Explanation
Given,
$$\lambda =\dfrac{hc}{eV}\ \ \ \ where,\ V=potential$$
$$\lambda \ \alpha \ \dfrac{1}{V}$$
$${{10}^{-10}}\ \alpha \ \dfrac{1}{150}\ ......\ (1)$$
$$\lambda \ \alpha \ \dfrac{1}{600}\ ......\ (2)$$
Divide (2) by (1)
$$ \dfrac{\lambda }{{{10}^{-10}}}=\dfrac{150}{600}=\dfrac{1}{4} $$
$$ \Rightarrow \lambda =0.25\times {{10}^{-10}}m\ =0.25\ {{A}^{o}} $$
For perfect gas,
Kinetic Energy of n molecule, $$K.E=n\left( \dfrac{1}{2}{{K}_{B}}T \right)$$
Where, $${{K}_{B}}$$ is Boltzmann constant
If there is no loss of energy.
Total kinetic energy of mixture is sum of each gas kinetic energy.
$$ {{n}_{total}}K.{{E}_{total}}={{n}_{1}}K.{{E}_{1}}+{{n}_{2}}K.{{E}_{2}}+{{n}_{3}}K.{{E}_{3}} $$
$$ \left( {{n}_{1}}+{{n}_{2}}+{{n}_{3}} \right)\left( \dfrac{1}{2}{{K}_{B}}T \right)={{n}_{1}}\left( \dfrac{1}{2}{{K}_{B}}{{T}_{1}} \right)+{{n}_{2}}\left( \dfrac{1}{2}{{K}_{B}}{{T}_{2}} \right)+{{n}_{3}}\left( \dfrac{1}{2}{{K}_{B}}{{T}_{3}} \right) $$
$$ T=\dfrac{{{n}_{1}}{{T}_{1}}+{{n}_{2}}{{T}_{2}}+{{n}_{3}}{{T}_{3}}}{{{n}_{1}}+{{n}_{2}}+{{n}_{3}}} $$
From ideas gas equation
$$ PV=nRT $$
$$ PV=\dfrac{m}{{{M}_{o}}}RT $$
$$ m=\dfrac{PV{{M}_{o}}}{RT}\ $$
In first event
$$ {{m}_{1}}=\dfrac{{{P}_{1}}V{{M}_{o}}}{R{{T}_{1}}} $$
$$ 6\ gram=\dfrac{PV{{M}_{o}}}{R\times 400}\ $$
$$ \dfrac{{{\operatorname{PVM}}_{o}}}{R}=400\times 6\ .\ .....\ (1) $$
In the second event when hole is made:
$$ {{m}_{2}}=\dfrac{{{P}_{2}}V{{M}_{o}}}{R{{T}_{2}}} $$
$$ {{m}_{2}}=\dfrac{\dfrac{P}{2}V{{M}_{o}}}{R\times 300}\ =\dfrac{1}{300\times 2}\dfrac{PV{{M}_{o}}}{R} $$
$$ {{m}_{2}}=\dfrac{1}{300\times 2}\times \left( 400\times 6 \right)=4\ gram $$
Leak out oxygen is $$\left( 6gram-4\,gram \right)\ =\ 2\,gram$$
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