Question

Calculate the specific volume of Helium using the compressibility factor. 500 kPa, 60 degrees Celsius

Answer 1

Explanation:

Formula for compressibility factor is as follows.

                     z = $\frac{P \times V_{m}}{R \times T}$

where,     z = compressibility factor for helium = 1.0005

               P = pressure

          $V_{m}$ = molar volume

                R = gas constant = 8.31 J/mol.K

                T = temperature

So, calculate the molar volume as follows.

                $V_{m} = \frac{z \times R \times T}{P}$

                             = $\frac{1.0005 \times 8.314 \times 10^{-3} m^{3}.kPa/mol K \times (60 + 273)K}{500 kPa}$

                             = 0.0056 $m^{3}/mol$

As molar mass of helium is 4 g/mol. Hence, calculate specific volume of helium as follows.

                    $V_{sp} = \frac{V_{m}}{M_{w}}$

                           = $\frac{0.0056 m^{3}/mol}{4 g/mol}$

                           = 0.00139 $m^{3}/g$

                           = 0.00139 $m^{3}/g \times \frac{1 g}{10^{-3}kg}$

                                 = 1.39 $m^{3}/kg$

Thus, we can conclude that the specific volume of Helium in given conditions is 1.39 $m^{3}/kg$.