Question

What volume is occupied by 28.2 g of argon gas at a pressure of 1.10 atm and a temperature of 326 K ?

Answer 1

Molar mass of Ar = 40 g.mol^-1.

Thus, 28.2 g of Ar = (28.2 / 40) mol = 0.705 mol

From ideal gas equation we know,

$pV=nRT$

$Or,V=\frac{nRT}{p}=\frac{0.705mol\times 0.0821 L.atm.K^{-1}.mol^{-1}\times 326K}{1.10atm}=17.154L$

So, the answer is 17.154 L.