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A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 × 10–7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 × 10–10 m).
Correct answer is '47.4 cm'. Can you explain this answer?
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A stream of oxygen molecules at 500 K exists from a pin-hole in an ove...
**Given Information:**
- Temperature of the oxygen molecules: 500 K
- Pressure outside the oven: 2.5 x 10^7 atm
- Radius of an oxygen molecule (r): 1.8 x 10^-10 m
**To find:**
The maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam of oxygen.
**Approach:**
To determine the maximum distance at which the slit must be placed, we need to consider the behavior of gas molecules and the conditions necessary for a collimated beam.
1. **Gas Behavior:**
- The gas molecules move randomly in all directions.
- The average distance traveled by the gas molecule is given by the mean free path.
2. **Mean Free Path:**
- Mean free path (λ) is defined as the average distance traveled by a gas molecule between successive collisions.
- It can be calculated using the following equation:
λ = [(2 * d^2 * N) / (π * r^2 * P)]
where:
- d is the diameter of the molecule
- N is the Avogadro's number (6.022 x 10^23)
- r is the radius of the molecule
- P is the pressure
3. **Conditions for a Collimated Beam:**
- For a collimated beam, the distance traveled by the molecule (L) should be much greater than the mean free path (λ).
4. **Calculation:**
- We need to find the maximum distance (L) at which the slit should be placed.
- Since L >> λ, we can assume L = 100 λ to ensure a collimated beam.
Substituting the given values into the equation for λ, we have:
λ = [(2 * (2 * r)^2 * N) / (π * r^2 * P)]
- Simplifying the equation:
λ = [(8 * r^2 * N) / (π * r^2 * P)]
λ = [(8 * N) / (π * P)]
- Substituting the given values for N and P:
λ = [(8 * 6.022 x 10^23) / (π * 2.5 x 10^7)]
- Calculating the value for λ:
λ ≈ 6.11 x 10^-16 m
- Finally, calculating the maximum distance (L):
L = 100 λ
L ≈ 100 * 6.11 x 10^-16 m
L ≈ 6.11 x 10^-14 m
**Answer:**
The maximum distance at which the slit must be placed to produce a collimated beam of oxygen is approximately 6.11 x 10^-14 m, which is equivalent to 6.11 x 10^-12 cm or 0.0611 nm.
- Temperature of the oxygen molecules: 500 K
- Pressure outside the oven: 2.5 x 10^7 atm
- Radius of an oxygen molecule (r): 1.8 x 10^-10 m
**To find:**
The maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam of oxygen.
**Approach:**
To determine the maximum distance at which the slit must be placed, we need to consider the behavior of gas molecules and the conditions necessary for a collimated beam.
1. **Gas Behavior:**
- The gas molecules move randomly in all directions.
- The average distance traveled by the gas molecule is given by the mean free path.
2. **Mean Free Path:**
- Mean free path (λ) is defined as the average distance traveled by a gas molecule between successive collisions.
- It can be calculated using the following equation:
λ = [(2 * d^2 * N) / (π * r^2 * P)]
where:
- d is the diameter of the molecule
- N is the Avogadro's number (6.022 x 10^23)
- r is the radius of the molecule
- P is the pressure
3. **Conditions for a Collimated Beam:**
- For a collimated beam, the distance traveled by the molecule (L) should be much greater than the mean free path (λ).
4. **Calculation:**
- We need to find the maximum distance (L) at which the slit should be placed.
- Since L >> λ, we can assume L = 100 λ to ensure a collimated beam.
Substituting the given values into the equation for λ, we have:
λ = [(2 * (2 * r)^2 * N) / (π * r^2 * P)]
- Simplifying the equation:
λ = [(8 * r^2 * N) / (π * r^2 * P)]
λ = [(8 * N) / (π * P)]
- Substituting the given values for N and P:
λ = [(8 * 6.022 x 10^23) / (π * 2.5 x 10^7)]
- Calculating the value for λ:
λ ≈ 6.11 x 10^-16 m
- Finally, calculating the maximum distance (L):
L = 100 λ
L ≈ 100 * 6.11 x 10^-16 m
L ≈ 6.11 x 10^-14 m
**Answer:**
The maximum distance at which the slit must be placed to produce a collimated beam of oxygen is approximately 6.11 x 10^-14 m, which is equivalent to 6.11 x 10^-12 cm or 0.0611 nm.
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A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 × 10–7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 × 10–10 m).Correct answer is '47.4 cm'. Can you explain this answer?
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A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 × 10–7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 × 10–10 m).Correct answer is '47.4 cm'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 × 10–7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 × 10–10 m).Correct answer is '47.4 cm'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 × 10–7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 × 10–10 m).Correct answer is '47.4 cm'. Can you explain this answer?.
A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 × 10–7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 × 10–10 m).Correct answer is '47.4 cm'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 × 10–7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 × 10–10 m).Correct answer is '47.4 cm'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 × 10–7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 × 10–10 m).Correct answer is '47.4 cm'. Can you explain this answer?.
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A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 × 10–7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 × 10–10 m).Correct answer is '47.4 cm'. Can you explain this answer?, a detailed solution for A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 × 10–7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 × 10–10 m).Correct answer is '47.4 cm'. Can you explain this answer? has been provided alongside types of A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 × 10–7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 × 10–10 m).Correct answer is '47.4 cm'. Can you explain this answer? theory, EduRev gives you an
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