Answer:
Let
\[{{N}_{1}}=\] number of molecules per unit volume.
\[A=\]
area
of the wall of the box.
\[{{\upsilon
}_{rms}}^{2}=\upsilon _{x}^{2}+\upsilon _{y}^{2}+\upsilon _{z}^{2}\].
Let
the molecule is moving along x-axis
\[\therefore
\] \[\upsilon _{x}^{2}=\frac{\upsilon _{rms}^{2}}{3}=\frac{1}{3}\times
\frac{3kT}{m}\]
\[(\because
\frac{1}{2}\,m\,\,\upsilon _{rms}^{2}=\frac{3}{2}kT)\]
\[=\frac{kT}{m}\] ? (i)
Average
number of molecules colliding with the wall of the box in time \[\Delta
t=\frac{{{N}_{i}}({{V}_{x}}\Delta t)A}{2}\]
\[=\frac{{{N}_{i}}\,A\Delta
t}{2}\sqrt{\frac{kT}{A}}\]
If these molecules
colliding the hole, they leave the box and the air molecules outside the box
enter the box. Average number of air molecules entering the box in time
\[\Delta
t=\frac{{{N}_{2}}A\Delta t}{2}\sqrt{\frac{kT}{m}}\] (if temperature inside and
outside the box is same)
\[\therefore
\] Net
number of molecules leaving the box \[=\frac{({{N}_{1}}-{{N}_{2}})A\Delta
t}{2}\sqrt{\frac{kT}{m}}\]
Let
after time \['t',\] the
pressure change to \[P_{1}^{'}\]. Since pressure \[\propto \] number of
particles colliding the wall, therefore, \[\frac{({{N}_{1}}-{{N}_{2}})At}{2}\sqrt{\frac{kT}{m}}=({{N}_{1}}-N_{1}^{'})V\]
or \[\frac{({{P}_{1}}-{{P}_{2}})At}{2}\sqrt{\frac{kT}{m}}=({{P}_{1}}-P_{1}^{'})V\]
or \[t=\frac{2({{P}_{1}}-P_{1}^{'})V}{({{P}_{1}}-{{P}_{2}})A}\sqrt{\frac{m}{kT}}=\frac{2(1.5-1.4)}{(1.5-1)}\]
\[\times
\frac{1}{1\times {{10}^{-8}}}\sqrt{\frac{46.7\times {{10}^{-27}}}{1.38\times
{{10}^{-27}}\times 300}}\]
\[=1.38\times
{{10}^{-5}}\,s\]
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