Blend Of Thermodynamics And Mechanics

Classical Mechanics Level 4

If nitrogen gas molecule goes straight up with its RMS (Root Mean Square) speed at 0 ∘ C { 0 }^{ \circ }\text{C} from the surface of the earth and there are no collisions with other molecules, then it will rise to an approximate height of __________ \text{\_\_\_\_\_\_\_\_\_\_} .

12 km \text{12 km} 12 m \text{12 m} 8 m \text{8 m} 8 km \text{8 km}

Swapnil Das by Swapnil Das , 20, India

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1 solution

Rajdeep Dhingra
Mar 4, 2016

First we calculate speed of the N 2 N_2 molecule.
We know that v r m s = 3 R T M v_{rms} = \sqrt{\frac{3RT}{M}} where R is Universal Gas Constant , T is the Temperature in Kelvin and M is the Molecular Mass in kg m o l − 1 \text{kg }{mol}^{-1} . For N 2 N_2 is 28 × 10 − 3 28 \times {10}^{-3} .

Plugging in the values we get v r m s 2 = 243184.5 m 2 s − 2 {v_{rms}}^2 = 243184.5 m^2s^{-2} .
To calculate maximum height, H , reached we know that H = v 2 2 g H = \frac{v^2}{2g} .
Plugging in values we get that H = 12166.3515 m H = \boxed{12166.3515 m} . Which is approximately 12 k m \boxed{12 km} .

Q . E . D . Q.E.D.

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Thank you.

Swapnil Das - 5 years, 3 months ago

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Welcome. I hope you liked the solution.

Rajdeep Dhingra - 5 years, 3 months ago

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Of course, I did.

Swapnil Das - 5 years, 3 months ago

I find this problem hard to understand.

Are you saying that the initial velocity is v r m s v_{rms} and then that nothing else matters? If so, that should be made more obvious. Currently, I read this question as "Nitrogen gas is released at the surface of the earth at 0 ∘ ^\circ C. Find the height to which it rises".

Calvin Lin Staff - 5 years, 3 months ago

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Yes sir it only depends on v r m s v_{rms} .

Rajdeep Dhingra - 5 years, 3 months ago

Great chance of marking the q ans as 12m because of not taking in consider mass in kg . One more thing that rms value has been very wisely picked since ht is proportional to square of velocity so in case u did nt mentioned RMS speed in q and rather used the term avg ht. Then even ans remained same and q had became a little tricky since we would take velocities of particle as v1 v2 v3 v4 and then found it proportional to vrms^2 and better u mention in q avg speed rather tha n temp to let it to the solvers to understand which velocity to take in account

aryan goyat - 4 years, 7 months ago

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