7) Concentration of n2 is equal c ppm. What will be concentration of n2 in mg/m3, if molecular mass 28 g/mol, temperature of air equal 60 ºC; pressure 740 mmHg.
Standard sources: T = 273 K; P = 101325 Pa = 760 mmHg; R = 8.31 J/(mol*K)
C(N2) = 50 ppm 50 ppm = (v/v) = 50 μL/ L = 50 ml/m3
Mr(N2) = 28 g/mol Now we need to find the weight of 50 mL of
T = 60 ºC nitrogen. For that purpose, we can use ideal
Pa = 740 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 (N2) = 28 g/moL
- Pressure and temperature are not given. But let’s imagine that we are in Almaty now. The pressure is 740 mmHg, temperature 60°C.
- We need to convert temperature to K: T = 273 + 60 = 333 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=0,0513 g = 51,3 mg
c(mg/m3) = 51,3 mg/m3
8) SO2 concentration in Almaty air now is 37 µg/m3. Convert this concentration to ppbV. Atmospheric pressure is 740 mmHg, temperature 25⁰C.
C(H2S) = 37 µg/m3
Pa = 740 mmHg
T = 25°C
перевести С → ppbV
0,12*10-6* C * M * P / T
M(SO2) = 64 g/mol
P = 101.325 kPa x 740 mmHg / 760 mmHg = 98,66 kPa
T = 273 + 25 = 298 K
0,12*10-6 * 37 µg/m3 * 64 g/mol * 98,66 kPa / 298 K = 94,07 ppb
9) H2S concentration in Almaty air now is 38 µg/m3. Convert this concentration to ppbV. Atmospheric pressure is 750 mmHg, temperature 30⁰C.
C(H2S) = 38 µg/m3
Pa = 750 mmHg
T = 30°C
перевести С → ppbV
0,12*10-6* C * M * P / T
M(H2S) = 34 g/mol
P = 101.325 kPa x 750 mmHg / 760 mmHg = 99,99 kPa
T = 273 + 30 = 303 K
0,12*10-6 * 38 µg/m3 * 34 g/mol * 99,99 kPa / 303 K = 51,16 ppb
10) CH4 concentration in Almaty air now is 39 µg/m3. Convert this concentration to ppbV. Atmospheric pressure is 730 mmHg, temperature 35⁰C.
C(H2S) = 39 µg/m3
Pa = 730 mmHg
T = 35°C
перевести С → ppbV
0,12*10-6* C * M * P / T
M(CH4) = 16 g/mol
P = 101.325 kPa x 730 mmHg / 760 mmHg = 97,33 kPa
T = 273 + 35 = 308 K
0,12*10-6 * 39 µg/m3 * 16 g/mol * 97,33 kPa / 308 K = 23,66 ppb
11. CO2 concentration in Almaty air now is 40 µg/m3. Convert this concentration to ppbV. Atmospheric pressure is 760 mmHg, temperature 15⁰C.
C(CO2) = 40 µg/m3
Pa = 760 mmHg
T = 15°C
перевести С → ppbV
0,12*10-6* C * M * P / T
M(CO2) = 44 g/mol
P = 101.325 kPa x 760 mmHg / 760 mmHg = 101,325 kPa
T = 273 + 15 = 288 K
0,12*10-6 * 40 µg/m3 * 44 g/mol * 101,325 kPa / 288 K = 74,305 ppb
12) H2 concentration in Almaty air now is 41 µg/m3. Convert this concentration to ppbV. Atmospheric pressure is 720 mmHg, temperature 10⁰C.
C(H2) = 41 µg/m3
Pa = 720 mmHg
T = 10°C
перевести С → ppbV
0,12*10-6* C * M * P / T
M(H2) = 2 g/mol
P = 101.325 kPa x 720 mmHg / 760 mmHg = 95,99 kPa
T = 273 + 10 = 283 K
0,12*10-6 * 41 µg/m3 * 2 g/mol * 95,99 kPa / 283 K = 3,337 ppb
13) NO2 concentration in Almaty air now is 42 µg/m3. Convert this concentration to ppbV. Atmospheric pressure is 770 mmHg, temperature 40⁰C.
C(NO2) = 42 µg/m3
Pa = 770 mmHg
T = 40°C
перевести С → ppbV
0,12*10-6* C * M * P / T
M(NO2) = 46 g/mol
P = 101.325 kPa x 770 mmHg / 760 mmHg = 102,658 kPa
T = 273 + 40 = 313 K
0,12*10-6 * 42 µg/m3 * 46 g/mol * 102,658 kPa / 313 K = 76,04 ppb
14) N2 concentration in Almaty air now is 43 µg/m3. Convert this concentration to ppbV. Atmospheric pressure is 740 mmHg, temperature 45⁰C.
C(N2) = 43 µg/m3
Pa = 740 mmHg
T = 45°C
перевести С → ppbV
0,12*10-6* C * M * P / T
M(N2) = 28 g/mol
P = 101.325 kPa x 740 mmHg / 760 mmHg = 98,659 kPa
T = 273 + 45 = 318 K
0,12*10-6 * 43 µg/m3 * 28 g/mol * 98,659 kPa / 318 K = 44,82 ppb
15) Sample bag (V = 1.00 L) was filled with 0.70 L of air having benzene concentration 56 µg/m3. Sampling was done at a temperature -10⁰C. Then, the bag was transported to the laboratory where the temperature was 25⁰C. What is the benzene concentration in the air inside a sampling bag stored in the lab?
С1 = 56 µg/m3
Т1 = -10ºС
Т2 = 25ºС
C(benzene) - ?
C(benzene) = С1Т1/Т2
Т1= 273 + (-10) = 263 К
Т2 = 273 + 25 = 298 К
C(benzene) = 56 µg/m3*263 К/ 298 К = 49 µg/m3
16) Sample bag (V = 1.00 L) was filled with 0.70 L of air having benzene concentration 57 µg/m3. Sampling was done at a temperature 0⁰C. Then, the bag was transported to the laboratory where the temperature was 25⁰C. What is the benzene concentration in the air inside a sampling bag stored in the lab?
С1 = 57 µg/m3
Т1 = 0ºС
Т2 = 25ºС
C(benzene) - ?
C(benzene) = С1Т1/Т2
Т1= 273 + 0 = 273 К
Т2 = 273 + 25 = 298 К
C(benzene) = 57 µg/m3*273 К/ 298 К = 52 µg/m3
17) Sample bag (V = 1.00 L) was filled with 0.70 L of air having benzene concentration 58 µg/m3. Sampling was done at a temperature 10⁰C. Then, the bag was transported to the laboratory where the temperature was 25⁰C. What is the benzene concentration in the air inside a sampling bag stored in the lab?
С1 = 58 µg/m3
Т1 = 10ºС
Т2 = 25ºС
C(benzene) - ?
C(benzene) = С1Т1/Т2
Т1= 273 + 10 = 283 К
Т2 = 273 + 25 = 298 К
C(benzene) = 58 µg/m3*283 К/ 298 К = 55 µg/m3
18) Sample bag (V = 1.00 L) was filled with 0.70 L of air having benzene concentration 59 µg/m3. Sampling was done at a temperature 20⁰C. Then, the bag was transported to the laboratory where the temperature was 25⁰C. What is the benzene concentration in the air inside a sampling bag stored in the lab?
С1 = 59 µg/m3
Т1 = 20ºС
Т2 = 25ºС
C(benzene) - ?
C(benzene) = С1Т1/Т2
Т1= 273 + 20 = 293 К
Т2 = 273 + 25 = 298 К
C(benzene) = 59 µg/m3*293 К/ 298 К = 58 µg/m3
19) Sample bag (V = 1.00 L) was filled with 0.70 L of air having benzene concentration 60 µg/m3. Sampling was done at a temperature 30⁰C. Then, the bag was transported to the laboratory where the temperature was 25⁰C. What is the benzene concentration in the air inside a sampling bag stored in the lab?
С1 = 60 µg/m3
Т1 = 30ºС
Т2 = 25ºС
C(benzene) - ?
C(benzene) = С1Т1/Т2
Т1= 273 + 30 = 303 К
Т2 = 273 + 25 = 298 К
C(benzene) = 60 µg/m3*303 К/ 298 К = 61 µg/m3
20) Sample bag (V = 1.00 L) was filled with 0.70 L of air having benzene concentration 65 µg/m3. Sampling was done at a temperature 40⁰C. Then, the bag was transported to the laboratory where the temperature was 25⁰C. What is the benzene concentration in the air inside a sampling bag stored in the lab?
С1 = 65 µg/m3
Т1 = 40ºС
Т2 = 25ºС
C(benzene) - ?
C(benzene) = С1Т1/Т2
Т1= 273 + 40 = 313 К
Т2 = 273 + 25 = 298 К
C(benzene) = 65 µg/m3*313 К/ 298 К = 68 µg/m3
21) Sample bag (V = 1.00 L) was filled with 0.70 L of air having benzene concentration 70 µg/m3. Sampling was done at a temperature 50⁰C. Then, the bag was transported to the laboratory where the temperature was 25⁰C. What is the benzene concentration in the air inside a sampling bag stored in the lab?
С1 = 70 µg/m3
Т1 = 50ºС
Т2 = 25ºС
C(benzene) - ?
C(benzene) = С1Т1/Т2
Т1= 273 + 50 = 323 К
Т2 = 273 + 25 = 298 К
C(benzene) = 70 µg/m3*323 К/ 298 К = 75,87 µg/m3
22) Solution of benzene (5.00 µL) in methanol with concentration 10 mg/mL was injected to calibrated bulb having volume 250 mL and filled with air. All injected solution were evaporated. What is the concentration of benzene in the air inside bulb (in µg/L)?
V1 = 5.00 µL
С1 = 11 mg/mL
V2 = 250 mL
C2
-
?
C1V1 = C2V2
C2 = 5.00 µL * 10 mg/mL / 250 mL = 0,2 mg/mL = 200 mg/L
23) Solution of benzene (6.00 µL) in methanol with concentration 12 mg/mL was injected to calibrated bulb having volume 250 mL and filled with air. All injected solution were evaporated. What is the concentration of benzene in the air inside bulb (in µg/L)?
V1 = 6.00 µL
С1 = 12 mg/mL
V2 = 250 mL
C2 - ?
C1V1 = C2V2
C2 = 6.00 µL * 12 mg/mL / 250 mL = 0,288 mg/mL = 288 mg/L
24) Solution of benzene (7.00 µL) in methanol with concentration 14 mg/mL was injected to calibrated bulb having volume 250 mL and filled with air. All injected solution were evaporated. What is the concentration of benzene in the air inside bulb (in µg/L)?
V1 = 7.00 µL
С1 = 14 mg/mL
V2 = 250 mL
C2 - ?
C1V1 = C2V2
C2 = 7.00 µL * 14 mg/mL / 250 mL = 0,392 mg/mL = 392 mg/L
25) Solution of benzene (8.00 µL) in methanol with concentration 16 mg/mL was injected to calibrated bulb having volume 250 mL and filled with air. All injected solution were evaporated. What is the concentration of benzene in the air inside bulb (in µg/L)?
V1 = 8.00 µL
С1 = 16 mg/mL
V2 = 250 mL
C2 - ?
C1V1 = C2V2
C2 = 8.00 µL * 16 mg/mL / 250 mL = 0,512 mg/mL = 512 mg/L
26) Solution of benzene (9.00 µL) in methanol with concentration 18 mg/mL was injected to calibrated bulb having volume 250 mL and filled with air. All injected solution were evaporated. What is the concentration of benzene in the air inside bulb (in µg/L)?
V1 = 9.00 µL
С1 = 18 mg/mL
V2 = 250 mL
C2 - ?
C1V1 = C2V2
C2 = 9.00 µL * 18 mg/mL / 250 mL = 0,648 mg/mL = 648 mg/L
27) Solution of benzene (10.00 µL) in methanol with concentration 20 mg/mL was injected to calibrated bulb having volume 250 mL and filled with air. All injected solution were evaporated. What is the concentration of benzene in the air inside bulb (in µg/L)?
V1 = 10.00 µL
С1 = 20 mg/mL
V2 = 250 mL
C2 - ?
C1V1 = C2V2
C2 = 10.00 µL * 20 mg/mL / 250 mL = 0,8 mg/mL = 800 mg/L
28) Solution of benzene (14.00 µL) in methanol with concentration 4 mg/mL was injected to calibrated bulb having volume 250 mL and filled with air. All injected solution were evaporated. What is the concentration of benzene in the air inside bulb (in µg/L)?
V1 = 14.00 µL
С1 = 4 mg/mL
V2 = 250 mL
C2 - ?
C1V1 = C2V2
C2 = 14.00 µL * 4 mg/mL / 250 mL = 0,224 mg/mL = 224 mg/L
29) “Zero” air is supplied at 100 mL/min rate. Benzene solution in methanol (C = 60 ng/µL) is supplied at 10 µL/h rate. Calculate benzene concentration in produced air.
Rair= 100 ml\min
Csol = 50 ng\ml
Rsol = 10 µL/h
C - ?
C = Ranalyte\Rair
Ranalyte = Rsol*Rair = 10 µL/h * 50 ng\ml = 500 ng\h
Rair= 100 ml\min= 6000 mg\h= 6 l\h
C = 500 ng\h|6 l\h = 83,33 ng\l
30) “Zero” air is supplied at 150 mL/min rate. Benzene solution in methanol (C = 60 ng/µL) is supplied at 15 µL/h rate. Calculate benzene concentration in produced air.
Rair= 150 ml\min
Csol = 60 ng\ml
Rsol = 15 µL/h
C - ?
C = Ranalyte\Rair
Ranalyte = Rsol*Rair = 15 µL/h * 60 ng\ml = 900 ng\h
Rair= 150 ml\min= 9000 mg\h= 9 l\h
C = 900 ng\h|9 l\h = 10,00 ng\l
31) “Zero” air is supplied at 200 mL/min rate. Benzene solution in methanol (C = 60 ng/µL) is supplied at 15 µL/h rate. Calculate benzene concentration in produced air.
Rair= 200 ml\min
Csol = 60 ng\ml
Rsol = 15 µL/h
C - ?
C = Ranalyte\Rair
Ranalyte = Rsol*Rair = 15 µL/h * 60 ng\ml = 900 ng\h
Rair= 200 ml\min= 12000 mg\h= 12 l\h
C = 900 ng\h|12 l\h = 75 ng\l
32) “Zero” air is supplied at 200 mL/min rate. Benzene solution in methanol (C = 60 ng/µL) is supplied at 20 µL/h rate. Calculate benzene concentration in produced air.
Rair= 200 ml\min
Csol = 60 ng\ml
Rsol = 20 µL/h
C - ?
C = Ranalyte\Rair
Ranalyte = Rsol*Rair = 20 µL/h * 60 ng\ml = 1200 ng\h
Rair= 200 ml\min= 12000 mg\h= 12 l\h
C = 1200 ng\h|12 l\h = 10 ng\l
33) “Zero” air is supplied at 250 mL/min rate. Benzene solution in methanol (C = 60 ng/µL) is supplied at 30 µL/h rate. Calculate benzene concentration in produced air.
Rair= 250 ml\min
Csol = 60 ng\ml
Rsol = 30 µL/h
C - ?
C = Ranalyte\Rair
Ranalyte = Rsol*Rair = 30 µL/h * 60 ng\ml = 1800 ng\h
Rair= 250 ml\min= 15000 mg\h= 15 l\h
C = 1800 ng\h|15 l\h = 120 ng\l
34) “Zero” air is supplied at 300 mL/min rate. Benzene solution in methanol (C = 60 ng/µL) is supplied at 20 µL/h rate. Calculate benzene concentration in produced air.
Rair= 300 ml\min
Csol = 60 ng\ml
Rsol = 20 µL/h
C - ?
C = Ranalyte\Rair
Ranalyte = Rsol*Rair = 20 µL/h * 60 ng\ml = 1200 ng\h
Rair= 300 ml\min= 18000 mg\h= 18 l\h
C = 1200 ng\h|18 l\h = 66,66 ng\l
35) “Zero” air is supplied at 300 mL/min rate. Benzene solution in methanol (C = 60 ng/µL) is supplied at 15 µL/h rate. Calculate benzene concentration in produced air.
Rair= 300 ml\min
Csol = 60 ng\ml
Rsol = 15 µL/h
C - ?
C = Ranalyte\Rair
Ranalyte = Rsol*Rair = 15 µL/h * 60 ng\ml = 900 ng\h
Rair= 300 ml\min= 18000 mg\h= 18 l\h
C = 900 ng\h|18 l\h = 50 ng\l
36) What volumes of carbon dioxide absorb the plant by normal condition? To grow a tree with the following parameters trunk diameter d=0.8 m, height h=15m, wood density p=0,08 g/сm3. Assume that all wood consists of carbon and that the tree trunk has a regular cylindrical shape.
РЕШЕНИЕ: Determine the mass of the tree. For this cross-sectional area equal to πr2. Multiply by the height h (the radius is equal to D/2=0,8/2=0,4 meter) and density ρ (ρ=0,08g/cm3=0,08*1000=80kg/m3).
I.e. m= πr2hp or 3,14*(0,4m)2*15m*80kg/m3=602.88kg.
The formation of wood from carbon dioxide is a reaction:
CO2=>C+O2.
Assuming in the equation that the mass m1 of carbon dioxide (CO2) is equal to the mass m2 of carbon (C) are equal and their molecular weights will be accordingly M1 and M2.
We use the ratio of the mass of the reacting substances and their molecular masses: m1/m2 = k1*M1/k2*M2.
Where m1 and m2 - the mass of the reacting substances; M1 and M2 - their molecular masses; k1 and k2 - their stoichiometric coefficients.
The atomic mass of oxygen is 16; carbon is 12 (from Mendeleev's table). Respectively, the molecular weight of CO2 (M1) is 16*2 + 12 = 44. The molecular weight of carbon is assumed equal to its atomic mass, i.e. M2=12. Using the formula, we get: m1=m2*M1/M2.
Substituting the data we get: m1=602,88kg*44/12 = 2210,56 kg.
It is known that under normal conditions, 1 mole of any gas occupies a volume of 22.4 liter. Since 1 mole of carbon dioxide has a mass of 0,044 kg or 44 g (The mass of 1 mole is numerically equal to the molecular weight), then, carbon dioxide, containing 2210,56 kg, per 22,4 liter, the required value is obtained:
VCO2=2210,56 kg*22,4 liter /0,044kg = 1125376 liter or 1125,376 m3.
The answer: the volume of carbon dioxide taken under normal conditions is 1125,376m3.
37) What volumes of carbon dioxide absorb the plant by normal condition? To grow a tree with the following parameters trunk diameter d=0.9 m, height h=17m, wood density p=0,5 g/сm3. Assume that all wood consists of carbon and that the tree trunk has a regular cylindrical shape. (Type of wood pine).
РЕШЕНИЕ: Determine the mass of the tree. For this cross-sectional area equal to πr2. Multiply by the height h (the radius is equal to D/2=0,9/2=0,45 meter) and density ρ (ρ=0,5g/cm3=0,5*1000=500kg/m3).
I.e. m= πr2hp or 3,14*(0,45m)2*17m*500kg/m3=5404.725kg.
The formation of wood from carbon dioxide is a reaction:
CO2=>C+O2.
Assuming in the equation that the mass m1 of carbon dioxide (CO2) is equal to the mass m2 of carbon (C) are equal and their molecular weights will be accordingly M1 and M2.
We use the ratio of the mass of the reacting substances and their molecular masses: m1/m2 = k1*M1/k2*M2.
Where m1 and m2 - the mass of the reacting substances; M1 and M2 - their molecular masses; k1 and k2 - their stoichiometric coefficients.
The atomic mass of oxygen is 16; carbon is 12 (from Mendeleev's table). Respectively, the molecular weight of CO2 (M1) is 16*2 + 12 = 44. The molecular weight of carbon is assumed equal to its atomic mass, i.e. M2=12. Using the formula, we get: m1=m2*M1/M2.
Substituting the data we get: m1=5404.725kg*44/12 = 19817,325 kg.
It is known that under normal conditions, 1 mole of any gas occupies a volume of 22.4 liter. Since 1 mole of carbon dioxide has a mass of 0,044 kg or 44 g (The mass of 1 mole is numerically equal to the molecular weight), then, carbon dioxide, containing 19817,325 kg, per 22,4 liter, the required value is obtained:
VCO2=19817,325 kg*22,4 liter /0,044kg = 10088820 liter or 10088,82 m3.
The answer: the volume of carbon dioxide taken under normal conditions is 10088,82m3.
38) What volumes of carbon dioxide absorb the plant by normal condition? To grow a tree with the following parameters trunk diameter d=0.6 m, height h=11m, wood density p=0,66 g/сm3. Assume that all wood consists of carbon and that the tree trunk has a regular cylindrical shape. (Type of wood nut).
РЕШЕНИЕ: Determine the mass of the tree. For this cross-sectional area equal to πr2. Multiply by the height h (the radius is equal to D/2=0,6/2=0,3 meter) and density ρ (ρ=0,66g/cm3=0,66*1000=660kg/m3).
I.e. m= πr2hp or 3,14*(0,3m)2*11m*660kg/m3=2051,676kg.
The formation of wood from carbon dioxide is a reaction:
CO2=>C+O2.
Assuming in the equation that the mass m1 of carbon dioxide (CO2) is equal to the mass m2 of carbon (C) are equal and their molecular weights will be accordingly M1 and M2.
We use the ratio of the mass of the reacting substances and their molecular masses: m1/m2 = k1*M1/k2*M2.
Where m1 and m2 - the mass of the reacting substances; M1 and M2 - their molecular masses; k1 and k2 - their stoichiometric coefficients.
The atomic mass of oxygen is 16; carbon is 12 (from Mendeleev's table). Respectively, the molecular weight of CO2 (M1) is 16*2 + 12 = 44. The molecular weight of carbon is assumed equal to its atomic mass, i.e. M2=12. Using the formula, we get: m1=m2*M1/M2.
Substituting the data we get: m1=2051,676kg*44/12 = 7522,812 kg.
It is known that under normal conditions, 1 mole of any gas occupies a volume of 22.4 liter. Since 1 mole of carbon dioxide has a mass of 0,044 kg or 44 g (The mass of 1 mole is numerically equal to the molecular weight), then, carbon dioxide, containing 7522,812 kg, per 22,4 liter, the required value is obtained:
VCO2=7522,812 kg*22,4 liter /0,044kg = 3829795,2 liter or 3829,7952 m3.
The answer: the volume of carbon dioxide taken under normal conditions is 3829,7952m3.
39) What volumes of carbon dioxide absorb the plant by normal condition? To grow a tree with the following parameters trunk diameter d=0.7 m, height h=11m, wood density p=0,7 g/сm3. Assume that all wood consists of carbon and that the tree trunk has a regular cylindrical shape. (Type of wood oak).
РЕШЕНИЕ: Determine the mass of the tree. For this cross-sectional area equal to πr2. Multiply by the height h (the radius is equal to D/2=0,7/2=0,35 meter) and density ρ (ρ=0,7g/cm3=0,7*1000=700kg/m3).
I.e. m= πr2hp or 3,14*(0,35m)2*11m*700kg/m3=2961,805kg.
The formation of wood from carbon dioxide is a reaction:
CO2=>C+O2.
Assuming in the equation that the mass m1 of carbon dioxide (CO2) is equal to the mass m2 of carbon (C) are equal and their molecular weights will be accordingly M1 and M2.
We use the ratio of the mass of the reacting substances and their molecular masses: m1/m2 = k1*M1/k2*M2.
Where m1 and m2 - the mass of the reacting substances; M1 and M2 - their molecular masses; k1 and k2 - their stoichiometric coefficients.
The atomic mass of oxygen is 16; carbon is 12 (from Mendeleev's table). Respectively, the molecular weight of CO2 (M1) is 16*2 + 12 = 44. The molecular weight of carbon is assumed equal to its atomic mass, i.e. M2=12. Using the formula, we get: m1=m2*M1/M2.
Substituting the data we get: m1=2961,805kg*44/12 = 10859,9517kg.
It is known that under normal conditions, 1 mole of any gas occupies a volume of 22.4 liter. Since 1 mole of carbon dioxide has a mass of 0,044 kg or 44 g (The mass of 1 mole is numerically equal to the molecular weight), then, carbon dioxide, containing 10859,9517 kg, per 22,4 liter, the required value is obtained:
VCO2=10859,9517kg*22,4liter/0,044kg =5528702,67 liter or 5528,70267m3.
The answer: the volume of carbon dioxide taken under normal conditions is 5528,70267 m3.
40) What volumes of carbon dioxide absorb the plant by normal condition? To grow a tree with the following parameters trunk diameter d=0.5 m, height h=9m, wood density p=0,725g/сm3. Assume that all wood consists of carbon and that the tree trunk has a regular cylindrical shape. (Type of wood pear).
РЕШЕНИЕ: Determine the mass of the tree. For this cross-sectional area equal to πr2. Multiply by the height h (the radius is equal to D/2=0,5/2=0,25 meter) and density ρ (ρ=0,725g/cm3=0,725*1000=725kg/m3).
I.e. m= πr2hp or 3,14*(0,25m)2*9m*725kg/m3=1280,53125kg.
The formation of wood from carbon dioxide is a reaction:
CO2=>C+O2.
Assuming in the equation that the mass m1 of carbon dioxide (CO2) is equal to the mass m2 of carbon (C) are equal and their molecular weights will be accordingly M1 and M2.
We use the ratio of the mass of the reacting substances and their molecular masses: m1/m2 = k1*M1/k2*M2.
Where m1 and m2 - the mass of the reacting substances; M1 and M2 - their molecular masses; k1 and k2 - their stoichiometric coefficients.
The atomic mass of oxygen is 16; carbon is 12 (from Mendeleev's table). Respectively, the molecular weight of CO2 (M1) is 16*2 + 12 = 44. The molecular weight of carbon is assumed equal to its atomic mass, i.e. M2=12. Using the formula, we get: m1=m2*M1/M2.
Substituting the data we get: m1=1280,53125kg*44/12=4695,28125kg.
It is known that under normal conditions, 1 mole of any gas occupies a volume of 22.4 liter. Since 1 mole of carbon dioxide has a mass of 0,044 kg or 44 g (The mass of 1 mole is numerically equal to the molecular weight), then, carbon dioxide, containing 4695,28125 kg, per 22,4 liter, the required value is obtained:
VCO2=4695,28125 kg*22,4 liter /0,044kg = 2390325 liter or 2390,325 m3.
The answer: the volume of carbon dioxide taken under normal conditions is 2390,325m3.