A closed vessel of volume 2.50 dm3 contains 100g of nitrogen (N2) and 100g of carbon dioxide (CO2) at 25 C. What are the partial pressures and mole fractions of each gas?
Can you check my answers.Thank you.
I got Partial Pressure for
N2=34.95 atm
CO2=22.24 atm
Mole fraction of N2=0.611 and CO2=0.389
You have used too many significant figures. You're allowed 3 places only so you should round those to 35 for pN2 and 22.2 for pCO2
mols fractions are OK.
To calculate the partial pressures and mole fractions of each gas, we need to use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in dm3)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
First, let's convert the volume from dm3 to L:
2.50 dm3 = 2.50 L
Now, let's calculate the number of moles for each gas:
For nitrogen (N2):
n(N2) = mass / molar mass(N2)
= 100g / 28 g/mol
= 3.57 mol
For carbon dioxide (CO2):
n(CO2) = mass / molar mass(CO2)
= 100g / 44 g/mol
= 2.27 mol
Next, let's calculate the partial pressures:
For nitrogen (N2):
P(N2) = (n(N2) * R * T) / V
= (3.57 mol * 0.0821 L.atm/mol.K * 298 K) / 2.50 L
= 34.95 atm
For carbon dioxide (CO2):
P(CO2) = (n(CO2) * R * T) / V
= (2.27 mol * 0.0821 L.atm/mol.K * 298 K) / 2.50 L
= 22.24 atm
So, your calculated partial pressures are correct.
Now, let's calculate the mole fractions:
For nitrogen (N2):
Mole fraction(N2) = n(N2) / (n(N2) + n(CO2))
= 3.57 mol / (3.57 mol + 2.27 mol)
= 0.611
For carbon dioxide (CO2):
Mole fraction(CO2) = n(CO2) / (n(N2) + n(CO2))
= 2.27 mol / (3.57 mol + 2.27 mol)
= 0.389
So, your mole fractions are also correct.
Therefore, your answers for the partial pressures and mole fractions are correct.
Please note that in the provided answer, the temperature was assumed to be given in Celsius and then converted to Kelvin by adding 273.15.