# PA lab - Adam Romman Partner Steven Chen Liquid-Vapour...

This preview shows pages 1–5. Sign up to view the full content.

Adam Romman Partner: Steven Chen Liquid-Vapour Phase Lab 6/22/2007 Data Analysis 1) Determination of System Volume Pc*Vc = Pb*Vb Let ‘c’ be for the cross volume and pressure, and ‘b’ be for both the cross and sample volume and pressure. Thus, Vb = Vc + Vs where Vs is only the sample volume. Pc*Vc = Pb*(Vc+Vs) 80.42*Vc = 51.17*(Vc+48.7) Vc = 85.2 Vb = 133.9 Volume (cm^3) Cross Volume (Vc) 85.2 Sample Volume (Vs) 48.7 System Volume (Vb) 133.9 pressure (torr) Cross Pressure (Pc) 80.42 Combined Pressure (Pb) 51.17 2) Experimental Step Dosing Pressure in Sample Volume (Pd) (torr) Initial Pressure in Cross Volume (Pi) (torr) Equilibrium Pressure in Cross and Sample Volume (Peq) (torr) 1 -0.09 34.56 10.79 2 10.79 41.87 22.68 3 22.68 49.68 33.39 4 33.39 56.85 42.67 5 42.67 63.44 51.02 6 51.02 70.09 58.73 7 58.73 77.7 66.27 8 66.27 84.63 73.97 9 73.97 91.02 81.07 10 81.07 98.97 88.34 3) Vj = Tstp * (Pd*Vs + Pi*Vc – Peq*Vb) ] T N2 *Pstp T N2 = bath temperature = 77K stp = standard temperature and pressure

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Tstp = 273.15 K Pstp = 760 torr V 1 = [273.15*(-.09*48.7 + 34.56*85.2 – 10.79*133.9)] /(77*760) = 6.979 cm Experimental Step Vj (cm^3) 1 6.98 2 4.93 3 4.04 4 3.53 5 3.04 6 2.76 7 2.83 8 2.49 9 2.34 10 2.57 4) Total Volume from step 2 = Volume from Step 1 + Volume from step 2 Vi2 = 6.979 + 4.928 = 15.951 cm Xi = Peq/Po Xi = relative pressure Peq = equilibrium pressure Po = vapor pressure of Argon at 77K – 181.1 torr X1 = 10.79/181.1 = 0.0596 Experimental Step Xi 1 0.0596 2 0.1252 3 0.1844 4 0.2356 5 0.2817 6 0.3243 7 0.3659 8 0.4084 9 0.4477 10 0.4878
Isotherm: Volume vs Relative Pressure 0 5 10 15 20 25 30 35 40 0.0 0.1 0.2 0.3 0.4 0.5 0.6 Xi Relative pressure Vi volume (cm^3 Figure 1 Isotherm. The figure shows a graph of volume adsorbed vs relative pressure, where relative pressure is equilibrium pressure / vapor pressure. The gas observed is argon and the adsorbent is zeolite. The temperature is at 77 K. 5) y = mx + b m = b = y = x = Graph y vs x, find trendline, and solve for m and b. x1 = 0.0596 y1 = 0.0596/ (6.979*(1-0.0596)) = 0.00908 System of equations: 1) m = 0.039659 = (C-1)/(C*Vm) 2) b = 0.006615 = 1/(C*Vm) C*Vm = 1/0.006615 = 151.169 0.039659 = (C-1) / 151.169 C = 6.995 0.006615 = 1 / (6.995*Vm) Vm = 21.61 cm C 7.00 Vm (cm^3) 21.61

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
BET isotherm y = 0.0397x + 0.0066 R 2 = 0.9934 0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.0 0.1 0.2 0.3 0.4 0.5 0.6 Xi Xi/ [Vi*(1-Xi)] (cm^- Figure 2. The figure is the BET isotherm for argon gas in zeolite. The gas observed is argon and the adsorbent is zeolite. The temperature is at 77 K. slope
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

PA lab - Adam Romman Partner Steven Chen Liquid-Vapour...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online