Mass of vaporized triethylamine : 2.606 g
Given
0.5 L triethylamine
P = 18.5 psi
T = 25 °C
Required
mass of vaporized triethylamine
Solution
Conversion :
P 18.5 psi = 1,26 atm
T = 25 +273 = 298 K
Ideal gas law :
PV=nRT
n = PV/RT
Input the value :
n = (1.26 atm x 0.5 L) /(0.08205 x 298)
n = 0.0258
MW triethylamine = 101 g/mol
Mass triethylamine :
= n x MW
= 0.0258 x 101 g/mol
= 2.606 g