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NEET-XII-Physics

24: Kinetic Theory of Gases

with Solutions - page 5
Qstn# iv-22 Prvs-QstnNext-Qstn
  • #22
    Hydrogen gas is contained in a closed vessel at 1 atm (100 kPa) and 300 K. (a) Calculate the mean speed of the molecules. (b) Suppose the molecules strike the wall with this speed making an average angle of 45° with it. How many molecules strike each square metre of the wall per second? (a) Calculate the mean speed of the molecules. (b) Suppose the molecules strike the wall with this speed making an average angle of 45° with it. How many molecules strike each square metre of the wall per second?
    Ans : Here,
    P = 105 Pa
    T = 300 K
    For H2
    M = 2×10-3 kg (a) `` \begin{array}{l}\left(\text{a}\right)\text{Mean speed is given by}\\ <v>=\sqrt{\frac{8RT}{\pi M}}\\ =\sqrt{\frac{8\times 8.3\times 300\times 7}{2\times {10}^{-3}\times 22}}\\ =1780{\text{ms}}^{-1}\\ (b) {\text{Let us consider a cubic volume of 1 m}}^{3}.\\ V=1{\text{m}}^{3}\\ \text{Momentum of 1 molecule normal to the striking surface before collision =}mu\,\mathrm{\,sin\,}{45}^{0}\\ \text{Momentum of 1 molecule normal to the striking surface after collision =}-mu\,\mathrm{\,sin\,}{45}^{0}\\ \text{Change in momentum of the molecule =}2mu\,\mathrm{\,sin\,}{45}^{0}=\sqrt{2}mu\\ \text{Change in momentum of n molecules =}2mnu\,\mathrm{\,sin\,}{45}^{0}=\sqrt{2}mnu\\ \text{Let}\Delta \text{t be the time taken in changing the momentum.}\\ \text{Force per unit area due to one molecule =}\frac{\sqrt{2}mu}{\Delta t}=\frac{\sqrt{2}mu}{\Delta t}\\ \text{Observed pressure due to collision by n molecules =}\frac{\sqrt{2}mnu}{\Delta t}={10}^{5}\\ n=\raisebox{1ex}{$\frac{\sqrt{2}mnu}{\Delta t}$}\!\left/ \!\raisebox{-1ex}{$\frac{\sqrt{2}mu}{\Delta t}$}\right.=\frac{{10}^{5}}{\sqrt{2}mu}\\ 6.0\times {10}^{23}\text{molecules}=\text{2}\times {\text{10}}^{-3}\text{kg}\\ \text{1 molecule =}\frac{\text{2}\times {\text{10}}^{-3}}{6\times {10}^{23}}=3.3\times {10}^{-27}\text{kg}\\ \Rightarrow n=\frac{{10}^{5}}{\sqrt{2}\times 3.3\times {10}^{-27}\times 1780}=1.2\times {10}^{28}\end{array}``
    Page No 35: (a) `` \begin{array}{l}\left(\text{a}\right)\text{Mean speed is given by}\\ <v>=\sqrt{\frac{8RT}{\pi M}}\\ =\sqrt{\frac{8\times 8.3\times 300\times 7}{2\times {10}^{-3}\times 22}}\\ =1780{\text{ms}}^{-1}\\ (b) {\text{Let us consider a cubic volume of 1 m}}^{3}.\\ V=1{\text{m}}^{3}\\ \text{Momentum of 1 molecule normal to the striking surface before collision =}mu\,\mathrm{\,sin\,}{45}^{0}\\ \text{Momentum of 1 molecule normal to the striking surface after collision =}-mu\,\mathrm{\,sin\,}{45}^{0}\\ \text{Change in momentum of the molecule =}2mu\,\mathrm{\,sin\,}{45}^{0}=\sqrt{2}mu\\ \text{Change in momentum of n molecules =}2mnu\,\mathrm{\,sin\,}{45}^{0}=\sqrt{2}mnu\\ \text{Let}\Delta \text{t be the time taken in changing the momentum.}\\ \text{Force per unit area due to one molecule =}\frac{\sqrt{2}mu}{\Delta t}=\frac{\sqrt{2}mu}{\Delta t}\\ \text{Observed pressure due to collision by n molecules =}\frac{\sqrt{2}mnu}{\Delta t}={10}^{5}\\ n=\raisebox{1ex}{$\frac{\sqrt{2}mnu}{\Delta t}$}\!\left/ \!\raisebox{-1ex}{$\frac{\sqrt{2}mu}{\Delta t}$}\right.=\frac{{10}^{5}}{\sqrt{2}mu}\\ 6.0\times {10}^{23}\text{molecules}=\text{2}\times {\text{10}}^{-3}\text{kg}\\ \text{1 molecule =}\frac{\text{2}\times {\text{10}}^{-3}}{6\times {10}^{23}}=3.3\times {10}^{-27}\text{kg}\\ \Rightarrow n=\frac{{10}^{5}}{\sqrt{2}\times 3.3\times {10}^{-27}\times 1780}=1.2\times {10}^{28}\end{array}``
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  • #22-a
    Calculate the mean speed of the molecules.
    Ans : `` \begin{array}{l}\left(\text{a}\right)\text{Mean speed is given by}\\ <v>=\sqrt{\frac{8RT}{\pi M}}\\ =\sqrt{\frac{8\times 8.3\times 300\times 7}{2\times {10}^{-3}\times 22}}\\ =1780{\text{ms}}^{-1}\\
  • #22-b
    Suppose the molecules strike the wall with this speed making an average angle of 45° with it. How many molecules strike each square metre of the wall per second?
    Ans : {\text{Let us consider a cubic volume of 1 m}}^{3}.\\ V=1{\text{m}}^{3}\\ \text{Momentum of 1 molecule normal to the striking surface before collision =}mu\,\mathrm{\,sin\,}{45}^{0}\\ \text{Momentum of 1 molecule normal to the striking surface after collision =}-mu\,\mathrm{\,sin\,}{45}^{0}\\ \text{Change in momentum of the molecule =}2mu\,\mathrm{\,sin\,}{45}^{0}=\sqrt{2}mu\\ \text{Change in momentum of n molecules =}2mnu\,\mathrm{\,sin\,}{45}^{0}=\sqrt{2}mnu\\ \text{Let}\Delta \text{t be the time taken in changing the momentum.}\\ \text{Force per unit area due to one molecule =}\frac{\sqrt{2}mu}{\Delta t}=\frac{\sqrt{2}mu}{\Delta t}\\ \text{Observed pressure due to collision by n molecules =}\frac{\sqrt{2}mnu}{\Delta t}={10}^{5}\\ n=\raisebox{1ex}{$\frac{\sqrt{2}mnu}{\Delta t}$}\!\left/ \!\raisebox{-1ex}{$\frac{\sqrt{2}mu}{\Delta t}$}\right.=\frac{{10}^{5}}{\sqrt{2}mu}\\ 6.0\times {10}^{23}\text{molecules}=\text{2}\times {\text{10}}^{-3}\text{kg}\\ \text{1 molecule =}\frac{\text{2}\times {\text{10}}^{-3}}{6\times {10}^{23}}=3.3\times {10}^{-27}\text{kg}\\ \Rightarrow n=\frac{{10}^{5}}{\sqrt{2}\times 3.3\times {10}^{-27}\times 1780}=1.2\times {10}^{28}\end{array}``
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