Kinetic Theory of Gases MCQs with Answers – Part 2 (NEET Physics)

Explanation: \(\textbf{Use directional pressure ratios:}\) For each direction, \[ P_x=nm\langle v_x^2\rangle, \quad P_y=nm\langle v_y^2\rangle, \quad P_z=nm\langle v_z^2\rangle \] Therefore the component-square averages are also in the ratio \[ \langle v_x^2\rangle:\langle v_y^2\rangle:\langle v_z^2\rangle=4:1:1 \] Their sum is \[ \langle v^2\rangle=6u^2 \] Hence the three component averages are \[ 4u^2,\quad u^2,\quad u^2 \] Thus for the anisotropic ensemble, \[ P_{x,\text{aniso}}=4nmu^2 \] \(\textbf{Compare with the isotropic ensemble:}\) For the same total mean-square speed under isotropy, \[ \langle v_x^2\rangle_{\text{iso}} = \frac13(6u^2)=2u^2 \] Hence \[ P_{x,\text{iso}}=2nmu^2 \] Therefore, \[ \frac{P_{x,\text{aniso}}}{P_{x,\text{iso}}}=2 \] \(\textbf{Final answer:}\) \[ \boxed{\frac{P_{x,\text{aniso}}}{P_{x,\text{iso}}}=2} \]

Explanation: For a smooth stationary wall, the component of velocity normal to the wall reverses. The tangential components remain unchanged. Therefore, \[ (3u,-4u,0)\rightarrow(-3u,-4u,0) \] The molecular speed before collision is \[ v_i=\sqrt{(3u)^2+(-4u)^2}=5u \] After collision, \[ v_f=\sqrt{(-3u)^2+(-4u)^2}=5u \] Thus the molecular kinetic energy is unchanged, as required for an elastic collision with a stationary massive wall. The normal molecular momentum changes from \[ +3mu \] to \[ -3mu \] Hence its momentum change has magnitude \[ |-3mu-3mu|=6mu \] The wall receives equal and opposite momentum of magnitude \(6mu\). \(\textbf{Final answer:}\) \[ \boxed{\left((-3u,-4u,0),6mu\right)} \]

Explanation: A molecule rebounding from a stationary wall reverses its velocity component normal to that wall. Therefore its momentum vector changes even if the magnitude of its velocity remains the same. Momentum of the molecule considered alone is consequently not conserved. The missing momentum is transferred to the wall. When the complete isolated molecule-wall system is considered, total momentum is conserved. For a very massive stationary wall, its acquired velocity is negligibly small even though it receives finite momentum. The molecular speed remains unchanged in the ideal elastic rebound, so the molecule's kinetic energy is unchanged to the approximation used. Elasticity therefore does not mean zero impulse. In fact, wall impulse is the central microscopic mechanism that produces gas pressure. Hence option D correctly distinguishes conservation for the subsystem from conservation for the complete collision system.

Explanation: A cube is convenient because its walls can be chosen perpendicular to Cartesian axes. That geometrical convenience is not an essential physical assumption of kinetic theory. For a general vessel, one can consider a sufficiently small wall element that is effectively planar. The normal component of molecular velocity relative to that local wall element is then used exactly as in the standard argument. The number of molecules reaching that patch in a short time is proportional to its area. The momentum transferred per unit time is therefore also proportional to that area. When force is divided by area to calculate pressure, the explicit area factor cancels. The resulting pressure depends on molecular number density, molecular mass and the appropriate mean-square velocity component. It does not depend on whether the complete vessel is cubic, spherical or irregular. Thus the Reason directly explains why the Assertion is correct.

Explanation: The derivation begins with a single collision because pressure ultimately comes from momentum transfer. For a molecule with normal velocity component \(v_x\), an elastic rebound transfers momentum magnitude \[ 2mv_x \] to the wall. The next task is to count how many appropriate molecules can reach the wall in a short interval. That count is then multiplied by the momentum transfer from one collision. The product gives the total momentum delivered during the selected time interval. Dividing by the interval gives the force on the wall. Dividing that force by wall area gives the directional pressure \[ P=nm\langle v_x^2\rangle. \] Only after this directional expression has been established is isotropy used. For an equilibrium isotropic gas, \[ \langle v_x^2\rangle=\frac13\langle v^2\rangle. \] Hence the final standard result is \[ P=\frac13nm\langle v^2\rangle. \] Therefore the correct sequence is I \(\rightarrow\) II \(\rightarrow\) III \(\rightarrow\) IV \(\rightarrow\) V \(\rightarrow\) VI.

Explanation: The slab of molecules capable of reaching the wall in time \(\Delta t\) has volume \[ Av_x\Delta t. \] It therefore contains \[ nAv_x\Delta t \] molecules of the stated velocity magnitude. However, in a random equilibrium gas, only half of that population is moving toward the selected wall. The correct collision count is therefore \[ N_{\rm hit}=\frac12nAv_x\Delta t. \] The momentum delivered by each elastic rebound is correctly written as \[ 2mv_x. \] Hence total momentum transferred is \[ Q= \left(\frac12nAv_x\Delta t\right)(2mv_x). \] Therefore, \[ Q=nAmv_x^2\Delta t. \] Using \[ P=\frac{Q}{A\Delta t}, \] we obtain \[ P=nmv_x^2 \] for that molecular velocity group. \(\textbf{Final answer:}\) The missing factor is the \(1/2\) selecting molecules moving toward the wall.

Explanation: \(\textbf{Construct the mean-square normal speed:}\) Pressure depends on \[ P=nm\langle v_x^2\rangle \] The signs of \(v_x\) do not matter after squaring. For population 1, \[ v_x^2=u^2, \] and its fractional population is \(0.60\). For population 2, \[ v_x^2=(2u)^2=4u^2, \] with fractional population \(0.40\). Therefore, \[ \langle v_x^2\rangle = 0.60u^2+0.40(4u^2) \] Thus \[ \langle v_x^2\rangle = 0.60u^2+1.60u^2 = 2.20u^2 \] Hence \[ P=nm(2.20u^2)=2.2nmu^2 \] The equal positive and negative populations make \(\langle v_x\rangle=0\), but they do not reduce the mean-square value. \(\textbf{Final answer:}\) \[ \boxed{P=2.2nmu^2} \]

Explanation: \(\textbf{Use the kinetic pressure relation:}\) \[ P=\frac13nmv_{\rm rms}^2 \] Therefore, \[ n=\frac{3P}{mv_{\rm rms}^2} \] Substituting the given data, \[ n= \frac{3(1.2\times10^5)} {(4.0\times10^{-26})(500)^2} \] Now, \[ (500)^2=2.5\times10^5 \] Hence, \[ (4.0\times10^{-26})(2.5\times10^5) =1.0\times10^{-20} \] Therefore, \[ n=\frac{3.6\times10^5}{1.0\times10^{-20}} =3.6\times10^{25}\,\text{m}^{-3} \] \(\textbf{Convert number density to total molecule number:}\) \[ N=nV \] Thus, \[ N=(3.6\times10^{25})(2.0\times10^{-3}) =7.2\times10^{22} \] \(\textbf{Final answer:}\) \[ \boxed{N=7.2\times10^{22}} \]

Explanation: \(\textbf{Infer mass density from pressure and RMS speed:}\) The kinetic pressure relation can be written as \[ P=\frac13\rho v_{\rm rms}^2 \] Therefore, \[ \rho=\frac{3P}{v_{\rm rms}^2} \] Substituting, \[ \rho = \frac{3(1.0\times10^5)} {(500)^2} = \frac{3.0\times10^5}{2.5\times10^5} \] Hence, \[ \rho=1.2\,\text{kg m}^{-3} \] \(\textbf{Obtain the mass of one molecule:}\) Since \[ \rho=nm, \] \[ m=\frac{\rho}{n} = \frac{1.2}{3.0\times10^{25}} =4.0\times10^{-26}\,\text{kg} \] \(\textbf{Convert to molar mass:}\) \[ M=mN_A \] Thus, \[ M=(4.0\times10^{-26})(6.0\times10^{23}) =2.4\times10^{-2}\,\text{kg mol}^{-1} \] Therefore, \[ M=24\,\text{g mol}^{-1} \] \(\textbf{Final answer:}\) \[ \boxed{M=24\,\text{g mol}^{-1}} \]

Explanation: \(\textbf{Interpret the graph slope:}\) The kinetic relation is \[ P=\frac13\rho v_{\rm rms}^2 \] Thus a graph of \(P\) against \(v_{\rm rms}^2\) has slope \[ s=\frac{\rho}{3} \] Therefore, \[ \frac{\rho_X}{\rho_Y} = \frac{s_X}{s_Y} = \frac32 \] \(\textbf{Convert density ratio to molecule-number ratio:}\) Because X and Y contain the same gas species, molecular mass \(m\) is the same. Hence \[ \rho=\frac{Nm}{V} \] Therefore, \[ \frac{N_X}{N_Y} = \frac{\rho_X}{\rho_Y} \frac{V_X}{V_Y} \] Substituting, \[ \frac{N_X}{N_Y} = \frac32\times\frac23 =1 \] Thus the two vessels contain equal numbers of molecules despite having different volumes and mass densities. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{\rho_X}{\rho_Y},\frac{N_X}{N_Y}\right) = \left(\frac32,1\right)} \]

Explanation: \(\textbf{Pressure change:}\) For the same gas species, \[ P=\frac13nmv_{\rm rms}^2 \] Hence, \[ \frac{P_2}{P_1} = \frac{n_2}{n_1} \left(\frac{v_{{\rm rms},2}}{v_{{\rm rms},1}}\right)^2 \] Substituting, \[ \frac{P_2}{P_1} = 0.80(1.50)^2 \] Thus, \[ \frac{P_2}{P_1} = 0.80(2.25) =1.80 \] \(\textbf{Check the molecule number:}\) Since \[ N=nV, \] \[ \frac{N_2}{N_1} = (0.80)(1.25)=1 \] Thus no net change in molecule number has occurred. \(\textbf{Compare total translational kinetic energies:}\) \[ E_{\rm tr}=N\left(\frac12mv_{\rm rms}^2\right) \] With the same \(N\) and \(m\), \[ \frac{E_{{\rm tr},2}}{E_{{\rm tr},1}} = (1.50)^2 =2.25 \] \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{P_2}{P_1},\frac{E_{{\rm tr},2}}{E_{{\rm tr},1}}\right) = (1.80,2.25)} \]

Explanation: Each component contributes to the total pressure according to \[ P_i=\frac13n_im_iv_{{\rm rms},i}^2. \] Therefore, \[ \frac{P_Y}{P_X} = \frac{n_Y}{n_X} \frac{m_Y}{m_X} \left(\frac{v_{{\rm rms},Y}}{v_{{\rm rms},X}}\right)^2. \] Substituting the given ratios, \[ \frac{P_Y}{P_X} = (2)(2)\left(\frac12\right)^2. \] Hence, \[ \frac{P_Y}{P_X} = 4\left(\frac14\right)=1. \] Thus X and Y make equal pressure contributions even though their number densities, molecular masses and RMS speeds are individually different. The larger number density and molecular mass of Y are exactly compensated by its smaller mean-square speed. The result follows from the full product \(nmv_{\rm rms}^2\), not from any one factor considered separately. \(\textbf{Final answer:}\) \[ P_Y=P_X. \]

Explanation: \(\textbf{Compare total translational energies:}\) For an ideal gas, \[ PV=\frac23E_{\rm tr} \] The two samples have equal \(P\) and equal \(V\), so \[ E_{{\rm tr},X}=E_{{\rm tr},Y} \] Average translational kinetic energy per molecule is \[ \overline K=\frac{E_{\rm tr}}{N} \] Therefore, \[ \frac{\overline K_X}{\overline K_Y} = \frac{N_Y}{N_X} = \frac12 \] \(\textbf{Convert energy ratio to RMS-speed ratio:}\) For each gas, \[ \overline K=\frac12mv_{\rm rms}^2 \] Hence, \[ \frac{\overline K_X}{\overline K_Y} = \frac{m_X}{m_Y} \frac{v_{{\rm rms},X}^2}{v_{{\rm rms},Y}^2} \] Substituting, \[ \frac12 = 3\frac{v_{{\rm rms},X}^2}{v_{{\rm rms},Y}^2} \] Thus, \[ \frac{v_{{\rm rms},X}^2}{v_{{\rm rms},Y}^2} = \frac16, \] and therefore \[ \frac{v_{{\rm rms},X}}{v_{{\rm rms},Y}} = \frac1{\sqrt6} \] \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{\overline K_X}{\overline K_Y},\frac{v_{{\rm rms},X}}{v_{{\rm rms},Y}}\right) = \left(\frac12,\frac1{\sqrt6}\right)} \]

Explanation: The kinetic pressure relation is \[ PV=\frac13Nm\langle v^2\rangle. \] The total translational kinetic energy is \[ E_{\rm tr} = N\left(\frac12m\langle v^2\rangle\right). \] Therefore, \[ PV = \frac23E_{\rm tr}. \] The Assertion is consequently correct. The quantity entering this pressure relation is specifically the translational kinetic energy associated with motion of molecular centres of mass. A polyatomic molecule can also possess rotational energy. When vibrational modes are active, vibrational energy can contribute as well. Those additional contributions belong to the complete internal energy but do not alter the derivation of \(PV=\frac23E_{\rm tr}\). Therefore \(E_{\rm tr}\) cannot generally be identified with the entire internal energy of every polyatomic ideal gas. The Reason is false, so option D is correct.

Explanation: Statement I is correct because kinetic theory describes a macroscopic gas as an enormous molecular population in continuous random motion. Statement II is also correct because equilibrium means statistical steadiness, not cessation of microscopic motion. Oppositely directed molecular velocities can cancel in their vector average even though individual molecular speeds remain non-zero. Statement III follows from the ideal-gas assumption that intermolecular interactions are negligible during ordinary free flight. With negligible resultant interaction force between collisions, Newton's first law gives approximately constant molecular velocity over each free-flight segment. Statement IV is incorrect because elasticity conserves the total kinetic energy of the colliding system. One molecule may gain kinetic energy while another loses kinetic energy during an elastic collision. Statement V is correct because isotropy is a statistical symmetry: \[ \langle v_x^2\rangle=\langle v_y^2\rangle=\langle v_z^2\rangle. \] It does not require \(v_x^2=v_y^2=v_z^2\) for every individual molecule. Therefore I, II, III and V are correct.

Explanation: For the same equilibrium gas, the molecular impact rate per unit wall area is the same at both patches. Therefore the expected number of impacts in a fixed time is proportional to wall area. Hence \[ \frac{N_2}{N_1}=\frac{4A}{A}=4 \] The statistical distribution of molecular velocities is also the same at the two patches. Thus the average momentum transfer per molecular impact is unchanged. Four times as many impacts therefore produce four times the momentum transfer per unit time: \[ \frac{F_2}{F_1}=4 \] Pressure is force divided by area: \[ P=\frac{F}{A} \] Hence \[ \frac{P_2}{P_1} = \frac{F_2/F_1}{A_2/A_1} = \frac44=1 \] The wall force depends on the chosen wall area, whereas pressure is an intensive local property of the equilibrium gas. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{N_2}{N_1},\frac{F_2}{F_1},\frac{P_2}{P_1}\right)=(4,4,1)} \]

Explanation: \(\textbf{Recover the number density from the collision count:}\) Only half of the molecules with the stated \(|v_x|\) move toward the selected wall. Thus \[ N_{\rm hit} = \frac12nAv_x\Delta t \] Substituting the data, \[ 1.5\times10^{21} = \frac12n(2.0\times10^{-3})(300)(1.0\times10^{-4}) \] The multiplying factor apart from \(n\) is \[ \frac12(2.0\times10^{-3})(300)(1.0\times10^{-4}) = 3.0\times10^{-5} \] Hence \[ n=\frac{1.5\times10^{21}}{3.0\times10^{-5}} =5.0\times10^{25}\,\text{m}^{-3} \] \(\textbf{Convert molecular impacts into pressure:}\) For this model distribution, \[ P=nmv_x^2 \] Therefore, \[ P=(5.0\times10^{25})(4.0\times10^{-26})(300)^2 \] Now \[ (5.0\times10^{25})(4.0\times10^{-26})=2.0, \] and \[ (300)^2=9.0\times10^4 \] Thus \[ P=1.8\times10^5\,\text{Pa} \] \(\textbf{Final answer:}\) \[ \boxed{(n,P)=\left(5.0\times10^{25}\,\text{m}^{-3},1.8\times10^5\,\text{Pa}\right)} \]

Explanation: \(\textbf{Force on the wall normal to the x-axis:}\) For two opposite walls separated by distance \(L_x\), the time between successive impacts on the same wall is \[ \Delta t_x=\frac{2L_x}{|v_x|} \] The impulse per impact is \[ 2m|v_x| \] Hence \[ F_x=\frac{m v_x^2}{L_x} \] Here \[ L_x=2l,\qquad |v_x|=2u \] Therefore, \[ F_x=\frac{m(2u)^2}{2l} =\frac{2mu^2}{l} \] \(\textbf{Force on the wall normal to the y-axis:}\) Similarly, \[ F_y=\frac{mv_y^2}{L_y} \] With \[ L_y=l,\qquad |v_y|=u, \] we get \[ F_y=\frac{mu^2}{l} \] Thus \[ \frac{F_x}{F_y}=2 \] Both the impulse per collision and the rate of collisions are encoded in the \(v_i^2/L_i\) dependence. \(\textbf{Final answer:}\) \[ \boxed{\frac{F_x}{F_y}=2} \]

Explanation: For a possibly anisotropic molecular ensemble, \[ P_x=nm\langle v_x^2\rangle, \] \[ P_y=nm\langle v_y^2\rangle, \] \[ P_z=nm\langle v_z^2\rangle \] Adding the three directional pressures gives \[ P_x+P_y+P_z = nm\langle v^2\rangle \] The stated randomisation does not change the number density or total translational kinetic energy. Therefore \(nm\langle v^2\rangle\) remains unchanged. Initially, \[ P_x+P_y+P_z=6+3+3=12\,\text{kPa} \] After isotropy is established, \[ P_x=P_y=P_z=P \] Hence \[ 3P=12\,\text{kPa}, \] so \[ P=4\,\text{kPa} \] The process redistributes the same total mean-square molecular motion equally among the three spatial directions. \(\textbf{Final answer:}\) \[ \boxed{P_f=4\,\text{kPa}} \]

Explanation: \(\textbf{Interpret the force-area graph:}\) Pressure is \[ P=\frac{F}{A} \] Therefore a graph of \(F\) against \(A\) has slope \[ s=P \] Hence \[ \frac{P_X}{P_Y}=\frac{s_X}{s_Y}=\frac32 \] \(\textbf{Connect pressure with translational kinetic energy:}\) For an ideal gas, \[ PV=\frac23E_{\rm tr} \] Thus \[ E_{\rm tr}=\frac32PV \] Therefore, \[ \frac{E_X}{E_Y} = \frac{P_X}{P_Y} \frac{V_X}{V_Y} \] Substituting the given ratios, \[ \frac{E_X}{E_Y} = \frac32\times2=3 \] The graph supplies the pressure ratio, while the volume ratio is required to convert pressure into total translational energy. \(\textbf{Final answer:}\) \[ \boxed{\frac{E_X}{E_Y}=3} \]