TS EAMCET · Chemistry · States of Matter
At \(300 \mathrm{~K}\) the average velocity of a gas is \(3 \times 10^2 \mathrm{~cm} / \mathrm{s}\). The average velocity (in \(\mathrm{cm} / \mathrm{s}\) ) of it at \(1200 \mathrm{~K}\) is
- A \(6 \times 10^2\)
- B \(4 \times 10^2\)
- C \(8 \times 10^2\)
- D \(1 \times 10^3\)
Answer & Solution
Correct Answer
(A) \(6 \times 10^2\)
Step-by-step Solution
Detailed explanation
Using this formula, \(\mu_{\mathrm{av}}=\sqrt{\frac{8 R T}{\pi M}}\) where, \(\mu_{\mathrm{av}}=\) average velocity of a gas \[ \begin{aligned} R & =\text { gas constant } \\ T & =\text { temperature } \\ M & =\text { mass } \\ \pi & =3.14 \end{aligned} \] At temperature…
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