Question

It is desired to inflate a baggie with a volume of 966 milliliters by filling it with argon gas at a pressure of 0.735 atm and a temperature of 299 K. How many grams of At gas are needed?

Answer 1

V = 966 ml = 0.966 L

P = 0.735 atm

T = 299 K

ideal gas equation is

PV = nRT

or

0.735 atm * 0.966 L = n * 0.0821 L-atm / mole-K * 299 K

or

n = 0.0289 mole.

mole of Ar gas = 0.0289 mole

and

grams of Ar gas needed = mole * molar mass = 0.0289 mole * 39.948 g / mole = 1.155 g