Question

A binary mixture of n-butane (\(C_4H_{10}\)) and n-pentane (\(C_5H_{12}\)) is under thermodynamic equilibrium at 180°F and 95 psia. The vapor pressures of pure C4H10 and pure C5H12 at 180°F are 160 psia and 54 psia, respectively. Assuming ideal solution behavior (i.e., Raoult's law and Dalton’s law are valid), the mole fraction of the n-butane in the gas phase is ............. (rounded to three decimal places).

Hint:

- For ideal mixtures, use Raoult’s law to calculate the partial pressures, and then apply Dalton’s law to find the total pressure and mole fractions.

Answer 1

According to Raoult's law, the mole fraction of n-butane in the gas phase is given by the equation: \[ y_{C4H10} = \frac{P_{C4H10}}{P_{\text{total}}} \] Where:
- \( P_{C4H10} \) is the partial vapor pressure of n-butane,
- \( P_{\text{total}} \) is the total vapor pressure of the mixture.
We can calculate the partial vapor pressure of n-butane using Raoult's law:
\[ P_{C4H10} = x_{C4H10} P^0_{C4H10} \] Where:
- \( x_{C4H10} \) is the mole fraction of n-butane in the liquid phase,
- \( P^0_{C4H10} \) is the vapor pressure of pure n-butane.
Similarly, for n-pentane: \[ P_{C5H12} = x_{C5H12} P^0_{C5H12} \] The total pressure is the sum of the partial pressures of the two components: \[ P_{\text{total}} = P_{C4H10} + P_{C5H12} \] After solving the system of equations, the mole fraction of n-butane in the gas phase is calculated to be approximately \( 0.645 \). Thus, the correct answer is \( \boxed{0.645} \).