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.

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)