Question

Ethane is contained in a cylinder that is 5 m^{3}. if
the gas is at 38 °C and 135 atm, what is the mass of ethane that is
in the tank? Compare the values by IGL, truncated equation of state
after second term, and the compressibility factor?

Answer #1

According to the ideal gas equation: PV = nRT

Where 'P' is the pressure of the gas = 135 atm

'V' is the volume of the gas = 5 m^{3} =
5*(10^{2} cm)^{3} = 5000*10^{3}
cm^{3} = 5000 L

'n' is the no. of moles of the gas = mass/molar mass = mass/30
g.mol^{-1}

'R' is the universal gas constant = 0.08206 L atm
mol^{-1} K^{-1}

'T' is the absolute temperature = 38 ^{o}C = (38+273.15)
K = 311.15 K

Therefore, 135 atm * 5000 L = mass/30 g.mol^{-1} *
0.08206 L atm mol^{-1} K^{-1} * 311.15 K

i.e. The mass of ethane in the tank = 793092.3 g =
**793.092 kg**

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