Answer:
(a)
\[{{V}_{rms}}=\]
\[=\sqrt{\frac{{{\mu
}_{1}}{{({{\upsilon }_{1}})}^{2}}+{{\mu }_{2}}\,{{({{\upsilon
}_{2}})}^{2}}+{{\mu }_{3}}{{({{\upsilon }_{3}})}^{2}}+{{\mu }_{4}}{{({{\upsilon
}_{4}})}^{4}}+\,{{\mu }_{5}}{{({{\upsilon }_{5}})}^{2}}}{{{\mu }_{1}}+\,{{\mu
}_{2}}+\,{{\mu }_{3}}+{{\mu }_{4}}+\,{{\mu }_{5}}}}\]
\[=\sqrt{\frac{10\,{{(200)}^{2}}+\,20\,{{(400)}^{2}}\,+40\,{{(600)}^{2}}+\,20\,{{(800)}^{2}}\,+10{{(1000)}^{2}}}{100}}\]
\[=\sqrt{\frac{4\times
\,{{10}^{5}}+32\,\times \,{{10}^{5}}+144\,\times {{10}^{5}}+128\,\times
{{10}^{5}}+\,100\times \,{{10}^{5}}}{10}}\]
\[=\,\sqrt{\frac{408\,\times
\,{{10}^{5}}}{10}}\,=-\,638.7\,m{{s}^{-1}}\]
Now
\[\frac{1}{2}\,m\upsilon _{rms}^{2}\,=\frac{3}{2}\,kT\]
\[\therefore
\]\[T\,=\,\frac{m\upsilon _{rms}^{2}}{3k}\,\,=\,\frac{3\times
\,{{10}^{-26}}\times \,{{(638.7)}^{2}}}{3\times \,1.38\,\times
\,{{10}^{-28}}}\]
\[=295.5\,K\]
(b)
Since 10% molecules with speed \[1000\text{ }m{{s}^{1}}\] escape, so now the
new \[{{\upsilon }_{rms}}\] is given by \[\upsilon {{'}_{rms}}\].
\[=\sqrt{\frac{10\,{{(200)}^{2}}+\,20{{(400)}^{2}}+40\,{{(600)}^{2}}+20\,(800)}{90}}\]
\[=585\text{
}m\text{ }{{s}^{1}}\]
and
new temperature, \[T\,=\,\frac{m\,{{(\upsilon {{'}_{rms}})}^{2}}}{3k}\]
\[=\,\frac{3\times
\,{{10}^{-26}}\times \,{{(585)}^{2}}}{3\times 1.38\,\times \,{{10}^{-23}}}\]
\[=247.989\simeq
248\text{ }K.\]
You need to login to perform this action.
You will be redirected in
3 sec