Question

¿ Que volumen ocupara 0,2 libras de argon a 80ºC y 780 mmHg?

Answer 1

V· P = n · R · T   (Ley de los gases ideales)

V = ?
P = 780 mmHg / 760 = 1.026 atm
T = 80 ºC + 273 = 353 K
m = 0.2 lb
g = lb  x  453.59
g = 0.2 x 453.59 = 90.718 g
R = 0.0821 (L atm / mol K)
Mm Ar = 39.94g/mol

1. calcular mol  mol = masa/Mm
mol = 90.718 g / 39.94 g/mol
mol = 2.271

2. calcular volumen  (ley de los gases ideales)

V · P = n · R · T  despejar volumen

V =  2.271 mol   ·   0.0821 (L atm / mol K)  ·  353 K
     ````````````````````````````````````````````````````````````````````
                                   1.026 atm

V = 64.14 L