A gas molecule at the surface of earth happens to have rms speed for the gas at $$0oC$$. Suppose it went straight up without colliding with other molecules, how high would it rise? Mass of the molecule is $$4.65$$×$$10$$−$$26kg$$. Boltzmann’s constant $$=1.38$$×$$10$$−$$23J$$ K−$$1$$.

Question

A gas molecule at the surface of earth happens to have rms speed for the gas at $$0oC$$. Suppose it went straight up without colliding with other molecules, how high would it rise? Mass of the molecule is  $$4.65$$×$$10$$−$$26kg$$. Boltzmann’s constant $$=1.38$$×$$10$$−$$23J$$  K−$$1$$.

Infinity Learn · Accepted Answer

$$vRMS=3kBTM=3$$×$$1.38$$×$$10$$−$$23$$×$$2734.65$$×$$10$$−$$26vRMS=493ms$$−$$1Particle$$ moves upwards under the influence of gravity and stops (temporarily) after some time.$$h=u22a=u22g=(493)22$$×$$10$$⇒$$h=1.22$$×$$104m$$

A gas molecule at the surface of earth happens to have rms speed for the gas at 0^oC. Suppose it went straight up without colliding with other molecules, how high would it rise? Mass of the molecule is $4.65 \times 10^{- 26} k g .$ Boltzmann’s constant $= 1.38 \times 10^{- 23} J K^{- 1}$ .

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A gas molecule at the surface of earth happens to have rms speed for the gas at 0oC. Suppose it went straight up without colliding with other molecules, how high would it rise? Mass of the molecule is 4.65×10−26kg. Boltzmann’s constant =1.38×10−23J K−1.

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A gas molecule at the surface of earth happens to have rms speed for the gas at 0^oC. Suppose it went straight up without colliding with other molecules, how high would it rise? Mass of the molecule is $4.65 \times 10^{- 26} k g .$ Boltzmann’s constant $= 1.38 \times 10^{- 23} J K^{- 1}$ .