States of Matter - Result Question 83
84. The average velocity at $T _1 K$ and the most probable at $T _2 K$ of $CO _2$ gas is $9.0 \times 10^{4} \hspace {1mm} cm s^{-1}$. Calculate the value of $T _1$ and $T _2$
(1990, 4M)
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Solution:
$u _{\text {av }}$ (average velocity) $=\sqrt{\dfrac{8 R T _1}{\pi M}}$
$\Rightarrow \dfrac{9 \times 10^{4}}{100} ms^{-1} =\sqrt{\dfrac{8 \times 8.314 T _1}{3.14 \times 44 \times 10^{-3}}} $
$\Rightarrow T _1 =1682.5 K$
Also, for the same gas
$ \dfrac{u _{av}}{u _{mps}} =\sqrt{\dfrac{8 R T _1}{\pi M}}: \sqrt{\dfrac{2 R T _2}{M}}=\sqrt{\dfrac{8 T _1}{\pi} \times \dfrac{1}{2 T _2}}=\sqrt{\dfrac{4 T _1}{\pi T _2}} $
$\Rightarrow \quad 1 =\sqrt{\dfrac{4 T _1}{\pi T _2}} $
$\Rightarrow T _2 =\dfrac{4 T _1}{\pi}=\dfrac{4 \times 1682.5}{3.14}=2142 K $
$ \text { Hence, } \quad T _1 =1682.5 K, T _2=2142 K$
BETA
