Question

1) 0.5 m² of air sample was collected. It is found that air sample is contaminated with 138.45ug of Benzene (C6H6). a) Which units would be the ideal way to report gaseous concentrations and why? b) Calculate the volume of benzene present in sample in terms of ppmy and ppby. Also, compute the volume fraction of benzene with respect to air sample. Assume - The air sample is at 25 °C and at 1 atmosphere pressure. Hint – The ideal gas law equation relates pressure, volume, temperature and mass of gaseous contaminant. In the equation mentioned below, Pair is in KPa, Vcontaminantis in L, R is 8.3144 -rand Tair is in kelvin (K). PairV contaminant = molescontaminant XRx Tair

Answer 1

The volume of the contaminent benzene is calculated with the help of ideal gas law equations and molar mass of benzene is taken as standard value. Please like.

aus Given volume of air Sample collected = 0.5m3 = 500L we know [Im² = 1000 LJ. mans of benzene found in air = 138.45ug. (a) To g repart gaseous concentrations ideal PPM (Parts Per million) unit as used because gas contain many particles which has very low concentrations That's why they are measured in parts per million), 6 contaminant benzene Also giren T = 25°c = 25+ 2734 = 299k R = Gas Constant = 8.314 L kPa) mol-k Pressure = tatm = 101.325 kPa. from the ideal gas law we have Pair Vcontanisant = moles, a XRx Tair 101.325 (kla) v = mass x 8,314 (kPa), 2 x 2984 a molar massa mol-x benzere = 138,45 mg 8.314 Kla-L x 298 78.11 glupl 101.325 kg benzene = 43.34 LX 10 - 0 = 43.34 ML air acts like an ideal gas then ratio of Pollutant volume to air volume are same as mole ratio independent of Presure & temperature At allutant volemple ratio As 1 2 & tempe

-3 Now we have PPmv = contaminant (mL) = 43.34 x 10 ml Sample Cm3) 0.5 (m3) [PPmV = 0.086 ] - and Ppbv = contaminant (ML) = 43.34 (ML) Sample (m3 0.5 (m3) [PPbv = 86.68] - and the volume fraction of benzene with air is given as: V benzene = 43.34X100 l = 8.66x108 2 500L Hence ratio is 8,66410-8 sm Vair

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