Question

The atmosphere is 0.9% Argon. Calculate the concentration of Argon in atmospheric air at a temperature of 286 K and pressure of 998 mBar

Answer 1

Since atmospheric pressure is 998 mBar and it contains 0.9% Argon,

Partial pressure of argon (P_Ar) = total pressure x fraction of argon

                                                  = 998 mBar x (0.9%)

                                                  = 998 mBar x (0.009)

                                                  = 8.982 mBar

                                                 = 0.008982 Bar

Since 1 atm = 1.01325 Bar

0.008982 Bar x ( 1 atm/1.01325 Bar) = 0.00886 atm

Ideal gas equation is,

PV = nRT

P is pressure, V is volume, n is number of moles, R is gas constant, and T is temperaure.

Rearrange the above expression as follows:

(n/V) = P/RT

Here n/V is the concentration.

(n/V) = (0.00886 atm)/(0.08206 atm L/mol K)x(286 K) = 3.78x10^-4 mol/L

So, the concentration of argon is 3.78x10^-4 mol/L.