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FTIR Gas Analysis Module Spectrum GX FTIR System – Perkin Elmer – Volume: 8.5 L – Wavenumber range: 10,000-400 cm -1 – Resolution: 0.5-64 cm -1 M-5-22-V variable pathlength long path gas cell – Infrared Analysis, Inc. Cell path length = # of passes * length of base path – Length of base path = 56 cm – # of passes = 4-64 (intervals of 4) – Path length = 2.24-35.84 m 1
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IR Spectrum of H 2 O 3 modes of vibration – Bent triatomic molecule k 1 = 3657 cm -1 k 2 = 1595 cm -1 k 3 = 3756 cm -1 O O H H H H O O H H H H O O H H H H K. Nakamoto. Infrared and Raman Spectra of Inorganic and Coordination Compounds, Part A: Theory and Applications in Inorganic Chemistry. Wiley Publishing, 6 th ed. (2009). k 1 = 3657 cm -1 k 2 = 1595 cm -1 k 3 = 3756 cm -1
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FTIR – Principles of Operation 3
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FTIR H 2 O Gas Analysis Monitor wavenumber range: 2,000-1,200 cm -1 – Monitoring k 2 vibration Resolution: 0.5 cm -1 Cell path length determines minimum detectable limit – Graph on next slide 4
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FTIR H 2 O Minimum Detectable Limit 5 Range provided to illustrate best case scenario (low) and worst case scenario (high), which is dependent on amount of H 2 O in background
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Minimum Detectable Limit for H Concentration Assumption: No source of hydrogen is available other than the sample: From the stoichiometric relationship we have: H 2 O => 2H + ½O. Detectable limit for H 2 O in gas (Ar+O 2 ) at equilibrium is => 0.015- 0.05 which is achieved by increasing the cell path length as high as ~ 20 m. After this (0.015- 0.05 ppm H 2 O) the possible detectable limit gets saturated (clear from the asymptotic shape of curve), i.e. after ~ 20 m of cell path length, the increasing cell path length can’t give any significant increase for the detectable limit of water. H 2 O => 2H + ½O…. From this equation, the detectable limit for H gave up by the sample in the gas (Ar+O 2 ) at equilibrium should be 0.03- 0.1 ppm. If we choose to use sample size of length of 6mm and diameter of 6mm, it must give up 0.03 – 0.1 ppm of hydrogen into the circulating gas (Ar+O 2 ) system So, the actual volume of H given up by the sample is …. = (0.03-0.1)*8.5 lit = (0.225- 0.85)*10 -6 lit of H Density of hydrogen = 0.08988 g/L. So, the mass of hydrogen to be given up by the sample is = (0.225 -0.85)* 10 -6 *ρ H g. H => (0.02- 0.07)* 10 -6 g. H Mass of 4340 sample with 6mm diameter and 6 mm length is = π (.3) 2 *0.6 * ρ 4340 => π (.3) 2 *0.6 * ρ 4340 = 3.14 * 0.09* 0.6 * 7.85 = 1.32 g So, concentration limit of the sample has to be => (0.02-0.07)* 10 -6 g H/1.32 g sample => (0.015 – 0.053)* 10 -6 So, minimum detectable limit for H conc. in terms of sample mass = 0.015-0.053 ppm (µg/g)
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