A rigid, closed vessel of volume V = 648 liters maintained at constant temperature T=365 K is loaded with 112 mol of n-hexane, 155 mol of n-heptane, and 14 mol of nitrogen. Calculate the pressure in the vessel and the composition of all phases present at equilibrium. Make the following three assumptions: a) n-hexane and n-heptane form an ideal liquid solution, b) the vapor phase is ideal, c) nitrogen has negligible solubility in the liquid phase at the conditions of interest. The vapor pressures of n-hexane and n-heptane at T=365 K are P1=0.199 MPa and P2=0.083 MPa, respectively. Their (liquid) molar volumes are V¯ 1=0.146 l/mol and V¯ 1=0.162 l/mol, respectively.

P = 2.92 atm

Explanation:

With the three assumptions in mind, the system consists of:

A liquid phase containing n-hexane and n-heptane, and A gaseous phase containing n-hexane vapor, n-heptane vapor, and nitrogen gas.

First we use PV=nRT to calculate the moles of n-hexane and n-heptane in the gaseous phase:

n-hexane:

P = 0.199 MPa ⇒ 0.199 * 1.869 = 1.964 atm

1.964 atm * 648 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 365 K n = 42.52 moles n-heptane:

P = 0.083 MPa ⇒ 0.083 * 1.869 = 0.155 atm

0.155 atm * 648 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 365 K n = 3.358 moles

So the gaseous phase consists of 42.52 moles of n-hexane, 3.358 moles of n-heptane, and 14 mol of nitrogen.

For the liquid phase, we calculate the remaining moles of n-hexane and n-heptane. Then we convert to liters, using their molar volumes:

n-hexane: n = 112 mol - 42.52 mol = 69.48 mol 69.48 mol * 0.146 L/mol = 10.14 L n-heptane: n = 155 mol - 3.358 mol = 151.642 mol 151.642 mol * 0.162 L/mol = 24.57 L

So the liquid phase occupies (10.14+24.57) = 34.71 L, and contains 69.48 mol of n-hexane and 151.64 mol of n-heptane.

Finally, to calculate the pressure in the vessel, we use PV=nRT:

P = ?

V = 648 - 34.71 = 613.29 L

n = 42.52 mol hexane + 3.35 mol heptane + 14 mol nitrogen = 59.87 mol

T = 365 K

P * 613.29 L = 59.87 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 365 K P = 2.92 atm