Density of a Compressed Sphere: Does Halving the Radius Double the Density?

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SUMMARY

The discussion centers on the relationship between the radius of a sphere and the density of air contained within it. When the radius of the sphere is halved, the density increases from an initial value of 1.4 kg/m³ to 11 kg/m³. This increase is due to the constant mass of air being compressed into a smaller volume, calculated using the formula D = m/v. The volume of a sphere is defined as V = (4/3)πr³, and halving the radius results in a volume reduction by a factor of 8, leading to the observed density increase.

PREREQUISITES
  • Understanding of basic physics concepts, specifically density and volume.
  • Familiarity with the formula for the volume of a sphere: V = (4/3)πr³.
  • Knowledge of the relationship between mass, density, and volume: D = m/v.
  • Basic algebra skills for manipulating equations and solving for variables.
NEXT STEPS
  • Explore the implications of compressibility in gases and how it affects density changes.
  • Study the concept of dimensional analysis and its application in physics problems.
  • Learn about the ideal gas law and how it relates to density and volume changes under different conditions.
  • Investigate the effects of temperature on gas density and how it interacts with volume changes.
USEFUL FOR

Students studying physics, educators teaching density and volume concepts, and anyone interested in understanding the principles of gas behavior under compression.

snubbly
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Air in a sphere has density "x" kg/m3

If radius is halved and air is compressed... does density double? ("2x" km/m3)
 
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One thing at a time. Suppose only one of these happened. Say the radius is halved. Does the density increase or decrease and by what factor? Once you figure that out, compress the air some more and see what happens to the final density relative to the initial.
 
Radius halved - density would increase.

I don't know by what factor though...

I have the Q&A ... initial is 1.4 kg/m3 and once radius is halved, density is 11 kg/m3.

I have to show working and I just can't figure it out.

Multiplying 1.4 by 7.9 gives me the answer, but I don't understand.
 
snubbly said:
Multiplying 1.4 by 7.9 gives me the answer, but I don't understand.
What's the magical 7.9? Or did you just do 11 kg/m3 / 1.4kg/m3
 
mg0stisha said:
What's the magical 7.9? Or did you just do 11 kg/m3 / 1.4kg/m3

yes, just the 11 / 1.4.

clueless.
 
What all are you given in the problem?
 
This is the question from the textbook:

Air enclosed in a sphere has density p = 1.4 kg/m3. What will the density be if the radius of the sphere is halved, compressing the air within?
 
What happens to the volume of a sphere if the radius is halved?
 
let say: D = Density
m = mass
v = volume

D1 = 1.4 kg/m3

mass is constant since the air will only be compressed.

D = m/v yielding m = Dv

Therefore, D1v1=D2v2

but v of sphere is (3*pi*r3)/4

and r2=r1/2

You can get it now.
 
Last edited:
  • #10
Minor correction: volume of a sphere is V = \frac{4\pi r^{3}}{3}
 
  • #11
Sorry. I interchanged 3 and 4.
 
  • #12
got your back ;)
 
  • #13
Thanks.
 
  • #14
A intuitive way to think about this: Think about a cube. If you cut the side lengths in half, what happens to volume? Halve one dimension and volume halves. Halve the second and volume decreases by a factor of 4. Halve the third and volume goes down how many times?

The same thing happens to a sphere.
 
  • #15
ideasrule said:
A intuitive way to think about this: Think about a cube. If you cut the side lengths in half, what happens to volume? Halve one dimension and volume halves. Halve the second and volume decreases by a factor of 4. Halve the third and volume goes down how many times?

The same thing happens to a sphere.

8.

Thank you, that really helped!
 

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