If I have 72 liters of argon gas held at a pressure of 3.4 atm and a temperature of 225k how many grams of argon do I have

516.77 grams of Argon gas is present

Explanation:

Using the gas formula

PV = nRT

number of moles (n) = mass / molar weight or mass

P = pressure = 3.4 atm

V = volume = 72 L

R = gas constant = 0.082 L atm mol^-1 K^-1

T = temperature = 225 K

MM = molar mass of Ar = 38.984 g/mol

PV = mRT / MM

m = PV MM / RT

m = 3.4 * 72 * 38.948 / 0.082 * 225

m = 9534.4704 / 18.45

m = 516.77 grams

the mass of Ar gas you have is 516.77 grams.