# How Do You Calculate the Average Force Exerted by a Gas Molecule in a Flask?

• dalitwil
In summary: Think ideal gas law to find the pressure, then do as dextercioby said.Ok so I used PV=nRT to find pressure:P*(.4L) = (0.075)*(8.31)*(300)=467.43PaThen I calculated the surface area by setting the volume equal to 3/4pi r^3:find the radius (=.45708) and finding surface A=4pi*r^2 =2.6Now calculating the F from F=PA gave me 1227. This was incorrect.
dalitwil
I have been having a hard time with the following question:

A 400 mL spherical flask contains 0.075 mol of an ideal gas at a temperature of 300 K. What is the average force-magnitude exerted on the walls of the flask by a single molecule?

I couldn't really start the problem because I have two unknowns: m and v. I need m to find v, and i need v to find F. Because it doesn't give me a molar mass, i am lost on how to find v, wondering if there is another approach I could take? Please help!

U can find the pressure (in Pa) and then compute the average force by multiplying the pressure in Pa with the surface area in m^{2}...U can't use kinetic theory,because computing the momentum transfer by a molecule would require knowledge of the average magnitude of velocity and the molecule's mass...

Daniel.

Think ideal gas law to find the pressure, then do as dextercioby said.

Ok so I used PV=nRT to find pressure:

P*(.4L) = (0.075)*(8.31)*(300)
=467.43Pa

Then I calculated the surface area by setting the volume equal to 3/4pi r^3:

find the radius (=.45708) and finding surface A=4pi*r^2 =2.6

Now calculating the F from F=PA gave me 1227. This was incorrect.
I also tried subsituting the volume with .0004m^3 (1L=10^-3m^3) and recalculating the P and A and it was still incorrect.

I don't understand what i did wrong, I understand the logic behind finding F this way, but my calculations are wrong. Can anyone please help me?

It can't be right.U got to be consistent with your units...Use SI-mKgs...

$$p=\frac{\nu RT}{V} \ [Pa]$$

$$p=\frac{0.075\cdot 10^{-3} \ \mbox{Kmol} \cdot 8314 \ \frac{\mbox{J}}{\mbox{Kmol}\cdot\mbox{K}} \cdot 300 \ \mbox{K}}{0.4\cdot 10^{-3} \ \mbox{m}^{3}}$$

and get that #.

Daniel.

## What is the definition of average force magnitude?

Average force magnitude refers to the average strength or intensity of a force over a specific period of time. It is typically measured in units of Newtons (N) or pounds-force (lbf).

## How is average force magnitude calculated?

To calculate average force magnitude, you divide the total force applied over a given period of time by the duration of that time period. The resulting value is the average force magnitude.

## What is the difference between average force magnitude and instantaneous force magnitude?

The average force magnitude is the overall strength of a force over a period of time, while instantaneous force magnitude is the strength of a force at a specific moment in time. Average force magnitude takes into account the entire duration of the force, while instantaneous force magnitude only considers a single point in time.

## How is average force magnitude used in physics?

Average force magnitude is an important concept in physics and is used in equations such as Newton's second law of motion (F=ma) and work-energy theorem (W=Fd). It helps to describe the overall effect of a force on an object over a period of time.

## Can average force magnitude be negative?

Yes, average force magnitude can be negative. This means that the force is acting in the opposite direction of the positive direction. For example, if a force is applied in the negative x-direction, its average force magnitude would be negative.

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