Air enclosed in a sphere has density ρ=1.4 kgm3. What will the density be if the radius of the sphere is halved, compressing the air within?

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Air enclosed in a sphere has density ρ=1.4 kgm3. What will the density be if the radius of the sphere is halved, compressing the air within?

pendukke · Accepted Answer

Given data:Density of air enclosed in a sphere is, ρ=1.4 kgm3.Let m be the mass of the enclosed air and V=43πr2 be the volume of the sphere. Here, r is the radius of sphere. Then density is,ρ=mVModified volume (V&#039;) is,V′=43π×(r2)3V′=18×43πr2V′=18×VThen new density is,ρ′=mVρ′=mV8ρ′=8ρρ′=8×1.4=11.2 kgm3Thus, the density if the radius of the sphere is halved is 11.2 kgm3.

fudzisako · Accepted Answer

Result:2·ρSolution:ρ=mVFor a sphere, the volume is given by the formula V=43πr3, where r is the radius.Given that the initial density is ρ=1.4kgm3, we can substitute the initial values into the density equation to find the mass:1.4=m43πr3To find the new density when the radius is halved, we can substitute r2 for r in the equation above:ρ′=m43π(r2)3Simplifying the equation, we get:ρ′=m43π(r2)3=8m43πr3=2·ρTherefore, the density will be doubled when the radius of the sphere is halved, compressing the air within.

xleb123 · Accepted Answer

Since we are dealing with a fixed amount of air enclosed in a sphere, the number of moles of gas (n) remains constant. Therefore, we can rewrite the ideal gas law as PV=constant.The density of a gas is defined as ρ=mV, where m is the mass of the gas and V is the volume. In this case, the mass of the gas remains constant, so we can write ρ1=mV1, where ρ1 is the initial density and V1 is the initial volume.When the radius of the sphere is halved, the volume of the sphere is reduced by a factor of 8 (since the volume of a sphere is proportional to the cube of its radius). Therefore, the new volume V2 is 18 times the initial volume V1.Now, let&#039;s find the new density ρ2 when the air is compressed within the sphere.We can write ρ2=mV2.From the ideal gas law, we know that PV=constant.Since the pressure P and the number of moles of gas n remain constant, we can write P1V1=P2V2, where P1 and P2 are the initial and final pressures, respectively.Dividing both sides of the equation by V2 gives us P1=P2(V1V2).Substituting the expression for V2, we have P1=P2(V118V1)=P2×8V1.Since PV=constant, we can write ρ1P1=ρ2P2.Substituting the expressions for ρ1 and P1 and rearranging the equation, we get ρ2=ρ1V18V1.Simplifying the equation further, we find ρ2=ρ18.Therefore, the density of the air will be ρ18 when the radius of the sphere is halved, compressing the air within.

Andre BalkonE · Accepted Answer

We know that the volume of a sphere is given by the formula:V=43πr3Let&#039;s consider the initial volume, V1, and the final volume, V2, after halving the radius:V1=43πr3V2=43π(r2)3Simplifying the equation for V2:V2=43π·r38V2=16πr3Now, we can calculate the final density, ρ2, by using the formula:ρ2=massvolumeThe mass of the air enclosed in both cases remains the same, so we can write:ρ1·V1=ρ2·V2Substituting the values for V1 and V2:ρ1·43πr3=ρ2·16πr3Canceling out the common terms:43ρ1=16ρ2To find ρ2, we can rearrange the equation:ρ2=43·61ρ1ρ2=8ρ1Therefore, the density of the air enclosed will be 8 times the initial density when the radius of the sphere is halved.

Air enclosed in a sphere has density ρ = 1.4\

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