A gaseous mixture consists of 6.95 g of N2, 4.75 g of O2, and 2.75 g of He. What volume does this mixture occupy at 23°C and 1.35 atm pressure?
Convert grams of the gases to mols. mol = grams/molar mass.
Add mols to find total mols = n, then use PV = nRT to solve for volume at the conditions listed. Don't forget T must be in kelvin.
hhj
To find the volume of the gaseous mixture, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin
First, let's find the number of moles for each gas in the mixture.
For nitrogen (N2):
1. Calculate the number of moles using the formula: moles = mass / molar mass.
The molar mass of N2 is 28 g/mol.
moles of N2 = 6.95 g / 28 g/mol ≈ 0.2482 mol
For oxygen (O2):
1. Calculate the number of moles using the formula: moles = mass / molar mass.
The molar mass of O2 is 32 g/mol.
moles of O2 = 4.75 g / 32 g/mol ≈ 0.1484 mol
For helium (He):
1. Calculate the number of moles using the formula: moles = mass / molar mass.
The molar mass of He is 4 g/mol.
moles of He = 2.75 g / 4 g/mol ≈ 0.6875 mol
Now, let's find the total number of moles in the mixture:
Total moles = moles of N2 + moles of O2 + moles of He
Total moles = 0.2482 mol + 0.1484 mol + 0.6875 mol ≈ 1.0841 mol
Next, let's convert the temperature to Kelvin:
°C to K Conversion: T(K) = T(°C) + 273.15
T(K) = 23°C + 273.15 ≈ 296.15 K
Now, we can plug the values into the ideal gas law equation and solve for the volume:
PV = nRT
V = (nRT) / P
V = (1.0841 mol * 0.0821 L·atm/(mol·K) * 296.15 K) / 1.35 atm
V ≈ 23.31 L
Therefore, the volume of the gaseous mixture at 23°C and 1.35 atm pressure is approximately 23.31 liters.