The mass density of propane at 3.50x106 Pa and 500.0 K is 48.07 kg m–3 . The mol

Question

The mass density of propane at 3.50x106 Pa and 500.0 K is 48.07 kg m–3 . The molar mass of propane is 42.0956 g mol–1

a. Using the mass density calculate the molar volume of propane.

b. Calculate the compressibility Z of propane from your answer to part a and the pressure (3.50x106 Pa) and temperature (500.0 K).

c. Using the van der Waals parameters a = 0.8779 m6 Pa mol–2 and b = 8.445x10–5 m3 mol–1, calculate the viral coefficients B and C for propane at 500 K.

d. Calculate the compressibility Z predicted by the truncated virial expansion: € Z =1+ B/V +C/V 2 at the molar volume computed in part (a).

e. Determine how well the virial equation predicts the compressibility by calculating the % difference in the values of Z obtained in (b) and (c). The % difference between two values, x and y, is defined as: | x y |/ 1/2(x + y) ×100

f. Is propane more or less compressible than an ideal gas at these temperature and pressure conditions?

Explanation / Answer

a) The molar volume of Propane can be calculated by using the ideal gas equation;

Vm = RT/P

Pressure = 3.5 * 106 Pa = 34.54 atm

Temperature = 500 K

R = 0.082095 L atm mol-1 K-1

Now, using the above equation,

Vm = (0.08205 * 500)/34.54

Or, Vm = 1.188 L/Mol

b)

Compressibility of Propane Gas; Z can be calculated by using the Viral Equation;

i.e                                                      Z = PVm /RT

or,                = (34.54 * 1.188)/(0.08205 * 500)

or, Z = 1.000195

or, Z = 1.0

c)

Viral Coefficient

                                                  PVm /RT = 1 + B/Vm

Or, B = 0

                                                  PVm /RT = 1 + B/Vm + C/ Vm2

Or, C = 0

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