NEET Practice Questions (MCQs) with Answers & Solutions

Real gases show ideal gas behaviour at high temperature and low pressures.

(c) According to Avogadro's hypothesis, Volume of a gas (V) $\propto$ number of moles (n)

Therefore, the ratio of the volumes of gases can be determined in terms of their moles. The ratio of volumes of H₂: O₂:methane (CH₄)
is given by

$$\begin{gathered} {\mathrm{V}}_{{\mathrm{H}}_2} :\hspace{0.17em}{\mathrm{V}}_{{\mathrm{O}}_2}: {\mathrm{V}}_{{\mathrm{CH}}_4} = {\mathrm{n}}_{{\mathrm{H}}_2} :\hspace{0.17em}{\mathrm{n}}_{{\mathrm{O}}_2} : {\mathrm{n}}_{{\mathrm{CH}}_4} \\ \Rightarrow {\mathrm{V}}_{{\mathrm{H}}_2} :\hspace{0.17em}{\mathrm{V}}_{{\mathrm{O}}_2}: {\mathrm{V}}_{{\mathrm{CH}}_4} = \frac{{\mathrm{m}}_{{\mathrm{H}}_2}}{{\mathrm{M}}_{{\mathrm{H}}_2}} : \frac{{\mathrm{m}}_{{\mathrm{O}}_2}}{{\mathrm{M}}_{{\mathrm{O}}_2}} : \frac{{\mathrm{m}}_{{\mathrm{CH}}_4}}{{\mathrm{M}}_{{\mathrm{CH}}_4}} \\ \mathrm{But} {\mathrm{m}}_{{\mathrm{H}}_2} = {\mathrm{m}}_{{\mathrm{O}}_2} = {\mathrm{m}}_{{\mathrm{CH}}_4} = \frac{\mathrm{m}}{2} :\hspace{0.17em}\frac{\mathrm{m}}{32}: \frac{\mathrm{m}}{16} = 16:1:2 \end{gathered}$$

The given problem is related to the concept of stiochiometry of chemical equations. Thus, we have to convert the given volumes into their moles and then, identify the limiting reagent [possessing minimum number of moles and gets completely used up in the reaction].

The limiting reagent gives the moles of product formed in the reaction.

              H₂(g) + Cl₂(g) $\to$2HCl (g)

Initial vol   22.4 L    11.2 L      2 mol

... 22.4 L volume at STP is occupied by,

              Cl₂ = 1 mole,

... 11.2 L volume will be occupied by,

             Cl₂ = 1x 11.2/22.4 mole = 0.5 mol

Thus, H₂(g) + Cl₂(g) $\to$2HCl (g)

(d) Easily liquefiable gases like NH₃, SO₂ etc. exhibit maximum deviation from ideal gas as for them Z<<< 1 .

${\mathrm{CH}}_4$  also exhibits deviation but it is less as compared to NH₃.

"CO="N₂ 

P_co = P_N2 

Given, P_co + P_N2 = 1atm 

or       2P_N2 = 1 atm

or        P_N2 = 0.5atm  

(a)

 $$\begin{gathered} \frac{{\mathrm{V}}_{\mathrm{A}}}{{\mathrm{V}}_{\mathrm{B}}} = \sqrt{\frac{{\mathrm{M}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{A}}}} \\ \mathrm{or} \frac{{\mathrm{V}}_{\mathrm{A}}}{{{\displaystyle \mathrm{t}}}_{\mathrm{A}}} \times \frac{{\mathrm{t}}_{\mathrm{B}}}{{\mathrm{V}}_{\mathrm{B}}} = \sqrt{\frac{{\displaystyle {\mathrm{M}}_{\mathrm{B}}}}{{\displaystyle {\mathrm{M}}_{\mathrm{A}}}}} \\ \mathrm{or} \frac{10}{20} = \sqrt{\frac{{\displaystyle {\mathrm{M}}_{\mathrm{B}}}}{{\displaystyle 49}}} \\ \mathrm{or} \frac{1}{4} = \frac{{\mathrm{M}}_{\mathrm{a}}}{49} \\ \therefore {\mathrm{M}}_{\mathrm{B}} = \frac{49}{4} = 12.25 \mathrm{u}. \end{gathered}$$

Average velocity = $\sqrt{\frac{8\mathrm{RT}}{\mathrm{\pi M}}}$

V_av $\mathrm{\alpha }$ $\sqrt{\mathrm{T}}$

or (V_av)₂ / (V_av )₁ = $\sqrt{\frac{2\mathrm{T}}{\mathrm{T}}}$ = 1.4

(D).  7.5 kcals

        The kinetic energy is given by $\mathrm{K}.\mathrm{E}.=\frac{3}{2}\times \mathrm{RT}$ per gm mole.

(B). The RMS velocity is given by $\mathrm{C} \mathrm{rms}=\sqrt{\left(\frac{3\mathrm{RT}}{\mathrm{M}}\right)}.$