Question

Que volumen ocupara 3,5 moles de amoniaco a -13°c y 4,9 atmosferas

Answer 1

n = 3.5 moles
T = -13 ºC + 273 = 260 K
R = 0.0821 (l x atm/mol x K)
P = 4.9 atm.

P x V  = n x R x T

V =  n x R x T / P

V = 3.5 moles x 0.0821 (l x atm/mol x K) x 260 k / 4,9  atm

V = 15.24 L