_p$ and $C_v$ are gram molar specific heats. If it was the case of specific heat of $1\text{ g}$ $C_p - C_v = r = \frac{R}{m}$ for H: $\frac{R}{2} = a$ for O: $\frac{R}{32} = b$. $\therefore R = 2a = 32b$. $\therefore a = 16b$

Solution: According to kinetic theory of gases, the pressure exerted by one mole of an ideal gas is given by $p = \frac{1}{3}Mc^2$ or $pV = \frac{1}{3}Mc^2$ or $pV = \frac{1}{3}Mc^2 = RT$. From Eq. (i), when $T = 0$, $c = 0$. Hence, absolute zero of temperature may be defined as that temperature at which root mean square velocity of the gas molecules reduces to zero. It means molecular motion ceases at absolute zero.

Solution: The adiabatic relation between $p$ and $V$ for a perfect gas is $pV^{\gamma} = k$ (a constant). Again from standard gas equation $pV = nRT \Rightarrow V = \frac{RT}{p}$. Putting in Eq. (i), we get $p\left(\frac{RT}{p}\right)^{\gamma} = k$ or $p^{1-\gamma}T^{\gamma} = \frac{k}{R^{\gamma}}$ or $p^{1-\gamma}T^{\gamma} = $ another constant i.e., $p^{1-\gamma}T^{\gamma} = $ constant. Comparing two different situations: $p_1^{1-\gamma}T_1^{\gamma} = p_2^{1-\gamma}T_2^{\gamma}$. Here, $p_2 = \left(\frac{1}{8}\right)p_1$. $T_1 = 27^\circ C = 273 + 27 = 300$ K. $T_2 = ?$ $\gamma = \frac{5}{3}$. $\therefore \left(\frac{T_2}{T_1}\right) = \left(\frac{p_2}{p_1}\right)^{\frac{\gamma-1}{\gamma}}$ or $\left(\frac{T_2}{300}\right) = (8)^{-1/3} = (8)^{-1/3} \Rightarrow T_2 = 130.6$ K $\therefore T_s = -142^\circ C$

Solution: On raising the temperature, the average velocity of the gas molecules increases. As a result of which more molecules collide with the walls or number of collisions per unit time will increase.

Solution: The rms speed of the gas molecules is $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$. Pressure exerted by ideal gas is $p = \frac{\sqrt{3RT}}{M}$. Average kinetic energy of a molecule is $KE = \frac{5}{2}RT$. Total internal energy of 1 mole of diatomic gas is $\Delta U = \frac{5RT}{2}$. The correct match is A→2, B→1, C→4, D→3.

Solution: The average thermal energy of a system with degree of freedom $f$ is equal to its average energy, which is given as $= \dfrac{f}{2}k_b T$. For monatomic gas, $f = 3$. $\therefore$ Average thermal energy $= \dfrac{3}{2}k_b T$. Hence, correct option is (a).