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Determine the thermodynamic state of 1,1,1,2-Tetrafluoroethane (R-134a) at 102 °C and 1000 kPa. Critical Pressure= 4.059...

Determine the thermodynamic state of 1,1,1,2-Tetrafluoroethane (R-134a) at 102 °C and 1000 kPa. Critical Pressure= 4.059 MPa, Critical Temperature = 101.06°C.

*I keep getting stuck with a quadratic equation and two volume values, when using the van der Waals equation of state (P=(RT/(V-b))-(a/V^2))....I am determining the thermodynamic state by finding if the compressibility factor (z) is less than or greater than 1, to find if attractive or repulsive forces dominate.

Solutions

Expert Solution

In order to find the value of Z as per the discussion added in the question we will make use of Compressibility chart rather than vander waal's equation


To solve this we will find Pr and Tr and than make use of graph to find Z

now Pr =P/Pc = 1000/4059 = 0.246

Tr =T/Tc = (102+273.15)/ (101.06+273.15) =1.0025

now we will locate this point to find the value of Z using the above chart


From above graph

Z = 0.38

Z< 1 hence attractive forces are dominating

Now since we know the value of Z we can use PV =n R T and find the value of V, once V is found we will have all the state point ( P, V,T) hence the thermodynamic state will become completely defined

P V = Z R T

1000 * V = 0.38*8.314*(102 +273.15)

V = 1.185 Liter/mol


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