NEET Practice Questions (MCQs) with Answers & Solutions

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

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

(C). More the ‘a’ value of the gas, more will be the inter molecular attraction between the gas

       molecules, therefore, easier will be the liquefaction.

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

$$\begin{gathered} \left(\mathrm{B}\right). {\mathrm{r}}_{{\mathbf{H}}_{\mathbf{2}}}=\frac{\mathrm{x}}{5} {\mathrm{r}}_{\mathbf{x}}=\frac{\mathrm{x}}{\mathrm{t}} \\ \therefore \frac{\mathrm{x}}{5}\times \frac{\mathrm{t}}{\mathrm{x}}=\sqrt{\frac{{\mathrm{M}}_{\mathbf{x}}}{{\mathrm{M}}_{{\mathbf{H}}_{\mathbf{2}}}}} \\ \mathrm{or} \mathrm{t}=\sqrt{\frac{{\mathrm{M}}_{\mathbf{x}}}{{\mathrm{M}}_{{\mathbf{H}}_{\mathbf{2}}}}}\times 5 \mathrm{or} \mathrm{t}=\sqrt{\frac{{\mathrm{M}}_{\mathbf{x}}}{2}}\times 5 \\ \mathrm{For} \mathrm{He} : {\mathrm{M}}_{\mathrm{x}}=4, \mathrm{t}=5\sqrt{2} \\ \mathrm{For} \mathrm{CO} : {\mathrm{M}}_{\mathrm{x}}=28, \mathrm{t}=5\sqrt{14} \\ \mathrm{For} {\mathrm{O}}_2 : {\mathrm{M}}_{\mathrm{x}}=32, \mathrm{t}=20 \\ \mathrm{For} {\mathrm{CO}}_2 : {\mathrm{M}}_{\mathrm{x}}=44, \mathrm{t}=5\sqrt{22} \end{gathered}$$

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

$$\begin{gathered} \left(\mathrm{A}\right). 48 \mathrm{gm}. \mathrm{C} = \frac{48}{12} = 4 \mathrm{moles} \mathrm{of} \mathrm{C}. \mathrm{From} 4 \mathrm{moles} \mathrm{of} \mathrm{C}, 4 \mathrm{moles} \mathrm{of} \mathrm{CO}\left(\mathrm{g}\right) \mathrm{and} 4 \mathrm{moles} \mathrm{of} {\mathrm{H}}_2\left(\mathrm{g}\right) \mathrm{is} \mathrm{obtained}. \\ \therefore \mathrm{total} \mathrm{moles} \mathrm{of} \mathrm{gas} \mathrm{produced} = 8 \\ \therefore \mathrm{volume} \mathrm{of} \mathrm{product} \mathrm{gas} \mathrm{mixture} \mathrm{at} \mathrm{STP} = 8 \times 22.4 = 179.2 \mathrm{lit}. \end{gathered}$$

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

(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,

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