Flexible plastic container with 0.86 g of 19.2 L helium gas. If you remove 0.205 grams of helium gas in continuous pressure and temperature, what is the new size?

This is a case of Avogadro's law, that is

Plugging in the values, we get

The requested answer was the new size reached by the flexible container (presumably a balloon) Removing 0.205 grams of helium brings the mass to 0.655 grams. By a simple proportion: 0.86 g : 0.655 g = 19.2 L : x L x= 14.62 The new size of the container is 14.62 L No need to calculate moles number. The above calculation holds true for every gas

To find the new size of the flexible plastic container after removing 0.205 grams of helium gas, we can use Boyle's Law, which states that the volume of a given amount of gas held at constant temperature varies inversely with the pressure.

Boyle's Law formula is:

[ P_1 \times V_1 = P_2 \times V_2 ]

Where:

• ( P_1 ) and ( P_2 ) are the initial and final pressures respectively (assuming constant pressure, this value remains the same).
• ( V_1 ) is the initial volume.
• ( V_2 ) is the final volume.

Given: Initial volume ( V_1 ) = 19.2 L Initial mass of helium = 0.86 g Mass of helium removed = 0.205 g

First, let's find the final mass of helium in the container: [ \text{Final mass} = \text{Initial mass} - \text{Mass removed} ] [ \text{Final mass} = 0.86 \text{ g} - 0.205 \text{ g} ] [ \text{Final mass} = 0.655 \text{ g} ]

Now, using the ideal gas law ( PV = nRT ), where ( n ) is the number of moles and ( R ) is the universal gas constant, we can rearrange it to: [ V = \frac{nRT}{P} ]

Assuming the temperature and pressure remain constant: The number of moles (( n )) can be found using the equation: [ n = \frac{\text{mass of gas}}{\text{molar mass of helium}} ]

Molar mass of helium (( \text{He} )) = 4.0026 g/mol

[ n = \frac{0.655 \text{ g}}{4.0026 \text{ g/mol}} ] [ n = 0.1636 \text{ mol} ]

Now, substituting the values into the rearranged ideal gas law: [ V_2 = \frac{n \times R \times T}{P} ]

Given that ( P_1 = P_2 ) (constant pressure), we can use: [ V_2 = \frac{n \times V_1}{n_1} ]

Where ( n_1 ) is the initial number of moles and ( V_1 ) is the initial volume.

[ V_2 = \frac{0.1636 \text{ mol} \times 19.2 \text{ L}}{0.86 \text{ g} / 4.0026 \text{ g/mol}} ] [ V_2 = \frac{3.143 \text{ mol} \cdot \text{L}}{0.215 \text{ mol}} ] [ V_2 = 14.60 \text{ L} ]

So, the new size of the flexible plastic container after removing 0.205 grams of helium gas is approximately 14.60 L.