Asked by Dia B
A toy balloon originally held 1.00 g of helium gas and had a radius of 10.0 cm.
During the night, 0.25 g of the gas effused from the balloon. Assuming ideal
gas behavior under these constant pressure and temperature conditions, what
was the radius of the balloon the next morning?
During the night, 0.25 g of the gas effused from the balloon. Assuming ideal
gas behavior under these constant pressure and temperature conditions, what
was the radius of the balloon the next morning?
Answers
Answered by
DrBob222
The balloon had 1.00g He initially and a radius of 10.0 cm. Calculate the volume of the sphere (V = (4/3)*pi*r^3) and convert 1.00 g He to mols. Use PV = nRT and solve for P (no T is given so assume a convenient number--then use that in all other calculations).
Then 0.25 g He is removed which leaves 0.75g. Convert to mols, use PV = nRT and solve for the new V (use P from the first calculation and T you assumed). Convert V to the new radius. Don't forget that V in PV = nRT is in L but if you use cm for the radius and (4/3)*pi*r^, that V is in cc.
Then 0.25 g He is removed which leaves 0.75g. Convert to mols, use PV = nRT and solve for the new V (use P from the first calculation and T you assumed). Convert V to the new radius. Don't forget that V in PV = nRT is in L but if you use cm for the radius and (4/3)*pi*r^, that V is in cc.
Answered by
sunil
9.08cm
Answered by
Vanditha raj
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