Question on Temperature/Kinetic Theory in AP Physics B

AI Thread Summary
The discussion revolves around calculating the net force on the sides of a heated cubic box filled with air. The user initially misapplied the pressure calculations by not accounting for the atmospheric pressure outside the box. The correct approach involves calculating the pressure difference between the inside and outside of the box, rather than simply adding the pressures. The final net force is determined by this pressure difference multiplied by the area of one side of the box. Properly applying these principles leads to the correct answer of 6400 N.
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Homework Statement



A Cubic Box of volume 3.9*10^-2 meters cubed is filled with air at atmospheric pressure at 20 degrees C. The box is closed and heated to 180 degrees C. What is the net force on each side of the box?

atmospheric pressure = 1 atm or 1.0*10^5 N/msquared


Homework Equations



Temperature K = C +273
P/T = P/T
P= F/A


The Attempt at a Solution



Okay i have no idea why i am getting this wrong. What i did was use P/T = P/T using the right temperatures and atm pressure in N/msquared for the pressure on the left side of the equation. on the right i am solving for P and used the 180+273 as the new temperature.

i got 154607.5 N/msquared

I then plugged that into P = F/A and cubed rooted the Volume given and squared that value to get area of one side of the box. and for the final answer got 17780 N.

the correct answer is 6400N

:confused:
 
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You forgot the air outside of the box.
 
am i supposed to add 1 atm to the answer i got? before plugging it into P=F/A
 
No, you use the formula P=F/A twice; once to calculate for the force from the inside (which you have done), then again to calculate the force from the outside (due to the air outside). Taking the difference of these will give the net force.
 
xopenx said:
am i supposed to add 1 atm to the answer i got? before plugging it into P=F/A

The air outside the box pushes on the sides in the opposite direction from the air inside. You can use cristo's suggestion, or equivalently find the pressure difference between the inside and outside before multiplying by the area. That would not come from adding the two pressures.
 
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