What will be the number of collision per second in a unit area?

[Mahfuz_Saim]

Homework Statement

One mole of oxygen gas is enclosed in a vessel at 27°C and 1atm pressure. The molecules are assumed to be moving with RMS seed. What will be the number of collision per second which the molecules make with the unit area of the vessel?

Relevant Equations

PV=nRT
v (rms) = √(M/(3RT))

By using PV=nRT formula, I have found the volume of the vessel. As far as I have learned to calculate the number of collision in a unit volume. So, it is being difficult for me to find the right way to solve.

I searched on the internet and have got this https://www.toppr.com/ask/question/if-n-is-the-number-of-collisions-per-unit-area-per-second-which-the-molecules/. But I am not understanding that properly. Please tell me how I can approach. Thanks.

 

[vela]

Your textbook likely has a derivation of the ideal gas law. Start there.

 

[Mahfuz_Saim]

I am not getting help from the textbook. If you can, please help.

 

[TSny]

I searched on the internet and have got this https://www.toppr.com/ask/question/if-n-is-the-number-of-collisions-per-unit-area-per-second-which-the-molecules/.

Can you identify the step or steps in the link that you do not understand?

 

[Mahfuz_Saim]

Please answer these questions in the picture. Thank you.

 

[TSny]

View attachment 264733
Please answer these questions in the picture. Thank you.

$P$ is pressure and $p$ is momentum. Recall that pressure is force per unit area: $\frac {F}{A}$ and force is the rate of change of momentum: $F = \frac {\Delta p}{\Delta t}$. So, this equation is just saying that pressure is force per unit area with force expressed as rate of change of momentum.

This is the same as the previous equation where $\Delta p$ in the previous equation has been replaced by an expression that represents the total change in momentum of $N$ molecules that strike the area $A$ during the time $\Delta t$.

$P$ is the pressure of the gas.

 

[Mahfuz_Saim]

Thank you for your help. Can you please help a little more?
The pressure is given in the question which is 1 atm equals to 101325 Pa. But I don't think that is the correct value of p. If you check the link again, you will see that they have considered P as a product of Avogadro number and another number(not understanding from where the number has come). What it really will be? And I think the value of m= 32*10^-3. Am I right? Please reply.

 

[TSny]

The pressure is given in the question which is 1 atm equals to 101325 Pa.

Yes. In the link, it appears that they are considering a situation where the pressure is 2 x 10⁵ Pa rather than 1 atm.

If you check the link again, you will see that they have considered P as a product of Avogadro number and another number(not understanding from where the number has come).

Avogadro's number occurs from converting molar mass M to the mass of a single molecule m.

And I think the value of m= 32*10^-3.

This would be the molar mass M of O₂. What are the units here?

The m that appears in the expression $\frac {P}{2mv_{rms}}$ is not the molar mass. It's the mass of a single molecule.

 

[Mahfuz_Saim]

But they have considered m= 28*10^-3. Maybe they are considering it for Nitrogen gas. That's why I am saying that it can be 32*10^-3 for oxygen.
I don't know whether it will be a good idea or not but can you tell me the value of each variable of the equation according to my question. Maybe I will understand it more easily. Thank you.

 

[TSny]

Yes, they are probably considering nitrogen.

How do you convert molar mass, M, to the mass of a single molecule, m?

Note: Homework helpers are not allowed to provide complete solutions.

 

[Mahfuz_Saim]

I think I have got it. They have used m=M/Avogadro number. Am I right?

 

[TSny]

I think I have got it. They have used m=M/Avogadro number. Am I right?

Yes!

 

[Mahfuz_Saim]

Thank you very much. It was my first question here and have learned a lot from it. Maybe I will ask some more question in the near future.