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Two moles of monoatomic ideal gas at 60ºC are mixed adiabatically with one mole of another monoatomic ideal gas at 12ºC. The final temperature of the mixture is
- a)36ºC
- b)44ºC
- c)48ºC
- d)72ºC
Correct answer is option 'B'. Can you explain this answer?
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Two moles of monoatomic ideal gas at 60ºC are mixed adiabatically...
°C and a pressure of 1 atm occupy a volume of 44.8 L. We can use the ideal gas law to calculate the number of molecules in the gas.
PV = nRT
Where P is the pressure (1 atm), V is the volume (44.8 L), n is the number of moles (2), R is the gas constant (0.0821 L · atm/mol · K), and T is the temperature in Kelvin (333 K, which is 60°C + 273.15).
Solving for the number of molecules (N):
N = n * NA
Where NA is Avogadro's number (6.022 x 10^23 molecules/mol).
Putting it all together:
PV = nRT
(1 atm) * (44.8 L) = (2 mol) * (0.0821 L · atm/mol · K) * (333 K)
NA = n * NA
N = (2 mol) * (6.022 x 10^23 molecules/mol)
N ≈ 7.23 x 10^23 molecules
Therefore, there are approximately 7.23 x 10^23 molecules in two moles of monoatomic ideal gas at 60°C and a pressure of 1 atm occupying a volume of 44.8 L.
PV = nRT
Where P is the pressure (1 atm), V is the volume (44.8 L), n is the number of moles (2), R is the gas constant (0.0821 L · atm/mol · K), and T is the temperature in Kelvin (333 K, which is 60°C + 273.15).
Solving for the number of molecules (N):
N = n * NA
Where NA is Avogadro's number (6.022 x 10^23 molecules/mol).
Putting it all together:
PV = nRT
(1 atm) * (44.8 L) = (2 mol) * (0.0821 L · atm/mol · K) * (333 K)
NA = n * NA
N = (2 mol) * (6.022 x 10^23 molecules/mol)
N ≈ 7.23 x 10^23 molecules
Therefore, there are approximately 7.23 x 10^23 molecules in two moles of monoatomic ideal gas at 60°C and a pressure of 1 atm occupying a volume of 44.8 L.
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Two moles of monoatomic ideal gas at 60ºC are mixed adiabatically with one mole of another monoatomic ideal gas at 12ºC. The final temperature of the mixture isa)36ºCb)44ºCc)48ºCd)72ºCCorrect answer is option 'B'. Can you explain this answer?
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Two moles of monoatomic ideal gas at 60ºC are mixed adiabatically with one mole of another monoatomic ideal gas at 12ºC. The final temperature of the mixture isa)36ºCb)44ºCc)48ºCd)72ºCCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two moles of monoatomic ideal gas at 60ºC are mixed adiabatically with one mole of another monoatomic ideal gas at 12ºC. The final temperature of the mixture isa)36ºCb)44ºCc)48ºCd)72ºCCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two moles of monoatomic ideal gas at 60ºC are mixed adiabatically with one mole of another monoatomic ideal gas at 12ºC. The final temperature of the mixture isa)36ºCb)44ºCc)48ºCd)72ºCCorrect answer is option 'B'. Can you explain this answer?.
Two moles of monoatomic ideal gas at 60ºC are mixed adiabatically with one mole of another monoatomic ideal gas at 12ºC. The final temperature of the mixture isa)36ºCb)44ºCc)48ºCd)72ºCCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two moles of monoatomic ideal gas at 60ºC are mixed adiabatically with one mole of another monoatomic ideal gas at 12ºC. The final temperature of the mixture isa)36ºCb)44ºCc)48ºCd)72ºCCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two moles of monoatomic ideal gas at 60ºC are mixed adiabatically with one mole of another monoatomic ideal gas at 12ºC. The final temperature of the mixture isa)36ºCb)44ºCc)48ºCd)72ºCCorrect answer is option 'B'. Can you explain this answer?.
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