If the mean free-path of gaseous molecule if 60 cm at a pressure of 1- 10-4mm mercury what will be its mean free-path when the pressure is increased to 1- 10-2 mm mercury-

The correct option is B 0.6 cm [ λ = RT /√( 2 )π d^ 2 NaP #160; amp; λ P= RT /√( 2 )π d^ 2 Na #160;; ...

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Byju's Answer

Standard XII

Chemistry

Mean Free Path, Collision Frequency

If the mean f...

Question

If the mean free-path of gaseous molecule if 60 cm at a pressure of $1 \times 10^{- 4} m m$ mercury what will be its mean free-path when the pressure is increased to $1 \times 10^{- 2}$ mm mercury.

.0 \times 10^{- 1} c m

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0.6 cm

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.0 \times 10^{- 2} c m

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.0 \times 10^{3} c m

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Solution

The correct option is B 0.6 cm
$\begin{matrix} λ = \frac{R T}{\sqrt{2} π d^{2} N a P} & λ P = \frac{R T}{\sqrt{2} π d^{2} N a} λ_{1} P_{1} = λ_{2} P_{2} 60 \times 10^{- 4} = λ_{2} \times 10^{- 2} λ_{2} = 60 \times 10^{- 2} c m λ_{2} = 0.6 c m \end{matrix}$

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If the mean free-path of gaseous molecule if 60 cm at a pressure of 1×10−4mm mercury what will be its mean free-path when the pressure is increased to 1×10−2 mm mercury.

The correct option is B 0.6 cmλ=RT√2πd2NaPλP=RT√2πd2Naλ1P1=λ2P260×10−4=λ2×10−2λ2=60×10−2cmλ2=0.6cm

If the mean free-path of gaseous molecule if 60 cm at a pressure of $1 \times 10^{- 4} m m$ mercury what will be its mean free-path when the pressure is increased to $1 \times 10^{- 2}$ mm mercury.

The correct option is B 0.6 cm
$\begin{matrix} λ = \frac{R T}{\sqrt{2} π d^{2} N a P} & λ P = \frac{R T}{\sqrt{2} π d^{2} N a} λ_{1} P_{1} = λ_{2} P_{2} 60 \times 10^{- 4} = λ_{2} \times 10^{- 2} λ_{2} = 60 \times 10^{- 2} c m λ_{2} = 0.6 c m \end{matrix}$