2}}}}}{{{U}_{He}}}=\frac{1}{2}=\sqrt{\frac{{{M}_{He}}}{{{M}_{S{{O}_{2}}}}}\frac{{{T}_{S{{O}_{2}}}}}{{{T}_{He}}}}=\sqrt{\frac{4}{64}.\frac{{{T}_{S{{O}_{2}}}}}{300}}\]                                 \[=\frac{4}{64}\frac{{{T}_{S{{O}_{2}}}}}{300}=\frac{1}{4}\];  \[{{T}_{S{{O}_{2}}}}={{1200}^{o}}K\]

"> 2}}}}}{{{U}_{He}}}=\frac{1}{2}=\sqrt{\frac{{{M}_{He}}}{{{M}_{S{{O}_{2}}}}}\frac{{{T}_{S{{O}_{2}}}}}{{{T}_{He}}}}=\sqrt{\frac{4}{64}.\frac{{{T}_{S{{O}_{2}}}}}{300}}\]                                 \[=\frac{4}{64}\frac{{{T}_{S{{O}_{2}}}}}{300}=\frac{1}{4}\];  \[{{T}_{S{{O}_{2}}}}={{1200}^{o}}K\]

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The temperature at which RMS velocity of \[S{{O}_{2}}\] molecules is half that of He molecules at 300 K is            [NTSE 1991]

Joint Entrance Exam - Main (JEE Main) Chemistry in Joint Entrance Exam - Main (JEE Main) 11 months ago

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\[\frac{{{U}_{S{{O}_{2}}}}}{{{U}_{He}}}=\frac{1}{2}=\sqrt{\frac{{{M}_{He}}}{{{M}_{S{{O}_{2}}}}}\frac{{{T}_{S{{O}_{2}}}}}{{{T}_{He}}}}=\sqrt{\frac{4}{64}.\frac{{{T}_{S{{O}_{2}}}}}{300}}\]                                 \[=\frac{4}{64}\frac{{{T}_{S{{O}_{2}}}}}{300}=\frac{1}{4}\];  \[{{T}_{S{{O}_{2}}}}={{1200}^{o}}K\]

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Posted on 07 Sep 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Chemistry in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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