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2.76dm of oxygen in container exact a pressure 0.825atm at 200K. what is the pressure exacted when temperature is reduce to have it's initial at a volume of 2.16dm?
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2.76dm of oxygen in container exact a pressure 0.825atm at 200K. what ...
To solve the problem, we can use the Ideal Gas Law and the principles of gas behavior under changing conditions of temperature and volume.
Understanding the Initial Conditions
- Initial Volume (V1): 2.76 dm³
- Initial Pressure (P1): 0.825 atm
- Initial Temperature (T1): 200 K
Understanding the Final Conditions
- Final Volume (V2): 2.16 dm³
- Final Temperature (T2): ? (We need to find the new pressure at this volume)
Using the Ideal Gas Law
We can apply the combined gas law, which is expressed as:
(P1 * V1) / T1 = (P2 * V2) / T2
Since we are not given a new temperature, we assume that the final temperature remains constant at 200 K for simplicity unless stated otherwise.
Calculating the New Pressure
Rearranging the formula to find P2:
P2 = (P1 * V1 * T2) / (V2 * T1)
Inserting the known values:
- P1 = 0.825 atm
- V1 = 2.76 dm³
- V2 = 2.16 dm³
- T1 = T2 = 200 K
Final Calculation
P2 = (0.825 atm * 2.76 dm³ * 200 K) / (2.16 dm³ * 200 K)
- The temperature cancels out, simplifying our calculation.
P2 = (0.825 atm * 2.76) / 2.16
P2 ≈ 1.058 atm
Conclusion
When the volume of the oxygen is reduced to 2.16 dm³ while keeping the temperature constant at 200 K, the pressure exerted will be approximately 1.058 atm.
Understanding the Initial Conditions
- Initial Volume (V1): 2.76 dm³
- Initial Pressure (P1): 0.825 atm
- Initial Temperature (T1): 200 K
Understanding the Final Conditions
- Final Volume (V2): 2.16 dm³
- Final Temperature (T2): ? (We need to find the new pressure at this volume)
Using the Ideal Gas Law
We can apply the combined gas law, which is expressed as:
(P1 * V1) / T1 = (P2 * V2) / T2
Since we are not given a new temperature, we assume that the final temperature remains constant at 200 K for simplicity unless stated otherwise.
Calculating the New Pressure
Rearranging the formula to find P2:
P2 = (P1 * V1 * T2) / (V2 * T1)
Inserting the known values:
- P1 = 0.825 atm
- V1 = 2.76 dm³
- V2 = 2.16 dm³
- T1 = T2 = 200 K
Final Calculation
P2 = (0.825 atm * 2.76 dm³ * 200 K) / (2.16 dm³ * 200 K)
- The temperature cancels out, simplifying our calculation.
P2 = (0.825 atm * 2.76) / 2.16
P2 ≈ 1.058 atm
Conclusion
When the volume of the oxygen is reduced to 2.16 dm³ while keeping the temperature constant at 200 K, the pressure exerted will be approximately 1.058 atm.
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2.76dm of oxygen in container exact a pressure 0.825atm at 200K. what is the pressure exacted when temperature is reduce to have it's initial at a volume of 2.16dm?
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2.76dm of oxygen in container exact a pressure 0.825atm at 200K. what is the pressure exacted when temperature is reduce to have it's initial at a volume of 2.16dm? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about 2.76dm of oxygen in container exact a pressure 0.825atm at 200K. what is the pressure exacted when temperature is reduce to have it's initial at a volume of 2.16dm? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2.76dm of oxygen in container exact a pressure 0.825atm at 200K. what is the pressure exacted when temperature is reduce to have it's initial at a volume of 2.16dm?.
2.76dm of oxygen in container exact a pressure 0.825atm at 200K. what is the pressure exacted when temperature is reduce to have it's initial at a volume of 2.16dm? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about 2.76dm of oxygen in container exact a pressure 0.825atm at 200K. what is the pressure exacted when temperature is reduce to have it's initial at a volume of 2.16dm? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2.76dm of oxygen in container exact a pressure 0.825atm at 200K. what is the pressure exacted when temperature is reduce to have it's initial at a volume of 2.16dm?.
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