4.0 L of He (g), 6.0 L of N2 (g), and 10. L of Ar (g), all at 0°C and 1.0 atm, are pumped into an evacuated 8.0 L rigid container, the final pressure in the container at 0°C is

The final pressure in the container at 0°C is 2.49 atm

Explanation:

We apply the Ideal Gases law to know the global pressure.

We need to know, the moles of each:

P He. V He = moles of He. R. 273K

(1atm. 4L) / R. 273K = moles of He → 0.178 moles

P N₂. V N₂ = moles of N₂. R. 273K

(1atm. 6L) / R. 273K = moles of N₂ → 0.268 moles

P Ar. V Ar = moles of Ar. R. 273K

(1atm. 10L) / R. 273K = moles of Ar → 0.446 moles

Total moles: 0.892 moles

P. 8L = 0.892 mol. R. 273K

P = (0.892. R. 273K) / 8L = 2.49 atm

R = 0.082 L. atm/mol. K