Let's break down the given problem and solve it step by step. **Given:** - Refrigerant 134a at -26°C - Volume of tank, \( V = 7 \, m^3 \) - Quality of mixture (x) = 60% = 0.6 - Specific volume of saturated liquid (\(v_f\)) = 0.0007265 \( m^3/kg \) - Specific volume of saturated vapor (\(v_g\)) = 0.18946 \( m^3/kg \) **Asked:** - Percentage of saturated liquid occupied in the tank --- ### **Step 1: Calculate the total mass in the tank** The specific volume of the mixture is: \[ v = v_f + x(v_g - v_f) \] \[ v = 0.0007265 + 0.6 \times (0.18946 - 0.0007265) \] \[ v = 0.0007265 + 0.6 \times 0.1887335 \] \[ v = 0.0007265 + 0.1132401 \] \[ v = 0.1139666 \, m^3/kg \] Total mass, \( m \): \[ m = \frac{V}{v} = \frac{7}{0.1139666} = 61.41 \, kg \] --- ### **Step 2: Mass of saturated liquid** Mass of saturated liquid, \( m_f \): \[ m_f = m \times (1-x) = 61.41 \times (1-0.6) = 61.41 \times 0.4 = 24.564 \, kg \] --- ### **Step 3: Volume occupied by saturated liquid** Volume by saturated liquid, \( V_f \): \[ V_f = m_f \times v_f = 24.564 \times 0.0007265 = 0.01785 \, m^3 \] --- ### **Step 4: Percentage of tank occupied by saturated liquid** \[ \text{Percentage occupied} = \frac{V_f}{V} \times 100 = \frac{0.01785}{7} \times 100 \approx 0.255\% \] --- ### **Answer:** The percentage of saturated liquid occupied in the tank is **0.25%** (rounded to two decimal places). So the correct option is: - **0.25%**