Please answer the following question and show work. STP for ambient air assumes P=760 mmg, T=298K. I used the P1V1/T1=P2V2/T2 and got 0.57 ug/m^3. Please confirm if this answer is correct, if not please let me know where I went wrong in the problem. Also, the cannot be expressed in ppm because we are talking about particle matter which is mass and not a volume(ppm). My question is, at the end, do i multiply my answer by 20 m^3 because per 10 ug of ozone we account for 1M^3 of air? Or was the 20m^3 volume concentration already accounted for wheni performed the cross multiplying. My work along with the question is attached below. STP in MalmBE EXPRESSED IN PPM BICconc.ugimPE ITSCANNUTPARTICULA TEMATTER (MASS)NOT VOLUME.whatwearetuld;V: 20 m3PMas conc.= 10 Ma Im²-p Im? of airPz: 760 mm Ha, Tz= 298 K V,=?T,=35°C+273.15 K: 308.15K, P;- 760 mm Hgug/m²-7p Im?2.5P.V.TzP.V.(760 mn Hal) (2D m?)(76)maital ( Vz)308,15298CAN'T EXPRESS4529600234,194IN PPM B/CエT3 PM Imas) NUT234.194(VVOLUME.V2=19.34 m³10 Mg/mNew PM, s cunc:-0.517 Malm?19.34m² You sampled 20 m^3 of air on a hot summer day when the temperature was 35 degrees C, and the atmospheric pressure was 760 mmHg. The resulting PM2.5 concentration was 10ug/m^3. What is the PM2.5 concentration at STP? Please express your answer as µg/m^3 and ppm of PM2.5.

Solution: 760mm of Hg is equal to 1 atmospheric pressure. Calculate the volume of air at STP.…

Answered: You sampled 20 m^3 of air on a hot…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Please answer the following question and show work. STP for ambient air assumes P=760 mmg, T=298K. I used the P1V1/T1=P2V2/T2 and got 0.57 ug/m^3. Please confirm if this answer is correct, if not please let me know where I went wrong in the problem. Also, the cannot be expressed in ppm because we are talking about particle matter which is mass and not a volume(ppm). My question is, at the end, do i multiply my answer by 20 m^3 because per 10 ug of ozone we account for 1M^3 of air? Or was the 20m^3 volume concentration already accounted for wheni performed the cross multiplying. My work along with the question is attached below.