Fnding the rms speed of hydrogen

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Homework Help Overview

The discussion revolves around calculating the root-mean-square (rms) speed of hydrogen molecules based on a known rms speed of nitrogen molecules at a specific temperature. Participants are exploring the relationship between molecular mass and rms speed in the context of gas behavior.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of the rms speed formula and the significance of molecular mass in determining the speed of hydrogen compared to nitrogen. There are attempts to derive temperature from the given rms speed of nitrogen and to calculate the rms speed of hydrogen using its molecular mass.

Discussion Status

Some participants have provided insights into the calculations and noted discrepancies in the results. There is an ongoing exploration of the factors affecting the rms speed, particularly the impact of molecular mass on the final outcome. No consensus has been reached regarding the correct approach or resolution of the problem.

Contextual Notes

Participants are working under the assumption that the temperature remains constant for both gases, and there is a reference to the atomic mass of hydrogen being significantly lighter than that of nitrogen, which may influence the calculations.

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Homework Statement



The rms speed of nitrogen molecules in air at some temperature is 493 m/s. What is the rms speed of hydrogen molecules in air at the same temperature?



Homework Equations



Vrms


The Attempt at a Solution

 
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Finding the rms speed of hydrogen

Homework Statement



The rms speed of nitrogen molecules in air at some temperature is 493 m/s. What is the rms speed of hydrogen molecules in air at some temperature?


Homework Equations



root-mean-square speedvrms= \sqrt{v<sup>2</sup>}=\sqrt{\frac{3kT}{m}}



The Attempt at a Solution



mnitrogen=\frac{28.0 g}{6.02 X 10<sup>23</sup>}=4.65 X 10-26

mhydrogen=\frac{2.0 g}{6.02 X 10<sup>23</sup>}= 3.32 X 10-27

493= \sqrt{\frac{(3)(1.38 X 10<sup>-23</sup>)(T)}{4.65 X 10<sup>-26</sup>}}
T= 233 K

Vrms of hydrogen= \sqrt{\frac{(3)(1.38 X 10<sup>-23</sup>(T)}{3.32 X 10<sup>-27</sup>}}=340.43 m/s

The answer is actually 1840 m/s.

What did I do wrong?
 
Wow, all that work and it didn't come out right!
Better to just think for a bit. The atomic mass for the H2 is lighter by a factor of 14.
So the 3kT/m will be 14 times larger for the hydrogen. And its square root will be sqrt(14) times larger.
 

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