Question

1. n-Pentane gas goes through an isenthalpic (∆H = 0) process known as throttling. The pentane gas starts at 170 bar and 375 K, and ends at 3.5 bar, where it is partially condensed.

b.) (20 pts) What is the sensible heat released from the initial state to the saturated vapor state at 3.5 bar? Note: at no point in this process is the gas behaving as an ideal gas.

c.) (10 pts) What is the heat of condensation (∆H_cond = -∆H_vap) at the temperature corresponding to a saturation pressure of 3.5 bar?

d.) (10 pts) What fraction of the vapor condenses? Recall, for the process, ∆H = 0.

ans:

B: Hr1=-22.2 kJ/mol; H2r=-1.016 kJ/mol; ∆Hig = -3.384 kJ/mol; ∆H = 1.7822 kJ/mol

C.) ∆Hcond = -23.061 kJ/mol; D.) xL = 0.773;

Answer 1

B) sensible heat = ?H- ?Hig- ?H2r- ?Hr1 = 1.78-(-3.3-1.106-22.2) = 28.38 KJ/mol

C) At 3.5 bar saturation pressure, Temperature = 65 degreeC = 338 K

Reference from Chart below (Note: 3.5 bar = 3.5*760 =2660 mmHg)

For Pentane, ? Hvap at saturation T is given by formula below

Using T=338 degree C, ?Hvap = 23.06 KJ/mol

Therefore, ?Hcon = -?Hvap = -23.06 KJ/mol

d) fraction of Gas condensed = -Specific heat/Heat of condensation = 23.06/28.38 = 0.81