## Step-by-Step Solution Let's break down the problem and solve it step by step. ### **Given Data** - Viscosity at room temperature (\(\eta_1\)): **1 Pa·s** - Room temperature (\(T_1\)): **Assume 25°C = 298 K** - Temperature at which to find viscosity (\(T_2\)): **100°C = 373 K** - Activation energy (\(Q\)): **10 kJ/mol = 10,000 J/mol** - Gas constant (\(R\)): **8.314 J/(mol·K)** --- ### **Step 1: The Viscosity-Temperature Relationship** Viscosity \(\eta\) changes with temperature according to the **Arrhenius equation**: \[ \eta = A \exp\left(\frac{Q}{RT}\right) \] Taking the ratio at two temperatures: \[ \frac{\eta_2}{\eta_1} = \exp\left( \frac{Q}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \right) \] --- ### **Step 2: Plug in the Values** - \(Q = 10,000\) J/mol - \(R = 8.314\) J/(mol·K) - \(T_1 = 298\) K (room temperature) - \(T_2 = 373\) K (100°C) \[ \frac{\eta_2}{\eta_1} = \exp\left( \frac{10,000}{8.314} \left(\frac{1}{373} - \frac{1}{298}\right) \right) \] --- ### **Step 3: Calculate the Exponent** #### **a) Calculate \( \frac{1}{T_2} - \frac{1}{T_1} \):** \[ \frac{1}{373} - \frac{1}{298} = 0.002681 - 0.003356 = -0.000675 \; \text{K}^{-1} \] #### **b) Calculate \( \frac{Q}{R} \):** \[ \frac{Q}{R} = \frac{10,000}{8.314} = 1202.6 \] #### **c) Multiply:** \[ 1202.6 \times (-0.000675) = -0.8118 \] --- ### **Step 4: Exponential Term** \[ \exp(-0.8118) = 0.444 \] --- ### **Step 5: Find the Final Viscosity** \[ \eta_2 = \eta_1 \times 0.444 = 1 \times 0.444 = 0.444 \text{ Pa·s} \] --- ## **Final Answer** \[ \boxed{0.44\ \text{Pa·s}} \] **(rounded to two decimal places)** --- ### **Summary Table** | Temperature (°C) | Temperature (K) | Viscosity (Pa·s) | |------------------|-----------------|------------------| | 25 | 298 | 1.00 | | 100 | 373 | 0.44 | --- **Explanation:** As temperature increases, the viscosity of the polymer decreases due to increased atomic/molecular motion, making it easier for the molecules to slide past each other.