Question

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 3.50 L ? (The temperature was held constant.)

Answer 1

Answer : The mass of helium added to the cylinder was, 1.5 grams

Explanation :

Avogadro's law : It is defined as the volume of gas is directly proportional to the number of moles of gas at constant pressure and temperature.

$V\propto n$

or,

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

where,

$V_1$ = initial volume of gas = 2.00 L

$V_2$ = final volume of gas = 3.50 L

$n_1$ = initial moles of gas = $\frac{\text{Mass of He}}{\text{Molar mass of He}}=\frac{2.00g}{4g/mol}=0.5mol$

$n_2$ = final temperature of gas = ?

Now put all the given values in the above equation, we get:

$\frac{2.00L}{0.5mol}=\frac{3.50L}{n_2}$

$n_2=0.875mol$

Now we have to calculate the mass of helium were added to the cylinder.

$\text{Mass of He}=\text{Moles of He}\times \text{Molar mass of He}$

$\text{Mass of He}=0.875mol\times 4g/mol=3.5g$

Mass of helium added = 3.5 - 2.00 = 1.5 g

Thus, the mass of helium added to the cylinder was, 1.5 grams