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Two moles of monoatomic ideal gas at 60°C are mixed adiabatically with one mole of another monoatomic ideal gas at 12°C. The final temperature of the mixture is
- a)36°C
- b)44°C
- c)48°C
- d)72°C
Correct answer is option 'B'. Can you explain this answer?
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Two moles of monoatomic ideal gas at 60°C are mixed adiabatically ...
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Two moles of monoatomic ideal gas at 60°C are mixed adiabatically ...
To find the final temperature, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in pascals)
V = Volume (in cubic meters)
n = Number of moles
R = Ideal gas constant (8.314 J/(mol·K))
T = Temperature (in kelvin)
We have:
n = 2 moles
R = 8.314 J/(mol·K)
Let's assume the initial temperature is T1 and the final temperature is T2.
At the initial state:
P1V1 = nRT1
At the final state:
P2V2 = nRT2
Since the number of moles and the gas constant are constant, we can rewrite the equations as:
P1V1 = RT1
P2V2 = RT2
We are given that the initial conditions are:
P1 = 60 atm
V1 = ?
To find V1, we need more information.
PV = nRT
Where:
P = Pressure (in pascals)
V = Volume (in cubic meters)
n = Number of moles
R = Ideal gas constant (8.314 J/(mol·K))
T = Temperature (in kelvin)
We have:
n = 2 moles
R = 8.314 J/(mol·K)
Let's assume the initial temperature is T1 and the final temperature is T2.
At the initial state:
P1V1 = nRT1
At the final state:
P2V2 = nRT2
Since the number of moles and the gas constant are constant, we can rewrite the equations as:
P1V1 = RT1
P2V2 = RT2
We are given that the initial conditions are:
P1 = 60 atm
V1 = ?
To find V1, we need more information.
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Two moles of monoatomic ideal gas at 60°C are mixed adiabatically with one mole of another monoatomic ideal gas at 12°C. The final temperature of the mixture isa)36°Cb)44°Cc)48°Cd)72°CCorrect answer is option 'B'. Can you explain this answer?
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Two moles of monoatomic ideal gas at 60°C are mixed adiabatically with one mole of another monoatomic ideal gas at 12°C. The final temperature of the mixture isa)36°Cb)44°Cc)48°Cd)72°CCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two moles of monoatomic ideal gas at 60°C are mixed adiabatically with one mole of another monoatomic ideal gas at 12°C. The final temperature of the mixture isa)36°Cb)44°Cc)48°Cd)72°CCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two moles of monoatomic ideal gas at 60°C are mixed adiabatically with one mole of another monoatomic ideal gas at 12°C. The final temperature of the mixture isa)36°Cb)44°Cc)48°Cd)72°CCorrect answer is option 'B'. Can you explain this answer?.
Two moles of monoatomic ideal gas at 60°C are mixed adiabatically with one mole of another monoatomic ideal gas at 12°C. The final temperature of the mixture isa)36°Cb)44°Cc)48°Cd)72°CCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two moles of monoatomic ideal gas at 60°C are mixed adiabatically with one mole of another monoatomic ideal gas at 12°C. The final temperature of the mixture isa)36°Cb)44°Cc)48°Cd)72°CCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two moles of monoatomic ideal gas at 60°C are mixed adiabatically with one mole of another monoatomic ideal gas at 12°C. The final temperature of the mixture isa)36°Cb)44°Cc)48°Cd)72°CCorrect answer is option 'B'. Can you explain this answer?.
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