Question

Balloons may contain tiny microscopic holes, which allow the gas particles within to escape via effusion. Suppose a 4.30 L balloon is filled with 1.21 g of He to a pressure of 1.68 atm. A second balloon is filled with 39.63 g of Xe to the same volume and pressure. From which balloon will atoms effuse more quickly?

Answer 1

Explanation:

It is known that according to Graham's law of effusion of gases.

              $\frac{R_{1}}{R_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$

where, $R_{1}$ = rate of effusion of gas 1 = rate of effusion of He

             $R_{2}$ = rate of effusion of gas 2 = rate of effusion of Xe

            $M_{1}$ = molar mass of gas 1 = molar mass He = 4.0026 g/mol

            $M_{2}$ = molar mass of gas 2 = molar mass Xe = 131.29 g/mol

Now, we will substitute the values into the above formula as follows.

           $\frac{R_{He}}{R_{Xe}} = \sqrt{\frac{M_{Xe}}{M_{He}}$

                    = $\sqrt{\frac{131.29 g/mol}{4.0026 g/mol}}$

                    = $\sqrt{32.8}$

                    = 5.73

Thus, we can conclude that helium effuses 5.73 times faster than Xenon.

Answer 2

Answer:

The balloon filled with Helium will effuse more quickly.

Explanation:

Step 1: Data given

Balloon 1: Has a volume 4.3L , a pressure of 1.68 atm and is filled with 1.21g of Helium

Balloon 2: Has also a volume of 4.3 L and a pressure of 1.68 atm but is filled with 39.63 grams of Xenon

Molar mass of He = 4g/mol

Molar mass of Xe = 131.3 g/mol

Since He is lighter than Xe, the effusion rate for He will be larger than that for Xe, which means the time of effusion for He will be smaller than that for Xe.

The balloon filled with Helium will effuse more quickly.