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

AI Thread Summary
To calculate the average force exerted by a gas molecule in a flask, start by using the ideal gas law (PV=nRT) to determine the pressure. The pressure is found to be approximately 467.43 Pa for 0.075 mol of gas at 300 K in a 400 mL flask. Next, calculate the surface area of the flask, which requires finding the radius from the volume. The force can then be computed using the formula F=PA, but it is crucial to ensure all units are consistent, particularly using SI units. The discussion emphasizes the importance of unit consistency and correct calculations to arrive at the correct average force.
dalitwil
Messages
23
Reaction score
0
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!
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top