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A cubic box of volume 6.15ร—10โˆ’2m3 is filled with air at atmospheric pressure at 15ยฐC. The box is closed and heated to 165ยฐC. What is the net force on each side of the box?

Short Answer

Expert verified

The force on each side of the box is 8207 N.

Step by step solution

01

Given data

The volume of the box isV=6.15ร—10โˆ’2m3.

The initial temperature isT1=15โˆ˜C=288K.

The final temperature isT2=165โˆ˜C=438K.

The initial pressure of the air inside the box isP1=1atm.

LetP2be the final pressure of the air inside the box.

LetLbe the sides of the cubic box.

Assume that there is no change in the volume of the box.

02

Concepts

The force on any side of the box equals the product of the pressure and area.

For this problem, you have to useP1T1=P2T2equation.

03

Calculation

Now, from the ideal gas law, you get:

cPV=nRTPT=nRV

The volume is constant in the process. Therefore,

cP1T1=P2T2P2=P1T2T1P2=(1atm)438K288KP2=1.52atm

Now, the extra pressure inside the box is:

cฮ”P=P2โˆ’P1=1.52atmโˆ’1atm=0.52atm=0.52ร—1.013ร—105Pa

Again, you get:

cL3=VL=V1/3L2=V2/3L2=(6.15ร—10โˆ’2m3)2/3

Therefore, the area of a side of the box is (6.15ร—10โˆ’2m3)2/3.

Now, the force on the side of the box is:

cF=P2L2=(0.52ร—1.013ร—105Pa)ร—(6.15ร—10โˆ’2m3)2/3=8207N

Hence, the force on each side of the box is 8207 N.

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Most popular questions from this chapter

Question 36:(III) A sealed test tube traps of air at a pressure of 1.00 atm and temperature of 18ยฐC. The test tubeโ€™s stopper has a diameter of 1.50 cm and will โ€œpop offโ€ the test tube if a net upward force of 10.0 N is applied to it. To what temperature would you have to heat the trapped air in order to โ€œpop offโ€ the stopper? Assume the air surrounding the test tube is always at a pressure of 1.00 atm.

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