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A sample of air analyzed at 25 °C and 1 atm pressure is reported to contain 0.04 ppm of SO2. Atomic mass of S = 32. O = 16. The equivalent SO2 concentration (in μg/m3) will be ________. (round off to the nearest integer)
Correct answer is between '102,108'. Can you explain this answer?
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A sample of air analyzed at 25 °C and 1 atm pressure is reported t...
°C and 1 atm pressure has a volume of 2.50 L. What will be the volume of the air sample if it is heated to 100°C while maintaining a constant pressure of 1 atm?
To solve this problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure is held constant. Mathematically, this can be represented as:
V1/T1 = V2/T2
Where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.
Given:
V1 = 2.50 L
T1 = 25°C = 25 + 273.15 = 298.15 K
T2 = 100°C = 100 + 273.15 = 373.15 K
P = 1 atm
Using the formula, we can substitute the values:
2.50 L / 298.15 K = V2 / 373.15 K
Cross-multiplying and solving for V2:
298.15 K * V2 = 373.15 K * 2.50 L
V2 = (373.15 K * 2.50 L) / 298.15 K
V2 = 3.13 L
Therefore, the volume of the air sample will be 3.13 L when heated to 100°C while maintaining a constant pressure of 1 atm.
To solve this problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure is held constant. Mathematically, this can be represented as:
V1/T1 = V2/T2
Where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.
Given:
V1 = 2.50 L
T1 = 25°C = 25 + 273.15 = 298.15 K
T2 = 100°C = 100 + 273.15 = 373.15 K
P = 1 atm
Using the formula, we can substitute the values:
2.50 L / 298.15 K = V2 / 373.15 K
Cross-multiplying and solving for V2:
298.15 K * V2 = 373.15 K * 2.50 L
V2 = (373.15 K * 2.50 L) / 298.15 K
V2 = 3.13 L
Therefore, the volume of the air sample will be 3.13 L when heated to 100°C while maintaining a constant pressure of 1 atm.
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A sample of air analyzed at 25 °C and 1 atm pressure is reported t...
∴ PV = nRT
∴ Pressure is same in both case (P1 = P2)
∴ V1/T1 = V2/T2
T1 = 273 K
T2 = 273 + 25 = 298 K
V1/273 = V2/298
T2 = 27T + 25 = 298
∴ 1 mole of gas occupies 22.4 lit of volume at 0°C temperature and 1 atm pressure.
∴ V2 = 24.45 lit = 1 mole of gas at 25°C
∴ 106 m3 of air = 0.04 m3 of SO2
= 40 lit of SO2
∴ 106 m3 of air = 40/24.45 = 1.636 mole of SO2
106 m3 of air = 1.636 x 64 g of SO2
1 m3 of air = 104.7 μ.g of
∴ Pressure is same in both case (P1 = P2)
∴ V1/T1 = V2/T2
T1 = 273 K
T2 = 273 + 25 = 298 K
V1/273 = V2/298
T2 = 27T + 25 = 298
∴ 1 mole of gas occupies 22.4 lit of volume at 0°C temperature and 1 atm pressure.
∴ V2 = 24.45 lit = 1 mole of gas at 25°C
∴ 106 m3 of air = 0.04 m3 of SO2
= 40 lit of SO2
∴ 106 m3 of air = 40/24.45 = 1.636 mole of SO2
106 m3 of air = 1.636 x 64 g of SO2
1 m3 of air = 104.7 μ.g of
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A sample of air analyzed at 25 °C and 1 atm pressure is reported to contain 0.04 ppm of SO2. Atomic mass of S = 32. O = 16. The equivalent SO2 concentration (in μg/m3) will be ________. (round off to the nearest integer)Correct answer is between '102,108'. Can you explain this answer?
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A sample of air analyzed at 25 °C and 1 atm pressure is reported to contain 0.04 ppm of SO2. Atomic mass of S = 32. O = 16. The equivalent SO2 concentration (in μg/m3) will be ________. (round off to the nearest integer)Correct answer is between '102,108'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A sample of air analyzed at 25 °C and 1 atm pressure is reported to contain 0.04 ppm of SO2. Atomic mass of S = 32. O = 16. The equivalent SO2 concentration (in μg/m3) will be ________. (round off to the nearest integer)Correct answer is between '102,108'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sample of air analyzed at 25 °C and 1 atm pressure is reported to contain 0.04 ppm of SO2. Atomic mass of S = 32. O = 16. The equivalent SO2 concentration (in μg/m3) will be ________. (round off to the nearest integer)Correct answer is between '102,108'. Can you explain this answer?.
A sample of air analyzed at 25 °C and 1 atm pressure is reported to contain 0.04 ppm of SO2. Atomic mass of S = 32. O = 16. The equivalent SO2 concentration (in μg/m3) will be ________. (round off to the nearest integer)Correct answer is between '102,108'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A sample of air analyzed at 25 °C and 1 atm pressure is reported to contain 0.04 ppm of SO2. Atomic mass of S = 32. O = 16. The equivalent SO2 concentration (in μg/m3) will be ________. (round off to the nearest integer)Correct answer is between '102,108'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sample of air analyzed at 25 °C and 1 atm pressure is reported to contain 0.04 ppm of SO2. Atomic mass of S = 32. O = 16. The equivalent SO2 concentration (in μg/m3) will be ________. (round off to the nearest integer)Correct answer is between '102,108'. Can you explain this answer?.
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