Question

A 75 L gas cylinder contains He gas at 16 atm at $27^\circ {\text{C}}$ . It is used to fill the balloon at $1.1{\text{ atm}}$ at $27^\circ {\text{C}}$ and volume of each balloon is 3 L is $1.1{\text{ atm}}$. Calculate the number of balloons filled by this gas.

Answer 1

We will use Boyle's law which is a relation between pressure and volume. The gas is allowed to fill in the balloon until the pressure becomes equal to $1.1$.

Complete step by step solution:
It is given that the volume of the cylinder is 75 L and the pressure of helium gas is 16 atm. Now this cylinder is used to fill the balloons which have the capacity of 3 L and the pressure of $1.1$ atm. The balloons can only be filled till the time the pressure of the cylinder is greater than $1.1$ atm. If the pressure of the cylinder becomes equal to that of $1.1{\text{ atm}}$ which is the same as that of a balloon, no gas will be filled because there will be no potential. Hence we will consider the initial pressure of gas as:
$16 - 1.1 = 14.9{\text{ atm}}$ .
The Boyle’s law give us the relation:
${{\text{P}}_1}{{\text{V}}_1} = {{\text{P}}_2}{{\text{V}}_2}$
Here ${{\text{P}}_1}$ is initial pressure, ${{\text{P}}_2}$ is final pressure, ${{\text{V}}_1}$ is initial volume and ${{\text{V}}_2}$ is final volume.
The initial pressure of gas is $14.9{\text{ atm}}$
The initial volume is 75 litres.
The final pressure of gas filled in balloon is $1.1{\text{ atm}}$
Now let us say that n balloons are filled and the volume of each balloon is 3 L. So the total final volume is $3 \times {\text{n L}}$ . Using Boyle’s law equation we will get:
$14.9 \times 75 = 1.1 \times 3 \times {\text{n}}$
Solving the above equation we will get:
${\text{n}} = 338$

Hence, 338 balloons will be filled.

Note: Boyle’s law states that the product of pressure and volume is always constant provided that temperature and number of moles are kept constant. Helium is the second lightest gas after hydrogen and hence is used in filling balloons.