Collisions of molecules: Calculating Collision Rates in a Gas Box

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SUMMARY

The discussion focuses on calculating the rate of collisions of nitrogen molecules in a 10 cm³ box at 20°C. Key equations include the Boltzmann constant (kb = 1.38 x 10^-23), molecular mass of N2 (28 u), and the formula for the rate of collisions: rate = (1/2)(N/V)Av. The solution involves determining the number of moles using Avogadro's number (6.02 x 10^23 mol^-1) and applying the ideal gas law to find the number of molecules in the given volume.

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  • Knowledge of kinetic theory of gases
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fredrick08
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[SOLVED] collisions of molecules..

Homework Statement


a 10cm^3 box contains nitrogen at 20degreeC. What is the rate of collisions on one wal of he box.


Homework Equations


kb=1.38x10^-23
1u=1.661x10^-27
molecular mass of N2=28u
avagadros number=6.02x10^23mol^-1
vrms=sqrt((3kbT/m))
rate of collisions=(1/2)(N/V)Av

The Attempt at a Solution


omg i got no idea where even start with this one... I am pretty sure i could do it if they gave me the mass of the gas or the number of molecules... but I am just stuck... y do they give me avagdros number? I am srry I am just really confused bout this one... anyone please help??
 
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n=N/Na?? but i don't know n?
 
Remember, an ideal gas (such as N2) has a molar volume of 22,400 cubic centimeters per mole at STP (Avogadro's Law).

We know the volume of the box, so we know how many moles of ideal gas there are at 20°C, so we know how many molecules there are.
 
Last edited:
oh ok ty
 
You're very welcome.
Once you're satisfied with the problem, please change the title to "[SOLVED] Collisions of molecules".
Good luck!
 

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