Questions: Determine the temperature, in K, of oxygen (O₂) at 250 bar and a specific volume of 0.003 m³/kg using generalized compressibility data and compare with the value obtained using the ideal gas model.

Solution Steps

First, we need to determine the reduced pressure \( P_r \) and reduced volume \( v_r \) using the critical properties of oxygen. The critical temperature \( T_c \) and critical pressure \( P_c \) for oxygen are:

\[ T_c = 154.6 \, \text{K}, \quad P_c = 50.43 \, \text{bar} \]

The reduced pressure \( P_r \) is given by:

\[ P_r = \frac{P}{P_c} = \frac{250 \, \text{bar}}{50.43 \, \text{bar}} = 4.958 \]

The specific volume \( v \) is given as \( 0.003 \, \text{m}^3/\text{kg} \). The critical specific volume \( v_c \) can be calculated using the critical properties:

\[ v_c = \frac{R T_c}{P_c} = \frac{0.2598 \, \text{kJ/(kg·K)} \times 154.6 \, \text{K}}{50.43 \, \text{bar}} = 0.7964 \, \text{m}^3/\text{kg} \]

The reduced volume \( v_r \) is then:

\[ v_r = \frac{v}{v_c} = \frac{0.003 \, \text{m}^3/\text{kg}}{0.7964 \, \text{m}^3/\text{kg}} = 0.0038 \]

Using the generalized compressibility chart, we locate the point corresponding to \( P_r = 4.958 \) and \( v_r = 0.0038 \). From the chart, we find the compressibility factor \( Z \).

Assuming the compressibility factor \( Z \approx 0.85 \) (this value is estimated based on typical generalized compressibility charts for the given reduced properties).

The real gas equation is given by:

\[ P v = Z R T \]

Solving for \( T \):

\[ T = \frac{P v}{Z R} \]

Substituting the known values:

\[ T = \frac{250 \, \text{bar} \times 0.003 \, \text{m}^3/\text{kg}}{0.85 \times 0.2598 \, \text{kJ/(kg·K)}} \]

Converting bar to kPa (1 bar = 100 kPa):

\[ T = \frac{250 \times 100 \, \text{kPa} \times 0.003 \, \text{m}^3/\text{kg}}{0.85 \times 0.2598 \, \text{kJ/(kg·K)}} \]

\[ T = \frac{75 \, \text{kPa·m}^3/\text{kg}}{0.2208 \, \text{kJ/(kg·K)}} \]

\[ T = 339.7 \, \text{K} \]

For the ideal gas model, the compressibility factor \( Z = 1 \). Thus, the temperature is:

\[ T_{\text{ideal}} = \frac{P v}{R} \]

Substituting the known values:

\[ T_{\text{ideal}} = \frac{250 \, \text{bar} \times 0.003 \, \text{m}^3/\text{kg}}{0.2598 \, \text{kJ/(kg·K)}} \]

Converting bar to kPa:

\[ T_{\text{ideal}} = \frac{250 \times 100 \, \text{kPa} \times 0.003 \, \text{m}^3/\text{kg}}{0.2598 \, \text{kJ/(kg·K)}} \]

\[ T_{\text{ideal}} = \frac{75 \, \text{kPa·m}^3/\text{kg}}{0.2598 \, \text{kJ/(kg·K)}} \]

\[ T_{\text{ideal}} = 288.6 \, \text{K} \]

Final Answer