Question

3. According to the publication NFPA58, a 20 lb (9.09 kg) propane (C3H8) tank at 76°F (25°C) would have a pressure of 150 psig (10.3bar). Determine the internal volume in Liters of the tank using i) van der Waals EOS, and ii) SRK EOS. For propane van der Waals constants are a= 9.385 bar L/mold and b = 0.09044 L/mol. In addition, Tis 369.9 K, PC is 42.Oatm and o is 0.152. BERNZOMATIC MONTAN

Answer 1

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The van der waal equation of state is given below

Pa RT T-6 a 72

27(RTC) 64PC

RTC b = SPC

Given

a = 9.385 bar L²/mol²

b=0.09044 L/mol

Given P = 10.3 bar = 10.167 atm

T = 25 C = 298.15 K

R =0.082 L atm K^-1mol^-1

Substituiting in Vander waal's equation of state, we get

V = 2.0764 L

Now we can use SRK equation of state

It is given by

RT Pey-b V(V+b)

Substituting and solving for V, we get

We get V = 2.0980L

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