Question

If 5.00 g helium gas is added to a 1.00 L balloon containing 1.00 g of helium gas, what is the new volume of the balloon? Assume no change in temperature or pressure. Enter your answer in the box provided. L

Answer 1

Explanation:

The given data is as follows.

                 $mass_{1}$ = 5 g,        $Volume_{1}$ = 1 L

                 $mass_{2}$ = 1 g,         $Volume_{1}$ = ?

No. of moles of helium present in 5 g helium gas are as follows.

                   No. of moles = $\frac{mass}{\text{molar mass}}$

                                         = $\frac{5 g}{4 g/mol}$  

                                         = 1.25 mol

No. of moles of helium present in 1 g helium gas are as follows.

                 No. of moles = $\frac{mass}{\text{molar mass}}$

                                         = $\frac{1 g}{4 g/mol}$  

                                         = 0.25 mol

According to the ideal gas equation, PV = nRT. And, since temperature and pressure are held constant. Therefore,

                      $\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$

                      $\frac{1 L}{1.25 mol} = \frac{V_{2}}{0.25 mol}$

                             $V_{2}$ = 0.2 L

Thus, we can conclude that the new volume of the balloon is 0.2 L.