Force exerted on walls of metal box

In summary, the problem involves calculating the force exerted on the walls of a metal box filled with air at different temperatures and pressures. Using the equation PV=nRT, the final pressure can be determined and then used to find the force using the equation P x (length)^2 x 6. However, there may be a discrepancy in the final calculation due to significant figures.
  • #1
Rubidium
17
0
force exerted on walls of metal box...please help!

1. Homework Statement
A cubic metal box that has 20-cm-long edges contains air at a pressure of 1.0 atm and a temperature of 300 K. The box is sealed so that the enclosed volume remains constant, and it is heated to a temperature of 400 K. Find the force due to the internal air pressure on each wall of the box.




2. Homework Equations
PV=nRT




3. The Attempt at a Solution
F=[tex]\left|[/tex][tex]\Delta[/tex][tex]\vec{p}[/tex]/[tex]\Deltat[/tex][tex]\right|[/tex]
v=[tex]\sqrt{3RT/M}[/tex]=[tex]\sqrt{3(8.314 J/molxK)(300K)/0.029kg/mol)}[/tex]=507.957 m/s
Delta[/tex][tex]\vec{p}[/tex]/[tex]=mv^2=(9423.9987g)(507.957m/s)=4.787x10^6
I just need delta t
 
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  • #2
Forget about v. You don't need to count individual particles, PV=nRT will give you the pressure. Pressure is force per unit area.
 
  • #3
How do I use all the information given in the problem to find that?
 
  • #4
PV=nRT. How many variables in this equation change when the temperature is varied in your problem?
 
  • #5
I think the pressure would change but the volume remains constant in the box. Is the pressure given in the problem negligible or should I be able to use it?
 
  • #6
Only P and T change, right. But you are given the initial P is 1atm. What's the final P? You certainly shouldn't neglect the pressure change!
 
  • #7
got it...thanks!
 
  • #8


I'm having severe trouble on this one:

I know that P1(T1)/T2 = P2

so P2 x (length)^2 x (6 sides) = Internal Pressure Force.

But I'm not getting the right values. Even if I just do P2 x (Length^2) i don't get the right answer.
 
  • #9


MeKnos said:
I'm having severe trouble on this one:

I know that P1(T1)/T2 = P2

so P2 x (length)^2 x (6 sides) = Internal Pressure Force.

But I'm not getting the right values. Even if I just do P2 x (Length^2) i don't get the right answer.

Show your whole solution. We can't really be expected to guess what you did wrong.
 
  • #10


101300 (400) / 300 = P2 = 135093

135093 Pa x (.20^2) x 6 = 32422 N

and the other way 135093 Pa (.2^2) = 5403 N

neither worked
 
  • #11


MeKnos said:
101300 (400) / 300 = P2 = 135093

135093 Pa x (.20^2) x 6 = 32422 N

and the other way 135093 Pa (.2^2) = 5403 N

neither worked

I don't know. I got 5403N as well. Significant figures issue??
 

Related to Force exerted on walls of metal box

1. What is force exerted on walls of metal box?

The force exerted on the walls of a metal box is the amount of pressure or weight that is applied to the walls from the inside or outside of the box.

2. How is force exerted on walls of metal box calculated?

The force exerted on the walls of a metal box can be calculated by multiplying the pressure applied to the walls by the area of the walls.

3. What factors affect the force exerted on walls of metal box?

The force exerted on the walls of a metal box can be affected by the weight and density of the contents inside the box, as well as the external forces acting on the box such as gravity or air pressure.

4. Can the force exerted on walls of metal box cause damage?

Yes, if the force exerted on the walls of a metal box exceeds the strength of the material, it can cause damage such as dents, bends, or even punctures in the walls.

5. How can the force exerted on walls of metal box be controlled?

The force exerted on the walls of a metal box can be controlled by properly distributing the weight and contents inside the box, as well as using materials and designs that can withstand the expected amount of force.

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