A mixture of two gases, neon and argon, is 80% neon by volume. Calculate the partial pressures of neon and argon if the total pressure inside the container is 4.00 atm.

0.80 atm Ne and 3.20 atm Ar

2.40 atm Ne and 1.60 atm Ar

1.60 atm Ne and 2.40 atm Ne

3.20 atm Ne and 0.80 atm Ar

1 answer

To find the partial pressures of neon (Ne) and argon (Ar) in a mixture where neon constitutes 80% of the gas by volume, we can use Dalton's Law of Partial Pressures. According to this law, the partial pressure of each gas in a mixture is equal to the total pressure multiplied by the mole fraction (or volume fraction) of the gas.

Given:

  • Total Pressure \( P_{total} = 4.00 , \text{atm} \)
  • Volume fraction of Ne = 80% = 0.80
  • Volume fraction of Ar = 20% = 0.20 (since 100% - 80% = 20%)

Now we can calculate the partial pressures:

  1. Partial pressure of neon (P_Ne): \[ P_{Ne} = P_{total} \times \text{Volume fraction of Ne} \] \[ P_{Ne} = 4.00 , \text{atm} \times 0.80 = 3.20 , \text{atm} \]

  2. Partial pressure of argon (P_Ar): \[ P_{Ar} = P_{total} \times \text{Volume fraction of Ar} \] \[ P_{Ar} = 4.00 , \text{atm} \times 0.20 = 0.80 , \text{atm} \]

Thus, the partial pressures are:

  • Neon: \( 3.20 , \text{atm} \)
  • Argon: \( 0.80 , \text{atm} \)

So, the correct answer is:

3.20 atm Ne and 0.80 atm Ar