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Planners of an experiment are evaluating the design of a sphere of radius r that is to be filled with helium (0°c, 1 atm pressure). ultrathin silver foil of thickness t will be used to make the sphere, and the designers claim that the mass of helium in the sphere will equal the mass of silver used. assuming that t is much less than r, calculate the ratio t/r for such a sphere.

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shinmin
The solution is:A mole of gas occupies 22.4 L A liter is 1000 cubic centimeters because 1 cubic centimeter = 1 mL
One Helium molecule (essentially one helium atom) has atomic mass 4 g/mol So for every 22400 cubic centimeters of volume, we have 4 grams of helium Density of helium = 4g / 22400 cm^3 = 1g / 5600 cm^3 
Volume of a sphere = (4/3)(pi)r^3 Volume of the outside sphere (the entire sphere) is (4/3)(pi)(R+T)^3 Volume of the inside sphere (the hollow region) is (4/3)(pi)R^3 
The difference for the volume of silver. (4/3)(pi)(R+T)^3 - (4/3)(pi)R^3 = (4/3)(pi)(3R^2T + 3RT^2 + T^3) 
The density of silver is 10.5g/cm^3 
So the mass of the silver is computed by:10.5*(4/3)(pi)(3R^2T + 3RT^2 + T^3) = (14*pi)(3R^2T + 3RT^2 + T^3) = (14pi)T(3R^2 + 3RT + T^2) 

Now for the mass of helium: volume x density = (4/3)(pi)R^3 (1/5600) = (pi/4200)R^3 

Set the two masses equal: (pi/4200)R^3 = (14pi)T(3R^2 + 3RT + T^2) R^3 = 58800*T(3R^2 + 3RT + T^2) R / T = 58800*(3R^2 + 3RT + T^2) / R^2 = 58800*( 3+T/R^2+(T/R)^2) 
then solve for xx = T / R 1/x = 58800*( 3+x/R+x^2) 1/58800 = x (3 + x/R + x^2) 1/58800 = 3x + x^3 x^3 + 3x - 1/58800= 0 x = ~ 5.66893x10^(-6)