Force exerted on walls of metal box

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The discussion focuses on calculating the force exerted on the walls of a cubic metal box containing air, initially at 1.0 atm and 300 K, and heated to 400 K. The relevant equation, PV=nRT, is used to determine the final pressure (P2) after heating, which is found to be 135093 Pa. Participants express confusion over the calculations, particularly regarding the correct application of pressure and area to find the total force on the walls. Despite attempts to compute the force using different methods, discrepancies in results arise, suggesting potential issues with significant figures or calculation errors. The conversation emphasizes the importance of accurately applying the ideal gas law and understanding how temperature affects pressure in a sealed container.
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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=\left|\Delta\vec{p}/\Delta<i>t</i>\right|
v=\sqrt{3RT/M}=\sqrt{3(8.314 J/molxK)(300K)/0.029kg/mol)}=507.957 m/s
Delta[/tex]\vec{p}/=mv^2=(9423.9987g)(507.957m/s)=4.787x10^6<br /> I just need delta t
 
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Forget about v. You don't need to count individual particles, PV=nRT will give you the pressure. Pressure is force per unit area.
 
How do I use all the information given in the problem to find that?
 
PV=nRT. How many variables in this equation change when the temperature is varied in your problem?
 
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?
 
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!
 
got it...thanks!
 


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.
 


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??
 
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