# Force exerted by a gas

1. Dec 2, 2007

### sun

[SOLVED] Force exerted by a gas

1. The problem statement, all variables and given/known data

A sealed cubical container 30.0 cm on a side contains three times Avogadro's number of molecules at a temperature of 23.0°C. Find the force exerted by the gas on one of the walls of the container.

2. Relevant equations
F=Nm/d(v^2)
P=F/A

3. The attempt at a solution
There always seems to be a missing variable to plug into the equations. Are there any hints that can get me pointed in the correct direction?

thanks

2. Dec 2, 2007

3. Dec 2, 2007

### sun

V of one mole of ideal gas=22.4x10^-3m^3
V of container=2.7x10^-5
T=296K
R=8.31j/mol*k
A=.54m^3
n=?

Last edited: Dec 2, 2007
4. Dec 2, 2007

### sun

This is what i've done:
V=m(He)/density(He))
m=.004Kg/(6.02x10^-23*3)=2.21x10^-27kg
V(He)=2.21x10^-27kg/1.79x10^-1Kg/m^3=1.23x10^-26m^3

Am i on the correct path? If so, all i need to figure out is how many moles there are, correct? Then i can solve for P.

5. Dec 2, 2007

### sun

does this essentially mean there are 3 moles?

6. Dec 2, 2007

### sun

any help would be appreciated :)

7. Dec 3, 2007

### sun

I'm not sure how to calculate the moles.

I tried V(f)/V(i)=n(f)T(f)/n(i)T(i)

when i solved for n(i) i got .0011mol. But i guess that isn't correct.

8. Dec 3, 2007

### sun

I also tried PV=NkT

would N equal 3*(6.02*10^23)?
If so, what i did was P(.027m^3)=1.81*10^24(1.381*10^-23)(296)
i got 2.74*10^5

Then i multiplied that by the area .54 and got 1.47*10^5N, but thats wrong.

9. Dec 3, 2007

### sun

argg still lost and confused

10. Dec 3, 2007

### Kurdt

Staff Emeritus
Yes 3 times Avogadro's number means there are 3 moles of the gas. You want to use the ideal gas law with the molar constant R. Thus:

$$PV = n R T$$

If you work out the pressure you can find the force exerted on one wall.

11. Dec 3, 2007

### sun

this is what i've done:
P(.027m^3)=(3mol)(8.31J/mol*K)(296K)

P=2.73*10^5

F=2.73*10^5(.54)=1.47*10^55 N, but that is incorrect. any idea as to what i may be doing wrong.

And thank you for your help.

12. Dec 3, 2007

### sun

do i divide 1.47*10^5 by 6, b/c there are 6 sides to the cube?

13. Dec 3, 2007

### Kurdt

Staff Emeritus
The question asks for the force on one side, and the 0.54 you've used is the total area of the box.