Say I have a sphere containing helium. What happens if I double the number of atoms of the gas but keep the pressure and temperature constant? I figured the volume would double. What about the radius? Say the radius is 1 m. The radius is proportional to `V^(1/3)` .

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If the number of atoms of helium gas are doubled, the moles of gas are also doubled. We can find the number of moles of a gas, by using the idea that each mole of a gas contains an Avogadro's number of atoms (= 6.023 x 10^23 atoms). The volume...

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If the number of atoms of helium gas are doubled, the moles of gas are also doubled. We can find the number of moles of a gas, by using the idea that each mole of a gas contains an Avogadro's number of atoms (= 6.023 x 10^23 atoms). The volume of a gas is directly proportional to the number of atoms (or moles of gas), if the pressure and temperature are held constant. Hence the volume of the sphere will increase too. Since the moles of helium are doubled, the volume of the sphere (keeping pressure and temperature constant) will be doubled as well.

The volume of a sphere of radius r is given as `4/3 pir^3`.

Since the radius of the sphere is given as 1 m, volume of this sphere is `4/3pi` . If the volume doubles (that is, becomes `8/3 pi`), the new radius will be `2^(1/3)` or 1.26 m.

Hope this helps. 

 

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