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a 5.0 gallon container contains 15.0 moles of propane gas at room temperature (20. °c). what is the pressure of the gas? (1.0 gal = 3.79 l)

Respuesta :

scme1702
We apply the ideal gas equation:
PV = nRT
V = 5 x 3.79/1000 m³
= 0.01895
T = 20 + 273 = 293 K
n = 15
R = 8.314
P = nRT/V
= (15 x 8.314 x 293)/(0.01895)
= 1,928,234 Pa ≈ 1.93 MPa

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