TS EAMCET · Chemistry · States of Matter
What is the approximate most probable velocity of oxygen? If the kinetic energy of one mole of oxygen is \(3741.3 \mathrm{~J}\).
- A \(\sqrt{43851} \mathrm{~J} \mathrm{~kg}{ }^1\)
- B \(\sqrt{48321} \mathrm{~J} \mathrm{~kg}{ }^1\)
- C \(\sqrt{155887} \mathrm{~J} \mathrm{~kg} 1\)
- D \(\sqrt{3950} \mathrm{~J} \mathrm{~kg}^1\)
Answer & Solution
Correct Answer
(C) \(\sqrt{155887} \mathrm{~J} \mathrm{~kg} 1\)
Step-by-step Solution
Detailed explanation
Kinetic energy \((\mathrm{KE})=\frac{3}{2} R T\) \[ \begin{aligned} \text { where, } R & =\text { gas constant } \\ T & =\text { temperature } \\ \therefore \quad R T & =\frac{2}{3}(\mathrm{KE}) \end{aligned} \] Most probable velocity (MP) \(=\sqrt{\frac{2 R T}{M}}\) where,…
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