A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 4.50 L

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

[tex]V\propto n[/tex]

or,

[tex]\frac{V_1}{V_2}=\frac{n_1}{n_2}[/tex]

As we know that:

[tex]Moles=\frac{Mass}{\text{Molar mass}}[/tex]

As the gas is same that is helium gas. So the molar mass will be same.

Thus the formula will be:

[tex]\frac{V_1}{V_2}=\frac{w_1}{w_2}[/tex]

where,

[tex]V_1[/tex] = initial volume of gas = 2.00 L

[tex]V_2[/tex] = final volume of gas = 4.50 L

[tex]w_1[/tex] = initial mass of gas = 2.00 g

[tex]w_2[/tex] = final mass of gas = ?

Now we put all the given values in this formula, we get

[tex]\frac{V_1}{V_2}=\frac{w_1}{w_2}[/tex]

[tex]\frac{2.00L}{4.50L}=\frac{2.00g}{w_2}[/tex]

[tex]w_2=4.50g[/tex]

Therefore, the mass of helium added to cylinder were, 4.50 grams.