V}_{rms}}=\SQRT{\frac{3RT}{M}}\] i.e., \[\frac{{{V}_{1}}}{{{V}_{2}}}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\] \[\frac{5\times {{10}^{4}}}{10\times {{10}^{4}}}=\frac{1}{2}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\] \[\therefore {{T}_{2}}=4{{T}_{1}}\]

"> V}_{rms}}=\SQRT{\frac{3RT}{M}}\] i.e., \[\frac{{{V}_{1}}}{{{V}_{2}}}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\] \[\frac{5\times {{10}^{4}}}{10\times {{10}^{4}}}=\frac{1}{2}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\] \[\therefore {{T}_{2}}=4{{T}_{1}}\]

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By how many folds the temperature of a gas would increase when the root mean square velocity of the gas molecules in a container of fixed volume is increased from \[5\times {{10}^{4}}cm/s\] to \[10\times {{10}^{4}}cm/s\]?

NEET Chemistry in NEET . 8 months ago

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[d] r.m.s. velocity \[{{V}_{rms}}=\SQRT{\frac{3RT}{M}}\] i.e., \[\frac{{{V}_{1}}}{{{V}_{2}}}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\] \[\frac{5\times {{10}^{4}}}{10\times {{10}^{4}}}=\frac{1}{2}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\] \[\therefore {{T}_{2}}=4{{T}_{1}}\]

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Posted on 13 Sep 2024, this text provides information on NEET related to Chemistry in NEET. 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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