A gas mixture with a total pressure of 745mmHg contains each of the following gases at the indicated partial pressures: CO2, 257 mmHg; Ar, 123mmHg; and O2, 160 mmHg. The mixture also contains helium gas.
What is the partial pressure of the helium gas?
What mass of helium gas is present in a 10.1L sample of this mixture at 263 K?
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.
Partial pressure of helium gas = Total pressure - (Partial pressure of CO2 + Partial pressure of Ar + Partial pressure of O2)
Partial pressure of helium gas = 745 mmHg - (257 mmHg + 123 mmHg + 160 mmHg)
Partial pressure of helium gas = 205 mmHg
Therefore, the partial pressure of the helium gas in the mixture is 205 mmHg.
To find the mass of helium gas present in a 10.1 L sample of the mixture, we can use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/K·mol)
T = temperature (in Kelvin)
First, let's convert the pressure from mmHg to atm:
205 mmHg = 205/760 atm (since 1 atm = 760 mmHg)
Next, we rearrange the Ideal Gas Law equation to solve for n (moles):
n = PV / RT
We are given:
P = 205/760 atm
V = 10.1 L
R = 0.0821 L·atm/K·mol
T = 263 K
Substituting these values into the equation, we get:
n = (205/760 atm) * (10.1 L) / (0.0821 L·atm/K·mol) * (263 K)
Simplifying:
n = 0.299 mol
Since helium has a molar mass of 4 g/mol, we can calculate the mass of helium gas:
Mass = moles * molar mass
Mass = 0.299 mol * 4 g/mol
Mass = 1.196 g
Therefore, the mass of helium gas present in a 10.1 L sample of the mixture at 263 K is approximately 1.196 grams.