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Four liters of a diatomic ideal gas ( λ
=1.4
) confined to a cylinder is subject to a closed cycle. Initially, the gas is at 1.00 atm and at 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressure. Finally, the gas is compressed isobarically to its original volume. Find the temperature at the end of the cycle- a)276 K
- b)285 K
- c)300 K
- d)332 K
Correct answer is option 'C'. Can you explain this answer?
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Four liters of a diatomic ideal gas (λ=1.4) confined to a cylin...
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Four liters of a diatomic ideal gas (λ=1.4) confined to a cylin...
Such as nitrogen or oxygen) initially at a pressure of 2 atm and a temperature of 300 K undergoes an isothermal expansion to a final volume of 10 liters. What is the final pressure of the gas?
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature
Since we are dealing with an ideal gas, we can assume that the number of moles of gas (n) remains constant throughout the process.
Given:
Initial pressure (P1) = 2 atm
Initial volume (V1) = 4 liters
Final volume (V2) = 10 liters
Initial temperature (T1) = 300 K
We need to find the final pressure (P2).
Using the ideal gas law equation, we can rewrite it as:
P1V1 = P2V2
Substituting the given values, we have:
2 atm * 4 liters = P2 * 10 liters
Simplifying the equation, we get:
8 atm = P2 * 10 liters
Dividing both sides of the equation by 10 liters, we get:
P2 = 8 atm / 10 liters
P2 = 0.8 atm
Therefore, the final pressure of the gas is 0.8 atm.
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature
Since we are dealing with an ideal gas, we can assume that the number of moles of gas (n) remains constant throughout the process.
Given:
Initial pressure (P1) = 2 atm
Initial volume (V1) = 4 liters
Final volume (V2) = 10 liters
Initial temperature (T1) = 300 K
We need to find the final pressure (P2).
Using the ideal gas law equation, we can rewrite it as:
P1V1 = P2V2
Substituting the given values, we have:
2 atm * 4 liters = P2 * 10 liters
Simplifying the equation, we get:
8 atm = P2 * 10 liters
Dividing both sides of the equation by 10 liters, we get:
P2 = 8 atm / 10 liters
P2 = 0.8 atm
Therefore, the final pressure of the gas is 0.8 atm.
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Four liters of a diatomic ideal gas (λ=1.4) confined to a cylinder is subject to a closed cycle. Initially, the gas is at 1.00 atm and at 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressure. Finally, the gas is compressed isobarically to its original volume. Find the temperature at the end of the cyclea)276 Kb)285 Kc)300 Kd)332 KCorrect answer is option 'C'. Can you explain this answer?
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Four liters of a diatomic ideal gas (λ=1.4) confined to a cylinder is subject to a closed cycle. Initially, the gas is at 1.00 atm and at 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressure. Finally, the gas is compressed isobarically to its original volume. Find the temperature at the end of the cyclea)276 Kb)285 Kc)300 Kd)332 KCorrect answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Four liters of a diatomic ideal gas (λ=1.4) confined to a cylinder is subject to a closed cycle. Initially, the gas is at 1.00 atm and at 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressure. Finally, the gas is compressed isobarically to its original volume. Find the temperature at the end of the cyclea)276 Kb)285 Kc)300 Kd)332 KCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Four liters of a diatomic ideal gas (λ=1.4) confined to a cylinder is subject to a closed cycle. Initially, the gas is at 1.00 atm and at 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressure. Finally, the gas is compressed isobarically to its original volume. Find the temperature at the end of the cyclea)276 Kb)285 Kc)300 Kd)332 KCorrect answer is option 'C'. Can you explain this answer?.
Four liters of a diatomic ideal gas (λ=1.4) confined to a cylinder is subject to a closed cycle. Initially, the gas is at 1.00 atm and at 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressure. Finally, the gas is compressed isobarically to its original volume. Find the temperature at the end of the cyclea)276 Kb)285 Kc)300 Kd)332 KCorrect answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Four liters of a diatomic ideal gas (λ=1.4) confined to a cylinder is subject to a closed cycle. Initially, the gas is at 1.00 atm and at 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressure. Finally, the gas is compressed isobarically to its original volume. Find the temperature at the end of the cyclea)276 Kb)285 Kc)300 Kd)332 KCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Four liters of a diatomic ideal gas (λ=1.4) confined to a cylinder is subject to a closed cycle. Initially, the gas is at 1.00 atm and at 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressure. Finally, the gas is compressed isobarically to its original volume. Find the temperature at the end of the cyclea)276 Kb)285 Kc)300 Kd)332 KCorrect answer is option 'C'. Can you explain this answer?.
Solutions for Four liters of a diatomic ideal gas (λ=1.4) confined to a cylinder is subject to a closed cycle. Initially, the gas is at 1.00 atm and at 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressure. Finally, the gas is compressed isobarically to its original volume. Find the temperature at the end of the cyclea)276 Kb)285 Kc)300 Kd)332 KCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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ample number of questions to practice Four liters of a diatomic ideal gas (λ=1.4) confined to a cylinder is subject to a closed cycle. Initially, the gas is at 1.00 atm and at 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressure. Finally, the gas is compressed isobarically to its original volume. Find the temperature at the end of the cyclea)276 Kb)285 Kc)300 Kd)332 KCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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