how many grams of Ne gas are needed?

Question

It is desired to inflate a baggie with a volume of 803 milliliters by filling it with neon gas at a pressure of 0.739 atm and a temperature of 298 K. How many grams of Ne gas are needed?  ____ g

Answer 1

Ideal Gas Law:

PV = nRT

P = pressure = 0.739 atm

V = volume in liters = 803 ml x 1 L/1000 ml = 0.803 L

n = moles = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = 298K

Solving for moles (n), we have...

n = PV/RT = (0.739)(0.803 L)/(0.0821 Latm/Kmol)(298K)

n = 0.0243 moles Ne

Grams Ne = 0.0243 moles Ne x 20.2 g/mole Ne = 0.491 g Ne