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

$$\begin{gathered} 4. \mathrm{Vander} \mathrm{Waal}\text{'}\mathrm{s} \mathrm{gas} \mathrm{constant} \mathrm{b} = 4 \times \mathrm{total} \mathrm{volume} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{molecules} \mathrm{of} \mathrm{the} \mathrm{gas} \mathrm{in} \mathrm{the} \mathrm{enclosure} \\ \mathrm{or} \mathrm{b}=4\times \mathrm{N}\times \frac{4}{3}\mathrm{\pi }{\left(\frac{\mathrm{d}}{2}\right)}^3 \\ =\frac{2}{3}{\mathrm{\pi Nd}}^3 \\ =\frac{2}{3}\times 3.14\times 6.02\times {10}^{23}\times {\left(2.94\times {10}^{\times -10}\right)}^3 \\ =32\times {10}^{-6}\frac{{\mathrm{m}}^3}{\mathrm{mol}} \end{gathered}$$

$$\begin{gathered} 2. {\left({\mathrm{C}}_{\mathrm{P}}\right)}_{\mathrm{mix}}=\frac{{\mathrm{\mu }}_1{\mathrm{C}}_{{\mathrm{P}}_1}+{\mathrm{\mu }}_2{\mathrm{C}}_{{\mathrm{P}}_2}}{{\mathrm{\mu }}_1+{\mathrm{\mu }}_2} \left({\mathrm{C}}_{{\mathrm{P}}_1}\left(\mathrm{He}\right)=\frac{5}{2}\mathrm{R} \mathrm{and} {\mathrm{C}}_{{\mathrm{P}}_2}\left({\mathrm{H}}_2\right)=\frac{7}{2}\mathrm{R}\right) \\ =\frac{1\times {\displaystyle \frac{5}{2}}\mathrm{R}+1\times {\displaystyle \frac{7}{2}}\mathrm{R}}{1+1}=3\mathrm{R} \\ =3\times 2 = 6 \mathrm{cal}/\mathrm{mol}°\mathrm{C} \\ \therefore \mathrm{Amount} \mathrm{of} \mathrm{heat} \mathrm{needed} \mathrm{to} \mathrm{raise} \mathrm{the} \mathrm{temperature} \mathrm{from} 0°\mathrm{C} \mathrm{to} 100 °\mathrm{C} \\ {\left(∆\mathrm{Q}\right)}_{\mathrm{P}}={\mathrm{\mu C}}_{\mathrm{P}}∆\mathrm{T} = 2\times 6\times 100=1200 \mathrm{cal} \end{gathered}$$

1. Average kinetic energy per molecule per degree of freedom $=\frac{1}{2} \mathrm{kT}$. Since both the gases are

    diatomic and at same temperature (300 K), both will have the same number of rotational degree of

    freedom i.e. two. Therefore, both the gases will have the same average rotational kinetic energy per

    molecule $\left(=2\times \frac{1}{2} \mathrm{kT}=\mathrm{kT}\right)$ . Thus $\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\frac{1}{1}$ .

$$\begin{gathered} 4. \mathrm{Total} \mathrm{internal} \mathrm{energy} \mathrm{of} \mathrm{system} ={\mathrm{U}}_{\mathrm{oxygen}}+{\mathrm{U}}_{\mathrm{argon}}={\mathrm{\mu }}_1\frac{{\mathrm{f}}_1}{2}\mathrm{RT} + {\mathrm{\mu }}_2\frac{{\mathrm{f}}_2}{2}\mathrm{RT} \\ = 2\frac{5}{2}\mathrm{RT} + 4\frac{3}{2}\mathrm{RT} \\ = 5 \mathrm{RT} + 6 \mathrm{RT} \\ = 11 \mathrm{RT} \left[\mathrm{As} {\mathrm{f}}_1=5 \left(\mathrm{for} \mathrm{oxygen}\right) \mathrm{and} {\mathrm{f}}_2=3\left(\mathrm{for} \mathrm{argon}\right)\right] \end{gathered}$$

$$\begin{gathered} 3. \mathrm{At} \mathrm{T} \mathrm{K}, \mathrm{pressure} \mathrm{of} \mathrm{gas} \left(\mathrm{P}\right) \mathrm{in} \mathrm{the} \mathrm{jar} \mathrm{pressure} = \mathrm{Total} \mathrm{pressure} - \mathrm{saturated} \mathrm{vapour} \mathrm{pressure} \\ \Rightarrow \mathrm{P}=\left(830-30\right)=800 \mathrm{mm} \mathrm{of} \mathrm{Hg} \\ \mathrm{New} \mathrm{temperature} \mathrm{T}\text{'}=\left(\mathrm{T}-\frac{\mathrm{T}}{100}\right)=\frac{99 \mathrm{T}}{100} \\ \mathrm{Using} \mathrm{Charle}\text{'}\mathrm{s} \mathrm{law} \frac{\mathrm{P}}{\mathrm{T}}=\frac{\mathrm{P}\text{'}}{\mathrm{T}} \Rightarrow \mathrm{P}\text{'}=\frac{\mathrm{PT}\text{'}}{\mathrm{T}} \\ =\frac{800\times 99 \mathrm{T}}{100 \mathrm{T}}= 792 \mathrm{mm} \mathrm{of} \mathrm{Hg} \\ \mathrm{Saturated} \mathrm{vapour} \mathrm{pressure} \mathrm{at} \mathrm{T}\text{'}=25 \mathrm{mm} \mathrm{of} \mathrm{Hg} \\ \therefore \mathrm{Total} \mathrm{pressure} \mathrm{in} \mathrm{the} \mathrm{jar} = \mathrm{Actual} \mathrm{pressure} \mathrm{of} \mathrm{gas} + \mathrm{Saturated} \mathrm{vapour} \mathrm{pressure} \\ = 792 + 25 = 817 \mathrm{mm} \mathrm{of} \mathrm{Hg}. \end{gathered}$$

$$\begin{gathered} 3. \mathrm{Given} \mathrm{R} = 8.3\frac{\mathrm{Joule}}{\mathrm{mol} \mathrm{K}}=\frac{8.3}{4.2}\frac{\mathrm{cal}}{\mathrm{mol} \mathrm{K}}=2\frac{\mathrm{cal}}{\mathrm{mol} \mathrm{K}} \\ {\mathrm{C}}_{\mathrm{V}}={\mathrm{C}}_{\mathrm{P}}-\mathrm{R}=\left(7.2-2\right) \approx 5 \mathrm{cal}/\mathrm{mol} \mathrm{l}°\mathrm{C} \\ \mathrm{At} \mathrm{constant} \mathrm{volume} \mathrm{required} \mathrm{heat} {\left(∆\mathrm{Q}\right)}_{\mathrm{V}}={\mathrm{\mu C}}_{\mathrm{V}}∆\mathrm{T}=5\times 5\times 10=250 \mathrm{cal} \end{gathered}$$

$$\begin{gathered} 4. {\left(∆\mathrm{Q}\right)}_{\mathrm{P}}={\mathrm{\mu C}}_{\mathrm{P}}∆\mathrm{T} \\ \Rightarrow 207=1\times {\mathrm{C}}_{\mathrm{P}}\times 10 \\ \Rightarrow {\mathrm{C}}_{\mathrm{P}}=20.7\frac{\mathrm{Joule}}{\mathrm{mol}‐\mathrm{K}}. \mathrm{Also} {\mathrm{C}}_{\mathrm{P}}-{\mathrm{C}}_{\mathrm{V}}=\mathrm{R} \\ \Rightarrow {\mathrm{C}}_{\mathrm{V}}={\mathrm{C}}_{\mathrm{P}}-\mathrm{R} =20.7-8.3=12.4\frac{\mathrm{Joule}}{\mathrm{mole}‐\mathrm{K}} \\ \mathrm{So}, {\left(∆\mathrm{Q}\right)}_{\mathrm{V}}={\mathrm{\mu C}}_{\mathrm{V}}∆\mathrm{T}=1\times 12.4\times 10=124 \mathrm{J}. \end{gathered}$$

3. At constant temperature $\mathrm{PV}$= constant  $\Rightarrow \mathrm{P}\propto \frac{1}{\mathrm{V}}$

(b) A derivation for mean free path of gas.

• If v is the velocity of the moving molecule then in one second it will collide with all moleculeswith in a distance σ between the centres.
• In one second it sweeps a volume πσ²v where any other molecule will collide with it.
• If n is the total number of molecules per unit volume, then nπσ²v is number of collisions a molecule suffers in one second.
• If v is the distance traversed by molecule in one second then mean free path is given by
       λ = total distance traversed in one second /no. of collision suffered by the molecules
        =v/πσ²vn
        =1/πσ²n
• This expression was derived with the assumption that all the molecules are at rest except the one which is colliding with the others.