Physics MCQs for NEET — Practice Questions with Answers

(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,

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

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

$$\begin{gathered} \left(\mathrm{A}\right). \mathrm{The} \mathrm{mean} \mathrm{free} \mathrm{path} \mathrm{of} \mathrm{gas} \mathrm{molecules}, \\ \mathcal{l}=\frac{1}{\sqrt{2}{\mathrm{\pi n\sigma }}^2}; \mathcal{l}\propto \frac{1}{{\mathrm{n\sigma }}^2}; \mathcal{l}\propto \frac{1}{{\mathrm{P\sigma }}^2} \\ 6990 \mathrm{cm} \propto \frac{1}{{10}^{-6}\times 4^2}; \mathrm{x} \mathrm{cm} \propto \frac{1}{{10}^{-3}\times 2^2} \\ \frac{6690}{\mathrm{x}}=\frac{{10}^6}{16}\times {10}^{-3}\times 4=250 \\ \therefore \mathrm{x}=\frac{6990}{250}=27.96 \mathrm{cm}\approx 28 \mathrm{cm}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{C}\right). \mathrm{As} \mathrm{per} \mathrm{equation} \mathrm{of} \mathrm{state} \\ \frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2} ; {\mathrm{P}}_2=\frac{{\mathrm{P}}_1}{2} ; {\mathrm{T}}_2=2{\mathrm{T}}_1 \\ \mathrm{or} {\mathrm{V}}_2=\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\times \frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}\times {\mathrm{V}}_1=\frac{2{\mathrm{T}}_1}{{\mathrm{T}}_1}\times \frac{{\mathrm{P}}_1}{{\mathrm{P}}_1/2}\times {\mathrm{V}}_1=4{\mathrm{V}}_1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{A}\right). {\mathrm{V}}_{\mathrm{C}}=3\mathrm{b}, \mathrm{assuming} \mathrm{the} \mathrm{gas} \mathrm{to} \mathrm{obey} \mathrm{van} \mathrm{der} \mathrm{waals}\text{'} \mathrm{equation}. \\ \therefore \mathrm{b}\left(\mathrm{the} \mathrm{covolume}\right)=\frac{0.072}{3}=0.024 \mathrm{litre} {\mathrm{mol}}^{–1} \\ \mathrm{b}=\frac{24 {\mathrm{cm}}^3}{6\mathrm{x}{10}^{23}} \mathrm{per} \mathrm{molecule}, \mathrm{where} {\mathrm{N}}_{\mathrm{A}}\approx 6\times {10}^{23} \\ \mathrm{b}=4\times {10}^{-23} {\mathrm{cm}}^3 \mathrm{per} \mathrm{molecule}=4\times \frac{4}{3} {\mathrm{\pi r}}^3. \\ \frac{4}{3}{\mathrm{\pi r}}^3={10}^{-23} ; {\mathrm{r}}^3=\frac{3}{4\mathrm{\pi }}\times {10}^{-23}; \mathrm{r}={\left(\frac{3}{4\mathrm{\pi }}\mathrm{x}{10}^{-23}\right)}^{\frac{\mathbf{1}}{\mathbf{3}}} \mathrm{cm} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{C}\right). 32 \mathrm{gm} {\mathrm{O}}_2=1 \mathrm{mole} {\mathrm{O}}_2 \\ 3 \mathrm{gm} \mathrm{hydrogen}=\frac{3}{2}=1.5 \mathrm{mole} {\mathrm{H}}_2 \\ \mathrm{Hence} \mathrm{total} \mathrm{moles} \mathrm{of} \mathrm{gas} \mathrm{present}=2.5 \\ \therefore \mathrm{volume} \mathrm{of} \mathrm{total} 2.5 \mathrm{moles} \mathrm{of} \mathrm{gas} \mathrm{mix} \mathrm{at} \mathrm{STP} \\ =2.5\times 22.4 = 56 \mathrm{lit}. \end{gathered}$$

(C).  Equal no. of molecules, that means equal no. of moles of gas present.

$$\begin{gathered} \therefore {\mathrm{n}}_{{\mathbf{O}}_{\mathbf{2}}}={\mathrm{n}}_{{\mathbf{N}}_{\mathbf{2}}}=\mathrm{x} \mathrm{say} \\ \therefore {\mathrm{P}}_1\mathrm{V}=\left({\mathrm{n}}_{{\mathbf{O}}_{\mathbf{2}}}+{\mathrm{n}}_{{\mathbf{N}}_{\mathbf{2}}}\right)\mathrm{RT}=2\times \mathrm{RT} \\ {\mathrm{P}}_2\mathrm{V}={\mathrm{n}}_{{\mathbf{O}}_{\mathbf{2}}\mathbf{ }}\mathrm{RT}=\mathrm{x} \mathrm{RT} \\ \therefore \frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}=2 \mathrm{or} {\mathrm{P}}_1=2{\mathrm{P}}_2 \\ \therefore \mathrm{when} \mathrm{nitrogen} \mathrm{is} \mathrm{removed}, \mathrm{final} \mathrm{pressure} \mathrm{will} \mathrm{be} \mathrm{P}/2. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{B}\right). \mathrm{Total} \mathrm{volume}=7 \mathrm{litre}. \mathrm{The} \mathrm{gas} \mathrm{in} \mathrm{the} 3 \mathrm{litre} \mathrm{vessel}, \mathrm{after} \mathrm{mixing}, \mathrm{will} \mathrm{have} \mathrm{a} \mathrm{partial} \mathrm{pressure} \mathrm{of} \\ \frac{3}{7}\times 202 \mathrm{kPa}=\frac{606}{7}\mathrm{kPa}. \\ \mathrm{Similarly}, \mathrm{the} 4 \mathrm{litre} \mathrm{vessel} \mathrm{gas} \mathrm{will} \mathrm{now} \mathrm{have} \mathrm{a} \mathrm{partial} \mathrm{pressure} \mathrm{of} \frac{4}{7}\times 101 \mathrm{kPa}=\frac{404}{7} \mathrm{kPa}. \\ \mathrm{The} \mathrm{total} \mathrm{pressure}=\frac{606}{7}+\frac{404}{7}=\frac{1010}{7} \\ =144\frac{2}{7}\approx 144 \mathrm{kPa}. \end{gathered}$$

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