2. a) Calculate the weight change if a full tank of oxygen(M60 aluminum cylinder

Question

2. a) Calculate the weight change if a full tank of oxygen(M60 aluminum cylinder, 1699 liters capacity -discharge at sea level and 20 deg C) is completely emptied.(10 Kg empty wt) Assume ideal gas laws hold and the tank is emptied at 20 deg C sea level. b) If the initial fill pressure was 2000 psi, what was the tank volume in liters. Under STPD(standard temperature-0 deg C, standard pressure-sea level, and dry) conditions 1 mole of gas occupies 22.4 liters of volume The molecular weight: oxygen-16, 2.2 lbs/Kg. 760 mm Hg=14.7 psi

Explanation / Answer

a) Since the tank is at sea level, P = 101325 atm

V = 1699 L = 1.699 m3

T = 20o C = 293 K

Since this obeys ideal gas law,

PV = nRT

Substuting, we get

n = PV/RT = (101325 Pa)(1.699 m3)/(8.314 J/mol-K x 293 K)

n = 70.669 moles

mass = n x MW = 70.699 moles x 32 g/mole = 2261.43 g = 2.261 kg

b) P = 2000 psi = 2000 psi x 1000 Pa/0.145038 psi = 13.7895 x 106 Pa

Assuming temperature here is also T= 293 K and it contains the same mass of oxygen as obtained previously i.e.

n= 70.669 moles

We get V = nRT/P = 70.669 x 8.314 x 293/(13.7895 x 106 ) = 0.01248 m3 = 12.48 L

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