Let's break down the problem and solve it step-by-step. --- ## **Given Data** - **Gas:** Carbon monoxide (CO) - **Mass (\(m\))**: .5 kg - **Initial Temperature (\(T_1\))**: -100°C = 173.15 K - **Initial Pressure (\(P_1\))**: 3.1 MPa - **Final Pressure (\(P_2\))**: 2.7 MPa - **Tank**: Closed and rigid (so volume is constant) - **Required**: 1. Volume of the tank (\(V\)) 2. Final temperature (\(T_2\)), **using compressibility charts**. --- ## **Step 1: Find the Initial Specific Volume (\(v_1\))** ### **1. Convert Initial Temperature to Kelvin** \[ T_1 = -100^\circ C + 273.15 = 173.15\,K \] ### **2. Calculate the Gas Constant for CO** - **Molar mass of CO (\(M\))**: 28.01 kg/kmol - **Universal gas constant (\(R_u\))**: 8.314 kJ/kmol·K \[ R = \frac{R_u}{M} = \frac{8.314}{28.01} = .297\, \text{kJ/kg·K} = 297\, \text{J/kg·K} \] --- ### **3. Find the Compressibility Factor (\(Z_1\))** We'll use the **compressibility chart** (generalized), which requires the **reduced pressure** and **reduced temperature**. - **Critical properties of CO:** - \(T_c = 132.9\,K\) - \(P_c = 3.50\,MPa\) #### **Calculate Reduced Properties** \[ P_{r1} = \frac{P_1}{P_c} = \frac{3.1}{3.5} = .886 \] \[ T_{r1} = \frac{T_1}{T_c} = \frac{173.15}{132.9} = 1.302 \] #### **Look up \(Z_1\) from Compressibility Chart** For \(P_r = .89\) and \(T_r = 1.30\), **interpolating from generalized compressibility charts** (or using tables): - \(Z_1 \approx .88\) --- ### **4. Calculate Specific Volume (\(v_1\))** The real-gas equation is: \[ v = \frac{ZRT}{P} \] Where: - \(v\) = specific volume (m³/kg) - \(Z\) = compressibility factor - \(R\) = specific gas constant (J/kg·K) - \(T\) = temperature (K) - \(P\) = pressure (Pa) \[ v_1 = \frac{Z_1 R T_1}{P_1} \] Convert \(P_1\) to Pa: \[ P_1 = 3.1 \text{ MPa} = 3.1 \times 10^6 \text{ Pa} \] Plug in values: \[ v_1 = \frac{.88 \times 297 \times 173.15}{3.1 \times 10^6} \] Calculate numerator: \[ .88 \times 297 \times 173.15 = 45,338.16 \] \[ v_1 = \frac{45,338.16}{3,100,000} \approx .0146\,\text{m}^3/\text{kg} \] --- ### **5. Calculate Volume of the Tank (\(V\))** \[ V = m \times v_1 = .5 \times .0146 = .0073\,\text{m}^3 \] --- ## **Step 2: Find the Final Temperature (\(T_2\))** **Since the tank is rigid, \(V\) is constant, so \(v_2 = v_1\).** --- ### **1. Calculate Reduced Pressure at Final State** \[ P_2 = 2.7\,\text{MPa} \] \[ P_{r2} = \frac{P_2}{P_c} = \frac{2.7}{3.5} = .771 \] --- ### **2. Use \(v_2 = v_1\) to Find \(T_2\)** \[ v_2 = v_1 = .0146\,\text{m}^3/\text{kg} \] Set up the real-gas equation for the final state: \[ v_2 = \frac{Z_2 R T_2}{P_2} \] Or, \[ T_2 = \frac{v_2 P_2}{Z_2 R} \] But we need \(Z_2\), which depends on \(T_{r2}\), which in turn depends on \(T_2\). **This requires an iterative approach.** --- ### **3. Iterative Solution** #### **First Guess: Assume \(Z_2 \approx .87\) (\(T_{r2} \approx 1.20\)), then update as needed.** \[ T_2 = \frac{.0146 \times 2,700,000}{.87 \times 297} \] \[ T_2 = \frac{39,420}{258.39} \approx 152.6\,K \] #### **Calculate Reduced Temperature** \[ T_{r2} = \frac{T_2}{T_c} = \frac{152.6}{132.9} = 1.15 \] From the chart, for \(P_r = .77\), \(T_r = 1.15\), \(Z_2 \approx .85\). #### **Recalculate \(T_2\) with new \(Z_2\):** \[ T_2 = \frac{.0146 \times 2,700,000}{.85 \times 297} \] \[ T_2 = \frac{39,420}{252.45} \approx 156.2\,K \] \[ T_{r2} = \frac{156.2}{132.9} = 1.18 \] From the chart, \(Z_2 \approx .86\). #### **One more iteration:** \[ T_2 = \frac{39,420}{.86 \times 297} = \frac{39,420}{255.42} \approx 154.3\,K \] \[ T_{r2} = \frac{154.3}{132.9} \approx 1.16 \] So, **final answer converges to:** --- ## **Final Answers** ### **(a) Volume of the tank:** \[ \boxed{.0073\,\text{m}^3} \] ### **(b) Final temperature:** \[ \boxed{154\,K} \] --- ### **Summary Table** | State | \(P\) (MPa) | \(T\) (K) | \(v\) (m³/kg) | \(Z\) | |---------|-------------|-----------|--------------|-------| | Initial | 3.1 | 173.2 | .0146 | .88 | | Final | 2.7 | 154 | .0146 | .86 | --- **Let me know if you have questions or want to see the compressibility chart or any calculation step!**