Question

In sampling for cresol (MW 108), 4.5 liters of workplace air from a copper smelting plant was collected in a Mylar bag. The worksite air temperature was 95^oF and the atmospheric pressure was 740 mmHg. Upon analysis at 71^oF laboratory with an atmospheric pressure of 765 mmHg, the concentration was found to be 348 mg/m³. What was the concentration at the worksite in ppm?

Answer 1

Lab conditions P = 765 mmHg = ( 765 / 760) atm = 1.007 atm

T = 71 F = ( 71-32) / 1.8 C = 217. C = 21.7 + 273K = 295 K

Under starting conditions P = 740 mmHg = ( 740/760) = 0.9737 atm ,  

T = 95 F = ( 95-32) / 1.8 = 35 C = 35 + 273 K = 308 K

we had formula PV = nRT

( n/V) = P/RT ( where n/V indicates mol / volume which we can say as concentration )

now ( concentration in lab / ( concentration at starting = ( P/T ) lab / ( P/T starting) = ( Plab x T starting) / ( P start x T lab)

348 mg/m3 / ( conc starting) = ( 1.007 atm x 308 K) / ( 0.9737 atm x 295K )

conc starting = 322 mg/m3

= ( 322 mg/m3) x ( 1m3 / 1000L) = 0.322 mg/L = ( 0.322 mg/L) x ( 1g/1000mg)

= 0.000322 g /L = 0.000322 g/L x ( 1L / 1000 ml) = 0.322 x 10^ -6 g/ml

W have  10^ -6 g/ml is equal to 1ppm

hence concentration in ppm = 0.322 ppm