For one mole of an ideal gas, increasing the temperature from `10^(@)C` to `20^(@)C`A. Increases the average kinetic energy by two timesB. Increases the rms velocity by `sqrt(2)` timesC. Increase the rms velocity by two timesD. Increase both the average kinetic energy and rms velocity, but not significantly.

For one mole of an ideal gas, increasing the temperature from `10^(@)C

1.	For one mole of an ideal gas, increasing the temperature from `10^(@)C` to `20^(@)C`A. Increases the average kinetic energy by two timesB. Increases the rms velocity by `sqrt(2)` timesC. Increase the rms velocity by two timesD. Increase both the average kinetic energy and rms velocity, but not significantly.
Answer» Correct Answer - D Initial temperature `= 283 K` Final temperature `= 293 K` `KE_(1)= (3RT_(1))/(2),KE_(2)=(3RT_(2))/(2)` `(KE_(1))/(KE_(2))=(T_(1))/(T_(2))= (283)/(293)== 0.96` `KE_(2)=1.04 KE_(1)` `v_(rms)=sqrt((3RT)/(M))implies (v_(1))/(v_(2))=sqrt((T_(1))/(T_(2)))impliessqrt((283)/(203))` `v_(2)=v_(1)sqrt((293)/(283))=1.02v_(1)` Both average K.E. and rms velocity increase but not significantly.

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