1) Concentration of h2s is equal 50 ppm. What will be concentration of h2s in mg/m3, if molecular mass 34 g/mol, temperature of air equal 10 ºC; pressure 680 mmHg.

Standard sources: T = 273 K; P = 101325 Pa = 760 mmHg; R = 8.31 J/(mol*K)

C(H2S) = 50 ppm 50 ppm = (v/v) = 50 μL/ L = 50 ml/m³

Mr(H2S) = 34 g/mol Now we need to find the weight of 50 mL of

T = 10 ºC hydrogen sulfide. For that purpose, we can use ideal

Pa = 680 mmHg gas law:

T = 273 K pV = mRT/M

P = 101325

Pa = 760 mmHg

R = 8.31 J/(mol*K)

V = 50 mL; R=8.31 L· kPa / (moL K); M (H₂S) = 34 g/moL

- Pressure and temperature are not given. But let’s imagine that we are in Almaty now. The pressure is 680 mmHg, temperature 10°C.

- We need to convert temperature to K: T = 273 + 10 = 283 K

- The pressure must be converted to kPa. We know that 760 mmHg = 101.325 kPa. P = 101.325 kPa x 680 mmHg / 760 mmHg = 90.66 kPa

m = *

m=0,0655 g = 65,5 mg

c(mg/m³) = 65,5 mg/m³

2) Concentration of ch4 is equal c ppm. What will be concentration of ch4 in mg/m3, if molecular mass 16 g/mol, temperature of air equal 15 ºC; pressure 690 mmHg.

Standard sources: T = 273 K; P = 101325 Pa = 760 mmHg; R = 8.31 J/(mol*K)

C(CH4) = 50 ppm 50 ppm = (v/v) = 50 μL/ L = 50 ml/m³

Mr(CH4) = 16 g/mol Now we need to find the weight of 50 mL of

T = 15 ºC hydrogen sulfide. For that purpose, we can use ideal

Pa = 690 mmHg gas law:

T = 273 K pV = mRT/M

P = 101325

Pa = 760 mmHg

R = 8.31 J/(mol*K)

V = 50 mL; R=8.31 L· kPa / (moL K); M (CH4) = 16 g/moL

- Pressure and temperature are not given. But let’s imagine that we are in Almaty now. The pressure is 690 mmHg, temperature 15°C.

- We need to convert temperature to K: T = 273 + 15 = 288 K

- The pressure must be converted to kPa. We know that 760 mmHg = 101.325 kPa. P = 101.325 kPa x 690 mmHg / 760 mmHg = 91.99 kPa

m = *

m=0,03075 g = 30,75 mg

c(mg/m³) = 30,75 mg/m³

3) Concentration of co2 is equal c ppm. What will be concentration of co2 in mg/m3, if molecular mass 44 g/mol, temperature of air equal 20 ºC; pressure 700 mmHg.

Standard sources: T = 273 K; P = 101325 Pa = 760 mmHg; R = 8.31 J/(mol*K)

C(CO2) = 50 ppm 50 ppm = (v/v) = 50 μL/ L = 50 ml/m³

Mr(CO2) = 44 g/mol Now we need to find the weight of 50 mL of

T = 20 ºC carbon dioxide. For that purpose, we can use ideal

Pa = 700 mmHg gas law:

T = 273 K pV = mRT/M

P = 101325

Pa = 760 mmHg

R = 8.31 J/(mol*K)

V = 50 mL; R=8.31 L· kPa / (moL K); M (CO2) = 44 g/moL

- Pressure and temperature are not given. But let’s imagine that we are in Almaty now. The pressure is 700 mmHg, temperature 20°C.

- We need to convert temperature to K: T = 273 + 20 = 293 K

- The pressure must be converted to kPa. We know that 760 mmHg = 101.325 kPa. P = 101.325 kPa x 700 mmHg / 760 mmHg = 93,33 kPa

m = *

m=0,0843 g = 84,3 mg

c(mg/m³) = 84,3 mg/m³