The mean free path of nitrogen molecules at a pressure of 1-0 atm and temperature 0- C is 0-8 - 10-7 m- If the number of density of molecules is then the molecular diameter isA- 3-2 nmB- 2-3 mmC- 3-2 -D- 3-2 - m

The correct option is C Mean free path λ =0.8×10^ 7m number of molecules per unit volume n=2.7×10^25 per m^3 Substituting these value in λ=1/√(2)π nd^2 we get ...

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Standard XII

Physics

Gas laws from KTG

The mean free...

Question

The mean free path of nitrogen molecules at a pressure of 1.0 atm and temperature $0^{\circ} C$ is $0.8 \times 10^{- 7} m$ . If the number of density of molecules is , then the molecular diameter is

3.2nm

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2.3 mm

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Solution

The correct option is B
Mean free path $λ = 0.8 \times 10^{-} 7 m$
number of molecules per unit volume $n = 2.7 \times 10^{2} 5 p e r m^{3}$
Substituting these value in $λ = \frac{1}{\sqrt{2} π n d^{2}}$ we get $d = \sqrt{1.04 \times 10^{- 10}} m = 3.2 ˙ A$

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The mean free path of nitrogen molecules at a pressure of 1.0 atm and temperature 0∘C is 0.8×10−7m . If the number of density of molecules is , then the molecular diameter is

The correct option is B Mean free path λ=0.8×10−7m number of molecules per unit volume n=2.7×1025perm3 Substituting these value in λ=1√2πnd2 we get d=√1.04×10−10m=3.2˙A

The mean free path of nitrogen molecules at a pressure of 1.0 atm and temperature $0^{\circ} C$ is $0.8 \times 10^{- 7} m$ . If the number of density of molecules is , then the molecular diameter is

The correct option is B
Mean free path $λ = 0.8 \times 10^{-} 7 m$
number of molecules per unit volume $n = 2.7 \times 10^{2} 5 p e r m^{3}$
Substituting these value in $λ = \frac{1}{\sqrt{2} π n d^{2}}$ we get $d = \sqrt{1.04 \times 10^{- 10}} m = 3.2 ˙ A$