Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the following experimental values of virial coefficients: B = −152.5 cm3·mol−1 C = −5800 cm6·mol−2 (b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)]. (c) The steam tables (App. E).

Explanation:

Given that:

the temperature = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

where; B = -   C = -5800

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

Multiplying through with V² ; we have

V = 2250.06  cm³ mol⁻¹

Z =

Z =

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

The compressibility is calculated as:

Z = 0.9386

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V  =  0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z =

Z =

Z = 0.588

possibly, yes.

explanation:

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