Question

Ethylene obeys the truncated virial equation-of-state: \[ \frac{PV}{RT} = 1 + \frac{B}{V}, \] where \( P \) is the pressure, \( V \) is the molar volume, \( T \) is the absolute temperature, and \( B \) is the second virial coefficient. The universal gas constant \( R = 83.14 \, \text{bar cm}^3 \text{mol}^{-1}\text{K}^{-1} \). At 340 K, the slope of the compressibility factor vs. pressure curve is \( -3.538 \times 10^{-3} \, \text{bar}^{-1} \). Let \( G^R \) denote the molar residual Gibbs free energy. At these conditions, the value of \( \left( \frac{\partial G^R}{\partial P} \right)_T \), in \( \text{cm}^3 \text{mol}^{-1} \), rounded off to 1 decimal place, is:

Hint:

For virial equations, use the compressibility factor relations and carefully handle partial derivatives with respect to pressure.

Answer 1

Given: \[ \frac{PV}{RT} = 1 + \frac{BP}{RT}. \] \[ R = 83.14 \, \text{bar} \, \text{cm}^3/\text{mol·K}. \] At \( 340 \, \text{K} \): \[ \left(\frac{\partial \left(\frac{Z}{P}\right)}{\partial P}\right)_T = -3.53 \times 10^{-3} \, \text{bar}^{-1}. \] Using the equation: \[ \frac{\partial \left(\frac{Z}{P}\right)}{\partial P} = \frac{1}{P^2} - \frac{B}{RT}. \] Rearranging: \[ \text{Slope} = \frac{\partial \left(\frac{Z}{P}\right)}{\partial P} = -\frac{B}{RT}. \] Substitute the values: \[ -3.53 \times 10^{-3} = -\frac{B}{RT}. \] Solving for \( B \): \[ B = -3.53 \times 10^{-3} \times 340 \, \text{bar·K} \times 83.14. \] \[ B = -100 \, \text{cm}^3/\text{mol}. \] Final Answer: \[ \boxed{B = -100 \, \text{cm}^3/\text{mol}.} \]