The atmosphere in a sealed diving bell contained oxygen and helium. If the gas mixture has 0.200 atm of oxygen and a total pressure of 3.00 atm, calculate the mass of helium in 10.0 L of the gas mixture at 40 degrees Celsius.

I assume the T of the mixture is 40 C.

If pO2 = 0.200 atm and Ptotal = 3.0 atm, then pHe must be 2.8 atm.
Use PV = nRT to solve for n = number of mols, then n = grams/atomic mass. You know n and atomic mass; solve for grams. Don't forget that T must be in kelvin.

To calculate the mass of helium in the gas mixture, we first need to find the partial pressure of helium.

Given:
Partial pressure of oxygen (PO2) = 0.200 atm
Total pressure (Ptotal) = 3.00 atm

The partial pressure of helium (PHe) can be calculated by subtracting the partial pressure of oxygen from the total pressure:

PHe = Ptotal - PO2
PHe = 3.00 atm - 0.200 atm
PHe = 2.80 atm

Now, let's convert the volume of the gas mixture from liters to moles.

Using the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)

Given:
Volume (V) = 10.0 L
Temperature (T) = 40 degrees Celsius = 273.15 K + 40 K = 313.15 K

So, according to the ideal gas law, we can rearrange the equation to find the number of moles:

n = PV / RT

nHe = PHe * V / RT
nHe = (2.80 atm) * (10.0 L) / (0.0821 L.atm/mol.K * 313.15 K)
nHe = 0.9002 mol

Now, we need to calculate the mass of helium using the molar mass of helium (4.00 g/mol):

Mass of helium (mHe) = nHe * molar mass of helium
mHe = 0.9002 mol * 4.00 g/mol
mHe = 3.60 g

Therefore, the mass of helium in 10.0 L of the gas mixture at 40 degrees Celsius is 3.60 grams.

To calculate the mass of helium in the given gas mixture, we need to use the Ideal Gas Law. The equation for the Ideal Gas Law is:

PV = nRT

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

First, let's find the number of moles of helium present in the gas mixture. We can do this by rearranging the Ideal Gas Law equation to solve for n:

n = PV / RT

Given:
Pressure of helium (PHelium) = Total Pressure - Pressure of Oxygen = 3.00 atm - 0.200 atm = 2.80 atm
Volume (V) = 10.0 L
Gas constant (R) = 0.0821 L•atm/mol•K
Temperature (T) = 40 degrees Celsius = 40 + 273.15 = 313.15 K

Now, let's substitute the values into the equation to find the number of moles of helium:

n = (2.80 atm) * (10.0 L) / (0.0821 L•atm/mol•K * 313.15 K)

Simplifying the equation gives:

n = 0.358 mol

Next, we can use the molar mass of helium (4.00 g/mol) to calculate the mass of helium:

Mass of helium = n * molar mass of helium
= 0.358 mol * 4.00 g/mol
= 1.43 g

Therefore, the mass of helium in 10.0 L of the gas mixture is 1.43 grams.

1.56 g He