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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 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.
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