Kinetic Theory of Gases MCQs for NEET — Physics Questions with Answers

$$\begin{gathered} \left(\mathrm{D}\right). {\mathrm{r}}_{{\mathbf{H}}_{\mathbf{2}}}=\frac{{\mathrm{V}}_{{\mathbf{H}}_{\mathbf{2}}}}{\mathrm{t}} ; {\mathrm{r}}_{{\mathbf{O}}_{\mathbf{2}}}=\frac{{\mathrm{V}}_{{\mathbf{O}}_{\mathbf{2}}}}{\mathrm{t}} \\ \therefore \frac{ {\mathrm{r}}_{{\mathbf{H}}_{\mathbf{2}}}}{{\mathrm{r}}_{{\mathbf{O}}_{\mathbf{2}}}}=\frac{{\mathrm{V}}_{{\mathbf{H}}_{\mathbf{2}}}}{{\mathrm{V}}_{{\mathbf{O}}_{\mathbf{2}}}}=\frac{1}{{\mathrm{n}}_{{\mathbf{O}}_{\mathbf{2}}}} \\ \mathrm{or} \frac{{\mathrm{r}}_{{\mathbf{O}}_{\mathbf{2}}}}{{\mathrm{r}}_{{\mathbf{H}}_{\mathbf{2}}}}={\mathrm{n}}_{{\mathbf{O}}_{\mathbf{2}}}=\frac{1}{\sqrt{16}}=\frac{1}{4} \\ \therefore {\mathrm{m}}_{{\mathbf{O}}_{\mathbf{2}}}=\frac{1}{4}\times 32=8 \mathrm{gm}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{B}\right). \mathrm{K}.\mathrm{E}.=\frac{3}{2}\mathrm{RT} \\ \therefore {\left(\mathrm{K}.\mathrm{E}.\right)}_1=\frac{3}{2}\times \mathrm{R}\times 400; {\left(\mathrm{K}.\mathrm{E}.\right)}_2=\frac{3}{2}\times \mathrm{R}\times 800 \\ \therefore \frac{{\left(\mathrm{K}.\mathrm{E}.\right)}_2}{{\left(\mathrm{K}.\mathrm{E}.\right)}_1}=2 \mathrm{or} {\left(\mathrm{K}.\mathrm{E}.\right)}_2=2{\left(\mathrm{K}.\mathrm{E}.\right)}_1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{B}\right). \mathrm{K}.\mathrm{E}.=\frac{1}{2}\mathrm{M}{\overline{\mathrm{C}}}^2 \\ \mathrm{Now} \mathrm{for} \mathrm{helium} \mathrm{atom}, \\ \mathrm{K}.\mathrm{E}.=\frac{1}{2}{\mathrm{M}}_{\mathrm{He}} {\overline{\mathrm{C}}}^2=\frac{1}{2}{\mathrm{M}}_{\mathrm{He}}\times \frac{3\mathrm{RT}}{{\mathrm{M}}_{\mathrm{He}}}=\frac{3}{2}\mathrm{RT} \\ \mathrm{Again} \mathrm{for} {\mathrm{H}}_2 \mathrm{molecules} \\ \mathrm{K}.\mathrm{E}.=\frac{1}{2}{\mathrm{M}}_{{\mathbf{H}}_{\mathbf{2}}} {\overline{\mathrm{C}}}^2=\frac{1}{2}\times {\mathrm{M}}_{{\mathbf{H}}_{\mathbf{2}}}\times \frac{3\mathrm{RT}}{{\mathrm{M}}_{{\mathbf{H}}_{\mathbf{2}}}}=\frac{3}{2}\mathrm{RT} \\ \therefore \mathrm{K}.\mathrm{E}. \mathrm{of} {\mathrm{H}}_2 \mathrm{molecules} \mathrm{is} \mathrm{same} \mathrm{as} \mathrm{it} \mathrm{is} \mathrm{for} {\mathrm{H}}_2 \mathrm{molecules}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{A}\right). \mathrm{K}.\mathrm{E}. \mathrm{for} {\mathrm{UF}}_6=\frac{1}{2}\times {\mathrm{M}}_{{\mathbf{UF}}_{\mathbf{6}}}\times \frac{3\mathrm{RT}}{{\mathrm{M}}_{{\mathbf{UF}}_{\mathbf{6}}}}=\frac{3}{2}\mathrm{RT} \\ \mathrm{K}.\mathrm{E}. \mathrm{for} {\mathrm{H}}_2=\frac{1}{2}\times {\mathrm{M}}_{{\mathbf{H}}_{\mathbf{2}}}\times \frac{3\mathrm{RT}}{{\mathrm{M}}_{{\mathbf{H}}_{\mathbf{2}}}}=\frac{3}{2}\mathrm{RT} \\ \therefore \mathrm{K}.\mathrm{E}. \mathrm{of} {\mathrm{UF}}_6 \mathrm{to} \mathrm{that} {\mathrm{H}}_2 \mathrm{is} 1 \mathbf{:}\mathbf{ }1. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{B}\right). \\ \mathrm{Gas} {\mathrm{C}}_{\mathrm{p}} \mathrm{in} \mathrm{cals}/\mathrm{mole} {\mathrm{C}}_{\mathrm{v}} \mathrm{in} \mathrm{cals}/\mathrm{mole} \\ \mathrm{Monoatomic} 5 3 \\ \mathrm{Diatomic} 7 5 \\ \mathrm{Triatomic} 8 6 \\ \mathrm{When} \mathrm{we} \mathrm{are} \mathrm{mixing} \mathrm{one} \mathrm{mole} \mathrm{of} \mathrm{each} \mathrm{gas}, \\ \mathrm{then} \mathrm{total} {\mathrm{C}}_{\mathrm{P}}=5+7+8=20 \mathrm{cals}/3 \mathrm{moles} \\ \mathrm{then} \mathrm{total} {\mathrm{C}}_{\mathrm{V}}=3+5+6=14 \mathrm{cals}/3 \mathrm{moles} \\ \therefore \frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{V}}}=\frac{20}{14}=1.428 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{A}\right). \left(\mathrm{P}+\frac{\mathrm{a}}{{\mathrm{V}}^2}\right) \mathrm{V}=\mathrm{RT} \mathrm{or} \mathrm{PV}+\frac{\mathrm{a}}{\mathrm{V}}=\mathrm{RT} \\ \mathrm{or} \frac{\mathrm{PV}}{\mathrm{RT}}+\frac{\mathrm{a}}{\mathrm{RTV}}=1 \mathrm{or} \mathrm{Z}=1-\frac{\mathrm{a}}{\mathrm{RTV}} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{D}\right). \mathrm{In} \mathrm{an} \mathrm{ideal} \mathrm{gas} \mathrm{internal} \mathrm{energy} =\frac{\mathrm{f}}{2} \mathrm{nRT} \\ \mathrm{U}=2\times \frac{5}{2}\times \mathrm{RT}+4\times \frac{3}{2} \mathrm{RT} = 11\mathrm{RT} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{B}\right). \mathrm{We} \mathrm{know} \mathrm{that} \\ {\mathrm{P}}_{\mathrm{A}}{\mathrm{V}}_{\mathrm{A}} = {\mathrm{n}}_{\mathrm{A}}\mathrm{RT}, {\mathrm{P}}_{\mathrm{B}}{\mathrm{V}}_{\mathrm{B}} = {\mathrm{n}}_{\mathrm{B}}\mathrm{RT} \\ \mathrm{and} {\mathrm{P}}_{\mathrm{f}} \left({\mathrm{V}}_{\mathrm{A}} + {\mathrm{V}}_{\mathrm{B}}\right) = \left({\mathrm{n}}_{\mathrm{A}} + {\mathrm{n}}_{\mathrm{B}}\right) \mathrm{RT} \\ {\mathrm{P}}_{\mathrm{f}} \left({\mathrm{V}}_{\mathrm{A}} + {\mathrm{V}}_{\mathrm{B}}\right) = {\mathrm{P}}_{\mathrm{A}}{\mathrm{V}}_{\mathrm{A}} + {\mathrm{P}}_{\mathrm{B}}{\mathrm{V}}_{\mathrm{B}} \\ \therefore {\mathrm{P}}_{\mathrm{f}} =\left(\frac{{\mathrm{P}}_{\mathrm{A}}{\mathrm{V}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}}{\mathrm{V}}_{\mathrm{B}}}{{\mathrm{V}}_{\mathrm{A}}+{\mathrm{V}}_{\mathrm{B}}}\right) \\ = \frac{1.4\times 0.1+0.7\times 0.15}{ 0.1+0.15} \mathrm{MPa} = 0.98 \mathrm{MPa} \end{gathered}$$

(A). The variation of pressure with height is given by the relation, ln $\frac{{\mathrm{P}}_0}{\mathrm{P}}=\left(\frac{\mathrm{g}}{\mathrm{RT}}\right) \mathrm{Mh},$ where M is

        the molecular weight of the gas,p_o  , the pressure at sea level and p, the pressure at a height ‘h’.

        For air, ln $2=\left(\frac{\mathrm{g}}{\mathrm{RT}}\right)\times 28.8\times 6$(units of M and h are mixed up but the same units are used in

        the next step also).

        For hydrogen, ln $4=\left(\frac{\mathrm{g}}{\mathrm{RT}}\right)\times 2\times \mathrm{h}.$ 

        Dividing one by the other,

$2=\frac{2}{28.8}\times \frac{\mathrm{h}}{6}; \mathrm{h}=6\times 28.8 \mathrm{km}=172.8 \mathrm{km}.$