NEET Practice Questions (MCQs) with Answers & Solutions

(C).  Do remember rms speed does not depend upon the pressure.

$$\begin{gathered} {\overline{\mathrm{C}}}_2=\sqrt{\frac{3\mathrm{R}\times 3\times 27}{\mathrm{M}}} \therefore \frac{\overline{{\mathrm{C}}_2}}{{\mathrm{C}}_1}=\sqrt{3} \\ {\overline{\mathrm{C}}}_2=\sqrt{\frac{3\mathrm{R}\times 27}{\mathrm{M}}} \mathrm{or} {\overline{\mathrm{C}}}_2=\sqrt{3} {\overline{\mathrm{C}}}_1=\sqrt{3}\times {10}^4 \mathrm{cm}/\sec \end{gathered}$$

$$\begin{gathered} \left(\mathrm{C}\right). \mathrm{At} \mathrm{a} \mathrm{given} \mathrm{temperature}, \mathrm{rate} \mathrm{of} \mathrm{effusion} \propto \mathrm{P} \mathrm{and} \propto \frac{1}{\sqrt{\mathrm{M}}} \\ \therefore \mathrm{rate} \mathrm{of} \mathrm{effusion} \mathrm{of} \mathrm{hydrogen} \mathrm{gas} \\ = 10 \mathrm{torr} {\min }^{-1}\times \frac{2000}{1000}\times \sqrt{\frac{4}{2}}=20\sqrt{2} \mathrm{torr} {\min }^{-1} \end{gathered}$$

(B).  Boyle temperature,

$$\begin{gathered} {\mathrm{T}}_{\mathrm{B}}=\frac{\mathrm{a}}{\mathrm{Rb}}=\frac{4 {\mathrm{lit}}^2 \mathrm{atm}. {\mathrm{mol}}^{-2}}{0.082 \mathrm{lit} \mathrm{atm} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1} \mathrm{x} 0.04 \mathrm{lit} {\mathrm{mol}}^{-1}} \\ =1219.5 = 1220 \mathrm{K}. \end{gathered}$$

(A). In the case of gases, the viscous resistance arises from the transport of molecules from one

       layer to another, with transfer of momenta, so that the fast moving molecules in one layer are

       slowed down, while the slow moving molecules in the other are accelerated. This momentum

       transfer increases with increase in temperature. A simple expression for the viscosity of a gas

       is $\mathrm{\eta }=\frac{1}{3} \overline{\mathrm{C}} \mathrm{d}\mathcal{l},$ where d is the density and $\mathcal{l}$ and the mean free path.

       d$\propto$ pressure and $\mathcal{l}$$\propto$$\frac{1}{\mathrm{pressure}}$ and so $\mathrm{\eta }$ is independent of pressure.

(A). Mol. wt. of $\mathrm{CO}$ = 28 Mol wt. of ${\mathrm{CO}}_2$ = 44

       Mol. wt. of ${\mathrm{N}}_2\mathrm{O}$ = 28 + 16 = 44

       Mol. wt. of ${\mathrm{NO}}_2$ = 14 + 32 = 46

       $\because {\mathrm{CO}}_2$ and ${\mathrm{N}}_2\mathrm{O}$ having same mol. wt. therefore, rate of diffusion for both the gases are same.

(A).  Using the expression $\mathrm{pV}=\mathrm{nRT},$ we have

$$\begin{gathered} \mathrm{n}=\frac{\mathrm{pV}}{\mathrm{RT}}=\frac{\left\{\left(7.6\times {10}^{-10}/760\right){\displaystyle \mathrm{atm}}\right\}{\displaystyle }{\displaystyle \left(1\mathrm{L}\right)}}{\left(0.0821 \mathrm{atm} \mathrm{L} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}\right){\displaystyle \left(273.15 \mathrm{K}\right)}} \\ =4.459\times {10}^{-14} \mathrm{mol} \\ \mathrm{N}={\mathrm{nN}}_{\mathrm{A}}=\left(4.459\times {10}^{-14}\mathrm{mol}\right)\left(6.023\times {10}^{23} {\mathrm{mol}}^{-1}\right) \\ =2.686\times {10}^{10}. \end{gathered}$$

(D).    The vander Waals constant ${\mathrm{T}}_{\mathrm{C}}, {\mathrm{P}}_{\mathrm{C}} \mathrm{and} {\mathrm{V}}_{\mathrm{C}}$ are related by the expression

${\mathrm{V}}_{\mathrm{C}}=3\mathrm{b} ; {\mathrm{T}}_{\mathrm{C}}=\frac{8\mathrm{a}}{27 \mathrm{Rb}} \mathrm{and} {\mathrm{P}}_{\mathrm{C}}=\frac{\mathrm{a}}{27 {\mathrm{b}}^2}$

(D). At the same temperature, oxygen and hydrogen molecules will have the same average

       energy; weight of ${\mathrm{H}}_2$ molecules is $\frac{1}{16}$of ${\mathrm{O}}_2$ molecules. So statements 2 and 4 are wrong.

$\left(\mathrm{D}\right). \frac{{\mathrm{n}}_{\mathrm{f}}}{{\mathrm{n}}_{\mathrm{i}}}=\frac{{\mathrm{P}}_{\mathrm{f}}}{{\mathrm{P}}_{\mathrm{i}}} ; {\mathrm{P}}_{\mathrm{f}}={\mathrm{P}}_{\mathrm{i}}\frac{{\mathrm{n}}_{\mathrm{f}}}{{\mathrm{n}}_{\mathrm{i}}} \mathrm{or} \left(2\right) \left(\frac{3}{2}\right)=3 \mathrm{atm}$

(C). In the expression, PV = RT $\left(\mathrm{A}+\frac{\mathrm{B}}{\mathrm{V}}+\frac{\mathrm{C}}{{\mathrm{V}}^2}+........\right),$, the first term viz., A within the brackets is

       the main term and the rest are correction terms for non-ideality. Therefore, A should be 1,

       since PV = RT per mole of an ideal gas. The van der Waals’ equation for one mole is

$\left(\mathrm{P}+\frac{\mathrm{a}}{{\mathrm{V}}^2}\right)\left(\mathrm{V}-\mathrm{b}\right)=$RT. Expanding,

$$\begin{gathered} \mathrm{PV}=\mathrm{RT}+\mathrm{bP}-\frac{\mathrm{a}}{\mathrm{V}}+\frac{\mathrm{ab}}{{\mathrm{V}}^2}\approx \mathrm{RT}+\mathrm{bP}-\frac{\mathrm{a}}{\mathrm{V}} \\ =\mathrm{RT}\left(1+\frac{\mathrm{bP}}{\mathrm{RT}}-\frac{\mathrm{a}}{\mathrm{RTV}}\right) \end{gathered}$$

          Applying the ideal gas equation in the correction term,

$\mathrm{PV}=\mathrm{RT}\left(1+\frac{\mathrm{b}}{\mathrm{V}}-\frac{\mathrm{a}}{\mathrm{RTV}}\right) =\mathrm{RT}\left[1+\frac{1}{\mathrm{V}}\left(\mathrm{b}-\frac{\mathrm{a}}{\mathrm{RT}}\right)\right].$

$\mathrm{Thus} \mathrm{A}=1; \mathrm{B}=\left(\mathrm{b}-\frac{\mathrm{a}}{\mathrm{RT}}\right)$