Solved: "Air enclosed in a sphere has density ρ=1.4kgm3...."

Sapewa

Answered question

2022-01-16

Air enclosed in a sphere has density

ρ = 1.4 k \frac{g}{m^{3}}

. What will the density be if the radius of the sphere is halved, compressing the air within?

braodagxj

Beginner2022-01-17Added 38 answers

Mass = density

\times

volume

m = ρ V

Since mass is constant
this implies

ρ V

is constant

ρ_{1} V_{1} = ρ_{2} V_{2}

ρ_{1} (\frac{4}{3}) π R^{3} = ρ_{2} (\frac{4}{3}) π {(\frac{R}{2})}^{3}

ρ_{1} = \frac{ρ_{2}}{8}

ρ_{2} = 8 ρ_{1} = 8 \times 1.4 = 11.2 k \frac{g}{m^{3}}

RizerMix

Expert2022-01-20Added 656 answers

Let m be the mass of the enclosed air and

V = \frac{4}{3} π r^{3}

be the volume of the sphere. Then,

ρ = \frac{m}{V}

Modified volume

V^{'} = \frac{4}{3} π (r / 2)^{3} = \frac{1}{8} \times \frac{4}{3} π r^{3} = \frac{V}{8}

New density

ρ^{'} = \frac{m}{V^{'}} = 8 \frac{m}{V} = 8 ρ = 8 \times 1.4 = 11.2 k g / m^{3}

Result:

11.2 k g / m^{3}

alenahelenash

Expert2022-01-23Added 556 answers

I tried to solve this example, but it didn’t work for me, I want to know the answer too.Thanks

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