Question

A 2.4 L balloon holds 3.7 mol He. If 1.6 mol He are added to the balloon, what is the new volume?

Answer 1

For this question, lets apply Avagadro's law
when Pressure and temperature are constant, the volume occupied is directly proportional to the number of moles of gases.
$\frac{V}{n} = k$
where V-volume, n-number of moles and k - constant 
Therefore at 2 instances 
$\frac{V1}{n1} = \frac{V2}{n2}$
where V1 and n1 are for 1st instance 
and V2 and n2 are for 2nd instance 
therefore 
$\frac{V1}{n1} = \frac{V2}{n2}$
V1 = 2.4 L
n1 = 3.7 mol
n2 = 3.7 + 1.6 = 5.3 mol
since more He moles are added at the 2nd instance its the sum of the moles.
V2 needs to be calculated $\frac{2.4}{3.7} = \frac{V2}{5.3}$
V2 = 2.4 x 5.3 / 3.7
     = 3.4 L
Answer is 1st option 3.4 L