Question

At 215 °C, the viscosity of a polystyrene of molecular weight 250 × 10³ g/mol is 8.0 × 10³ Pa·s. The critical molecular weight of polystyrene, \( M_c = 35 \times 10^3 \) g/mol. For a similar polystyrene of molecular weight 500 × 10³ g/mol, the viscosity (rounded off to nearest integer) will be \(\underline{\hspace{2cm}}\) × 10³ Pa·s.

Hint:

To estimate viscosity for different molecular weights, use the equation \( \eta = K M^n \) and the ratio of viscosities for different molecular weights.

Answer 1

Correct Answer: 62

The viscosity \( \eta \) of a polymer is related to its molecular weight by the empirical equation:
\[ \eta = K M^n \] where \( M \) is the molecular weight and \( K \) and \( n \) are constants. The relationship for the viscosity of two polymers with different molecular weights is:
\[ \frac{\eta_2}{\eta_1} = \left( \frac{M_2}{M_1} \right)^n \] Given:
- \( \eta_1 = 8.0 \times 10^3 \) Pa·s, \( M_1 = 250 \times 10^3 \) g/mol,
- \( M_2 = 500 \times 10^3 \) g/mol,
- \( M_c = 35 \times 10^3 \) g/mol, and
- Using \( n = 0.8 \) (for polystyrene),
\[ \frac{\eta_2}{8.0 \times 10^3} = \left( \frac{500 \times 10^3}{250 \times 10^3} \right)^{0.8} \] Solving:
\[ \eta_2 \approx 8.0 \times 10^3 \times (2)^{0.8} \approx 8.0 \times 10^3 \times 1.741 \approx 13.9 \times 10^3 \, \text{Pa·s}. \] Thus, the viscosity will be approximately \( 14 \times 10^3 \) Pa·s.