A gas mixture with a total pressure of 770 contains each of the following gases at the indicated partial pressures: CO2 , 249mmHg ; , Ar 114mmHg ; and , O2 174mmHg . The mixture also contains helium gas.
1. What is the partial pressure of the helium gas?
2. What mass of helium gas is present in a 11.2 L sample of this mixture at 281 K?
Ptotal = PCO2 + PAr + PO2 + PHe.
Only 1 unknown; i.e., PHe.
2. I would use PV = nRT, solve for n = number of moles He, then moles = grams/molar mass will solve for grams.
Post your work if you get stuck.
To find the partial pressure of helium gas in the mixture, we need to subtract the sum of the partial pressures of CO2, Ar, and O2 from the total pressure of the gas mixture.
1. Partial pressure of helium gas:
First, let's find the sum of the partial pressures of CO2, Ar, and O2:
Partial pressure of CO2 = 249 mmHg
Partial pressure of Ar = 114 mmHg
Partial pressure of O2 = 174 mmHg
Sum of partial pressures = Partial pressure of CO2 + Partial pressure of Ar + Partial pressure of O2
= 249 mmHg + 114 mmHg + 174 mmHg
= 537 mmHg
Now, we can find the partial pressure of helium gas by subtracting the sum of partial pressures from the total pressure of the gas mixture:
Partial pressure of helium gas = Total pressure of the gas mixture - Sum of partial pressures
= 770 mmHg - 537 mmHg
= 233 mmHg
Therefore, the partial pressure of the helium gas in the mixture is 233 mmHg.
2. Mass of helium gas present:
To find the mass of helium gas in the mixture, we need to use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 atm·L/mol·K)
T = Temperature (in Kelvin)
Given:
Volume (V) = 11.2 L
Temperature (T) = 281 K
First, we need to convert the partial pressure of helium gas from mmHg to atm:
Partial pressure of helium gas = 233 mmHg = 233 mmHg * 1 atm / 760 mmHg
= 0.307 atm
Next, we rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Now we can plug in the values to calculate the number of moles of helium gas:
n = (0.307 atm) * (11.2 L) / (0.0821 atm·L/mol·K) * (281 K)
Simplifying the equation:
n = (3.4279 L·atm) / (22.9141 atm·L/mol·K)
n ≈ 0.1493 mol
Finally, we can find the mass of helium gas using the molar mass of helium, which is approximately 4 g/mol:
Mass of helium gas = Number of moles * Molar mass
= 0.1493 mol * 4 g/mol
= 0.5972 g
Therefore, the mass of helium gas present in a 11.2 L sample of this gas mixture at 281 K is approximately 0.5972 grams.