Question

(AP Chem)

A rigid 2.5L sealed vessel contains O2, N2, and Ar with partial pressures of 0.438 atm, 0.284 atm, and 0.625 atm respectively. Find the mole fraction of each gas.

Answer 1

Mole fraction of O2 = 0.325

Mole fraction of N2 = 0.211

Mole fraction of Ar = 0.464.

Step-by-step explanation:

Calculate the Total Pressure

The total pressure (
`\sf P_{\textsf{total}}`) is the sum of the partial pressures of the three gases:

`\sf P_{\textsf{total}} = P_{\textsf{O2}} + P_{\textsf{N2}} + P_{\textsf{Ar}}`

`\sf P_{\textsf{total}} = 0.438 \, \textsf{atm} + 0.284 \, \textsf{atm} + 0.625 \, \textsf{atm}`

`\sf P_{\textsf{total}} = 1.347 \, \textsf{atm}`

Calculate the Mole Fraction of Each Gas

Using the Ideal Gas Law:

`\sf PV = nRT`

where:

• 
  `\sf P` is the pressure (in atm),
• 
  `\sf V` is the volume (in liters),
• 
  `\sf n` is the number of moles (in mol),
• 
  `\sf R` is the ideal gas constant (in L·atm/mol·K),
• 
  `\sf T` is the temperature (in K).

Given:

• 
  `\sf P = P_{\textsf{total}} = 1.347 \, \textsf{atm}`
• 
  `\sf V = 2.5 \, \textsf{L}`
• 
  `\sf T = 273.15 \, \textsf{K}` (assuming room temperature)
• 
  `\sf R = 0.08206 \, \textsf{L•atm/mol•K}`

Solving for
`\sf n`:

`\sf n = (PV)/(RT)`

`\sf n = \frac{(1.347 \, \textsf{atm})(2.5 \, \textsf{L})}{(0.08206 \, atm/ mol·K)(273.15 \, \textsf{K})}`

`\sf n = 0.100 \, \textsf{mol}`

Mole Fractions:

`\sf X_{\textsf{O2}} = \frac{n_{\textsf{O2}}}{n_{\textsf{total}}} = \frac{0.438 \, \textsf{atm}}{1.347 \, \textsf{atm}} = 0.325`

`\sf X_{\textsf{N2}} = \frac{n_{\textsf{N2}}}{n_{\textsf{total}}} = \frac{0.284 \, \textsf{atm}}{1.347 \, \textsf{atm}} = 0.211`

`\sf X_{\textsf{Ar}} = \frac{n_{\textsf{Ar}}}{n_{\textsf{total}}} = \frac{0.625 \, \textsf{atm}}{1.347 \, \textsf{atm}} = 0.464`

Therefore, the mole fraction of O2 is 0.325, the mole fraction of N2 is 0.211, and the mole fraction of Ar is 0.464.