States of Matter 1 Question 83
84. The average velocity at $T_{1} \mathrm{~K}$ and the most probable at $T_{2} \mathrm{~K}$ of $\mathrm{CO}{2}$ gas is $9.0 \times 10^{4} \mathrm{~cm} \mathrm{~s}^{-1}$. 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) $=\sqrt{\frac{8 R T_{1}}{\pi M}}$
$$ \begin{aligned} \Rightarrow & & \frac{9 \times 10^{4}}{100} \mathrm{~ms}^{-1} & =\sqrt{\frac{8 \times 8.314 T_{1}}{3.14 \times 44 \times 10^{-3}}} \ \Rightarrow & & T_{1} & =1682.5 \mathrm{~K} \end{aligned} $$
Also, for the same gas
$$ \begin{array}{rlrl} & & \frac{u_{\mathrm{av}}}{u_{\mathrm{mps}}} & =\sqrt{\frac{8 R T_{1}}{\pi M}}: \sqrt{\frac{2 R T_{2}}{M}}=\sqrt{\frac{8 T_{1}}{\pi} \times \frac{1}{2 T_{2}}}=\sqrt{\frac{4 T_{1}}{\pi T_{2}}} \ \Rightarrow \quad 1 & =\sqrt{\frac{4 T_{1}}{\pi T_{2}}} \ \Rightarrow & T_{2} & =\frac{4 T_{1}}{\pi}=\frac{4 \times 1682.5}{3.14}=2142 \mathrm{~K} \ & \text { Hence, } \quad T_{1} & =1682.5 \mathrm{~K}, T_{2}=2142 \mathrm{~K} \end{array} $$
