Ammonia has a molecular weight of 17-031- How much (kg) of the substan.docx

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Ammonia has a molecular weight of 17.031. How much (kg) of the substance is contained in a tank or volume 5 m^3 at 100 kPa and 300 k. Calculate mass (kg) if the substance b considered to be an ideal gas and real gas. and the percent difference. Solution The Equation is ; P * V = Z* n* R * T P = 100 kpa V = 5 m Â´3 T = 300 K R = 8.31 J / K * mol n = m / M m = mass M = Molar mass = PM * 1 gr / mol = 17.031 gr / mol Z = Compressibility factor so : P * V = Z* ( m/M )* R * T m = ( P * V * M ) / R *T * Z m = ( ( 100 kpa ) * ( 5 m3 ) * ( 17.031 gr / mol ) / ( 8.31 J / K * mol ) * ( 300 K ) * Z m = 3415.56 gr * Z If is a ideal gas , Z = 1, so : m = 3.42 kg If is a real gas , Z will be less to 1, so : m = 3.42 kg * Z will be less than 3.42 kg his difference will be: ideal mass = mi = 3.42 kg real mass = mr = 3.42 kg * Z Diferrence = mi - mr = ( 3.42 kg -Â Â 3.42 kg * Z ) = 3.42 kg * ( 1 - Z ) .

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If is a ideal gas , Z = 1, so :
m = 3.42 kg
If is a real gas , Z will be less to 1, so :
m = 3.42 kg * Z will be less than 3.42 kg
his difference will be:
ideal mass = mi = 3.42 kg
real mass = mr = 3.42 kg * Z
Diferrence = mi - mr = ( 3.42 kg -Â Â 3.42 kg * Z ) = 3.42 kg * ( 1 - Z )

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