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 $9.0 \times 10^{4} cm s^{-1}$. Calculate the value of $T _1$ and $T _2$
(1990, 4M)
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Answer:
Correct Answer: 84. (12.15)
Solution:
- $u _{\text {av }}$ (average velocity) $=\sqrt{\frac{8 R T _1}{\pi M}}$
$$ \begin{aligned} \Rightarrow & & \frac{9 \times 10^{4}}{100} ms^{-1} & =\sqrt{\frac{8 \times 8.314 T _1}{3.14 \times 44 \times 10^{-3}}} \ \Rightarrow & & T _1 & =1682.5 K \end{aligned} $$
Also, for the same gas
$$ \begin{array}{rlrl} & & \frac{u _{av}}{u _{mps}} & =\sqrt{\frac{8 R T _1}{\pi M}}: \sqrt{\frac{2 R T _2}{M}}=\sqrt{\frac{8 T _1}{\pi} \times \frac{1}{2 T _2}}=\sqrt{\frac{4 T _1}{\pi T _2}} \ \Rightarrow \quad 1 & =\sqrt{\frac{4 T _1}{\pi T _2}} \ \Rightarrow & T _2 & =\frac{4 T _1}{\pi}=\frac{4 \times 1682.5}{3.14}=2142 K \ & \text { Hence, } \quad T _1 & =1682.5 K, T _2=2142 K \end{array} $$
