A flexible plastic container contains 0.833 g of helium gas in a volume of 18.3 L. If 0.203 g of helium is removed at constant pressure and temperature, what will be the new volume?

Let n = moles and V = Volume

n ∝ V => n(1)/n(2) = V(1)/V(2)
n(1) = (0.833 gms / 4 gms/mol) = 0.2083 mol
n(2) = [(0.833 – 0.203) gms / 4 gms/mol) = 0.1575 mol
V(1) = 18.3 Liters
V(2) = X
(0.2083 mol / 0.1575 mol) = (18.3 Liters / X)
X = [(0.1575 mol)(18.3 Liters)/ (0.5083 mol)] = 13.8 Liters

To determine the new volume, we need to use the ideal gas law equation, which is:

PV = nRT

Where:
P = pressure (constant)
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature (constant)

First, let's find the number of moles of helium gas using the given mass and molar mass of helium. The molar mass of helium is approximately 4 g/mol.

Number of moles (n) = mass (m) / molar mass (M)

n = 0.833 g / 4 g/mol
n = 0.20825 mol

Since we are removing 0.203 g of helium gas, the new mass of helium gas will be:

New mass = initial mass - removed mass
New mass = 0.833 g - 0.203 g
New mass = 0.63 g

Next, let's find the new number of moles of helium gas:

New number of moles (n') = new mass (m') / molar mass (M)

n' = 0.63 g / 4 g/mol
n' = 0.1575 mol

Since the pressure (P) and temperature (T) remain constant, we can rewrite the ideal gas law equation as:

V * (n/T) = V' * (n'/T)

Where:
V' = new volume

Now, let's solve for the new volume:

V' = (V * (n/T)) / (n'/T)

V' = (V * n) / n'

Substituting the given values:

V' = (18.3 L * 0.20825 mol) / 0.1575 mol
V' ≈ 24.15 L

Therefore, the new volume of the flexible plastic container will be approximately 24.15 L.