Ideal Gas law Power Point #24

Given:

V = 2.20 L

P = 102 kPa

T = 37^oC + 271 = 300 K

Molar mass: 29 g/mol

Required:  M

Calculation:

n = PV/RT

n = (102 kPa x 2.20 L)/((8.31 LkPa/Kmol)x 300 K)

n = 224.4 LkPa/(2493 LkPa/mol)

n = 0.09 mol

M = moles x molar mass

M = 0.09 mol x 29 g/mol = 2.60 g

Paraphrase: The lung holds 2.60 grams of air.

Analysis: It makes sense because it’s not far from how much much air an average child’s lungs can hold, which is around 2.20 g.