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 4.50 L ? (The temperature was held constant.)

Answer 1

Answer: 2.5 grams

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\textMolar mass}}=\frac{2g}{4g/mol}=0.5moles$

Avogadro's Law: This law states that volume is directly proportional to the number of moles of the gas at constant pressure and temperature.

$V\propto n$   (At constant temperature and pressure)  

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

where,

$V_1$ = initial volume of gas = 2.00 L

$V_2$ = final volume of gas = 4.50 L

$n_1$ = initial number of moles = 0.5 moles

$n_2$ = final number of moles = ?

Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{2.00}{0.5}=\frac{4.50}{n_2}$

$n_2=1.125moles$

Thus moles of helium added to the cylinder = (1.125-0.5)= 0.625 moles

Mass of helium added =$moles\times {\text {Molar mass}}=0.625\times 4=2.5grams$

2.5 grams of helium were added to the cylinder if the volume was changed from 2.00 L to 4.50 L.