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October 26, 2014

October 26, 2014

Posted by **Dia B** on Wednesday, September 12, 2012 at 2:50pm.

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?

- Chemistry -
**DrBob222**, Wednesday, September 12, 2012 at 3:13pmThe 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.

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