4) Concentration of so2 is equal c ppm. What will be concentration of so2 in mg/m3, if molecular mass 64 g/mol, temperature of air equal 30 ºC; pressure 710 mmHg.

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

C(SO2) = 50 ppm 50 ppm = (v/v) = 50 μL/ L = 50 ml/m³

Mr(SO2) = 64 g/mol Now we need to find the weight of 50 mL of

T = 30 ºC carbon dioxide. For that purpose, we can use ideal

Pa = 710 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 (SO2) = 64 g/moL

- Pressure and temperature are not given. But let’s imagine that we are in Almaty now. The pressure is 710 mmHg, temperature 30°C.

- We need to convert temperature to K: T = 273 + 30 = 303 K

- The pressure must be converted to kPa. We know that 760 mmHg = 101.325 kPa. P = 101.325 kPa x 710 mmHg / 760 mmHg = 94,66 kPa

m = *

m=0,01203 g = 120,3 mg

c(mg/m³) = 120,3 mg/m³

5) Concentration of h2 is equal c ppm. What will be concentration of h2 in mg/m3, if molecular mass 2 g/mol, temperature of air equal 40 ºC; pressure 720 mmHg.

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

C(H2) = 50 ppm 50 ppm = (v/v) = 50 μL/ L = 50 ml/m³

Mr(H2) = 2 g/mol Now we need to find the weight of 50 mL of

T = 40 ºC hydrogen. For that purpose, we can use ideal

Pa = 720 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 (H₂) = 2 g/moL

- Pressure and temperature are not given. But let’s imagine that we are in Almaty now. The pressure is 720 mmHg, temperature 40°C.

- We need to convert temperature to K: T = 273 + 40 = 303 K

- The pressure must be converted to kPa. We know that 760 mmHg = 101.325 kPa. P = 101.325 kPa x 720 mmHg / 760 mmHg = 95,99 kPa

m = *

m=0,0381 g = 38,1 mg

c(mg/m³) = 38,1 mg/m³

6) Concentration of no2 is equal c ppm. What will be concentration of no2 in mg/m3, if molecular mass 46 g/mol, temperature of air equal 50 ºC; pressure 730 mmHg.

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

C(NO2) = 50 ppm 50 ppm = (v/v) = 50 μL/ L = 50 ml/m³

Mr(NO2) = 46 g/mol Now we need to find the weight of 50 mL of

T = 50 ºC hydrogen. For that purpose, we can use ideal

Pa = 730 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 (NO₂) = 2 g/moL

- Pressure and temperature are not given. But let’s imagine that we are in Almaty now. The pressure is 760 mmHg, temperature 50°C.

- We need to convert temperature to K: T = 273 + 50 = 313 K

- The pressure must be converted to kPa. We know that 760 mmHg = 101.325 kPa. P = 101.325 kPa x 760 mmHg / 760 mmHg = 101,325 kPa

m = *

m=0,08956 g = 89,59 mg

c(mg/m³) = 89,59 mg/m³