Let's solve this step by step using the **ideal gas law**: ## Step 1: Write out the Ideal Gas Law The **ideal gas law is: \[ PV = nRT \] For mass-based calculations, it's often written as: \[ PV = mRT \] Where: - \(P\) = pressure (kPa) - \(V\) = volume (m³) - \(m\) = mass (kg) - \(R\) = specific gas constant for air (\(R = .287 \, \text{kJ/kg·K}\)) - \(T\) = temperature (K) --- ## Step 2: Find the Volume of the Tank Initial conditions: - \( m_1 = 2 \, \text{kg} \) - \( P_1 = 400 \, \text{kPa} \) - \( T_1 = 20^\circ\text{C} = 293.15 \, \text{K} \) From the ideal gas law: \[ V = \frac{m_1RT_1}{P_1} \] Plug in the values: \[ V = \frac{2 \times .287 \times 293.15}{400} \] \[ V = \frac{2 \times .287 \times 293.15}{400} \] \[ = \frac{168.4491}{400} \] \[ = .4211 \, \text{m}^3 \] --- ## Step 3: Write the Final State Equation After adding air: - \( m_2 = 2 + .6 = 2.6 \, \text{kg} \) - \( T_2 = 32^\circ\text{C} = 305.15 \, \text{K} \) - \( V \) (same as above, since the tank is rigid) Use the ideal gas law for the final state: \[ P_2V = m_2RT_2 \] \[ P_2 = \frac{m_2RT_2}{V} \] --- ## Step 4: Solve for the Final Pressure Plug in all known values: \[ P_2 = \frac{2.6 \times .287 \times 305.15}{.4211} \] Calculate numerator: \[ 2.6 \times .287 \times 305.15 = 228.095 \] So, \[ P_2 = \frac{228.095}{.4211} \] \[ P_2 = 541.7 \, \text{kPa} \] --- ## Step 5: Final Answer (Three Significant Figures) \[ \boxed{542\ \text{kPa}} \] --- ### **Summary Table** | Step | Value | |----------------|----------------| | Initial mass | 2 kg | | Added mass | .6 kg | | Final mass | 2.6 kg | | Initial temp | 293.15 K | | Final temp | 305.15 K | | Initial pressure | 400 kPa | | Final pressure | **542 kPa** | --- **Final Answer:** > The pressure in the tank after adding .6 kg of air and heating to 32°C is **542 kPa** (to three significant figures).