Ar fills a 8.40 L container at 9.20x10^3 torr and 18 degrees Celsius. How many mol of Ar are in the container

Question

could some one show me how to work this problem out please

i just cant seem to figure it out

Answer 1

From the Ideal Gas Law

P2V2/T2 = P1V1/T1

V2 = V1(P1/P2)(T2/T1)

At Standard Temperature and Pressure,

ρ = (1 mol/22.4 L)

(8.40 L)(9200 torr/760 torr)(293.15 °K/291.15 °K)(1 mol/22.4 L) ≈ 4.57065674 mol ≈ 4.57 mol