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, we first need to know the total pressure of the gas mixture and the partial pressures of the other gases present.
In this case, the total pressure of the gas mixture is given as 745 mmHg, and we are provided with the partial pressures of CO2 (257 mmHg), Ar (123 mmHg), and O2 (160 mmHg).
To calculate the partial pressure of helium gas, we can subtract the sum of the partial pressures of the other gases from the total pressure of the mixture:
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)
= 745 mmHg - 540 mmHg
= 205 mmHg
Therefore, the partial pressure of the helium gas is 205 mmHg.
To find the mass of helium gas present in the 10.1L sample of the mixture at 263 K, we can use the ideal gas law and the molar mass of helium gas.
The ideal gas law is given by the equation:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, we need to calculate the number of moles of helium gas using the ideal gas law:
n = PV / RT
where P is the partial pressure of helium gas, V is the volume, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.
Plugging in the values:
n = (Partial pressure of helium gas) * (Volume) / (R * Temperature)
n = (205 mmHg) * (10.1L) / (0.0821 L·atm/mol·K * 263 K)
Simplifying the units:
n = (205 mmHg) * (10.1 L) / (0.0821 * 263)
Now, we can calculate the mass of helium gas using the molar mass of helium, which is approximately 4 g/mol:
Mass = Number of moles * Molar mass
Mass = n * Molar mass
= (n) * (4 g/mol)
Substituting the value of n:
Mass = [(205 mmHg) * (10.1 L) / (0.0821 * 263)] * 4 g/mol
Evaluating this expression will give you the mass of helium gas present in the 10.1 L sample of the mixture at 263 K.