At sea level, a nitrogen molecule in the air has an average translational kinetic energy of 6.20 x 10^-21 J.  Its mass is 4.70 x 10^-26 kg.If the molecule could shoot straight up without striking...

At sea level, a nitrogen molecule in the air has an average translational kinetic energy of 6.20 x 10^-21 J.  Its mass is 4.70 x 10^-26 kg.

If the molecule could shoot straight up without striking other air molecules, how high would it rise and what was the molecule's intial speed?

Asked on by shayaan

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

This can be done using the law of energy conservation.

At the moment of shoot the molecule only has kinetic energy. When it reach the highest level kinetic energy will become zero (velocity is zero) and the potential energy will be there.

 

Let us assume the sea level as the datum level.

By law of energy conservation

Initial energy of the system = Final energy of the system

Kinetic energy = Potential energy

`6.2xx10^(-21) = 4.7xx10^(-26)xx9.81xxH`

                   `H = 13447m`

 

Kinetic energy `= 6.2xx10^(-21)`

`1/2xx 4.7xx10^(-26)xxV^2 = 6.2xx10^(-21)`

                                      `V = 513.64`

 

So the Nitrogen mole will reach a maximum height of 13447m and its initial velocity will be 513.64m/s.

 

Note:

According to the values these are very much unrealistic.We cant expect a N mole to go 13000+ height with a shoot. Here we have to assume a environment free of air pressure,wind and we have to assume that N mole has not loss its energy by any mean other than converting kinetic energy to potential energy.

Sources:

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