The root mean square speed of molecules of nitrogen gas is v at a certain temperature. When the temperature is doubled, the molecules dissociate into individual atoms. Calculate the factor by which root mean square speed of the individual atoms increases.

The root mean square speed of molecules of nitrogen gas is v at a cert

1.	The root mean square speed of molecules of nitrogen gas is v at a certain temperature. When the temperature is doubled, the molecules dissociate into individual atoms. Calculate the factor by which root mean square speed of the individual atoms increases.
Answer» `u_(rms)=sqrt((3RT)/(M))` `(u_(N))/(u_(N_(2)))=sqrt((T_(N))/(T_(N_(2)))xxM_(N_(2))/(M_(N_(2))))" or "(u_(N))/(v)=sqrt((2T)/(T)xx(28)/(14))=sqrt(4)=2` or `u_(N)=2v` Thus, speed of individual atoms becomes two times.

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The root mean square speed of molecules of nitrogen gas is v at a certain temperature. When the temperature is doubled, the molecules dissociate into individual atoms. Calculate the factor by which root mean square speed of the individual atoms increases.

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