mole fraction is equal to partial pressure divided by the total pressure
Total pressure= sum of partial pressure
use PV=nRT where n= number of moles
equation become PV/RT=n make sure to convert the temperature to Kelvin and to use 0.0821 for R. V=volume, P=total pressure
The partial pressure of CH4(g) is 0.185 atm and that of O2(g) is 0.300 atm in a mixture of the two gases.
a) What is the mole fraction of each gas in the mixture?
b) If the mixture occupies a volume of 11.5 L at 65 degress C, calculate the total number of moles of gas in the mixture.
answered below.
a) To determine the mole fraction of each gas in the mixture, we can use the formula:
Mole fraction = Partial pressure / Total pressure
Given that the partial pressure of CH4(g) is 0.185 atm and the partial pressure of O2(g) is 0.300 atm, we need to find the total pressure. Since the total pressure is the sum of the partial pressures, we can calculate it as follows:
Total pressure = Partial pressure of CH4(g) + Partial pressure of O2(g) = 0.185 atm + 0.300 atm = 0.485 atm
Now we can calculate the mole fraction of CH4(g) and O2(g) using the formula above:
Mole fraction of CH4(g) = Partial pressure of CH4(g) / Total pressure = 0.185 atm / 0.485 atm ≈ 0.381
Mole fraction of O2(g) = Partial pressure of O2(g) / Total pressure = 0.300 atm / 0.485 atm ≈ 0.619
Therefore, the mole fraction of CH4(g) in the mixture is approximately 0.381, and the mole fraction of O2(g) is approximately 0.619.
b) To calculate the total number of moles of gas in the mixture, we can use the ideal gas law equation:
PV = nRT
Given that the volume (V) is 11.5 L, the temperature (T) is 65 degrees C (convert to Kelvin: 65 + 273 = 338 K), and the gas constant (R) is 0.0821 L·atm/(mol·K), we need to find the total number of moles (n). Rearranging the equation, we have:
n = PV / RT
Plugging in the values:
n = (0.485 atm) * (11.5 L) / (0.0821 L·atm/(mol·K) * 338 K) ≈ 1.69 moles
Therefore, the total number of moles of gas in the mixture is approximately 1.69 moles.