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A box containing 2 moles of a diatomic ideal gas at temperature T0 is connected to another identical box containing 2 moles of a monoatomic ideal gas at temperature 5T0. There are no thermal losses and the heat capacity of the boxes and vibrational degree of freedom are negligible. Find the final temp, of the mixture
- a)2.5 T0
- b)2.0 T0
- c)2.66 T0
- d)1.5 T0
Correct answer is option 'A'. Can you explain this answer?
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A box containing 2 moles of a diatomic ideal gas at temperature T0 is ...
Heat gained = heat lost
nCΔT=nCvΔT
nCΔT=nCvΔT
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A box containing 2 moles of a diatomic ideal gas at temperature T0 is ...
Final Temperature of the Mixture
Given:
- A box containing 2 moles of a diatomic ideal gas at temperature T0
- Another identical box containing 2 moles of a monoatomic ideal gas at temperature 5T0
- No thermal losses and negligible heat capacity of the boxes and vibrational degree of freedom
To find:
The final temperature of the mixture
Solution:
The final temperature of the mixture can be found using the principle of energy conservation and the ideal gas law.
1. Energy Conservation:
According to the principle of energy conservation, the total energy of the system before mixing is equal to the total energy of the system after mixing.
2. Total Energy before Mixing:
The total energy before mixing is the sum of the energies of the individual gases in their respective boxes.
- For the diatomic ideal gas:
The internal energy of an ideal gas is given by the equation U = (3/2) nRT, where U is the internal energy, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Therefore, the total energy of the diatomic ideal gas before mixing is (3/2) * 2 * R * T0 = 3RT0.
- For the monoatomic ideal gas:
The internal energy of a monoatomic ideal gas is given by the equation U = (3/2) nRT, where U is the internal energy, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Therefore, the total energy of the monoatomic ideal gas before mixing is (3/2) * 2 * R * 5T0 = 15RT0.
3. Total Energy after Mixing:
After mixing, the gases will reach a common final temperature. Let's assume this final temperature is Tf.
- For the diatomic ideal gas:
The final internal energy of the diatomic ideal gas is (3/2) * 2 * R * Tf = 3RTf.
- For the monoatomic ideal gas:
The final internal energy of the monoatomic ideal gas is (3/2) * 2 * R * Tf = 3RTf.
Therefore, the total energy of the system after mixing is 6RTf.
4. Equating Total Energies:
According to the principle of energy conservation, the total energy before mixing is equal to the total energy after mixing.
3RT0 + 15RT0 = 6RTf
18RT0 = 6RTf
Dividing both sides by 6R:
3T0 = Tf
Therefore, the final temperature of the mixture is T0/3.
Answer:
The correct option is (a) 2.5 T0.
Given:
- A box containing 2 moles of a diatomic ideal gas at temperature T0
- Another identical box containing 2 moles of a monoatomic ideal gas at temperature 5T0
- No thermal losses and negligible heat capacity of the boxes and vibrational degree of freedom
To find:
The final temperature of the mixture
Solution:
The final temperature of the mixture can be found using the principle of energy conservation and the ideal gas law.
1. Energy Conservation:
According to the principle of energy conservation, the total energy of the system before mixing is equal to the total energy of the system after mixing.
2. Total Energy before Mixing:
The total energy before mixing is the sum of the energies of the individual gases in their respective boxes.
- For the diatomic ideal gas:
The internal energy of an ideal gas is given by the equation U = (3/2) nRT, where U is the internal energy, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Therefore, the total energy of the diatomic ideal gas before mixing is (3/2) * 2 * R * T0 = 3RT0.
- For the monoatomic ideal gas:
The internal energy of a monoatomic ideal gas is given by the equation U = (3/2) nRT, where U is the internal energy, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Therefore, the total energy of the monoatomic ideal gas before mixing is (3/2) * 2 * R * 5T0 = 15RT0.
3. Total Energy after Mixing:
After mixing, the gases will reach a common final temperature. Let's assume this final temperature is Tf.
- For the diatomic ideal gas:
The final internal energy of the diatomic ideal gas is (3/2) * 2 * R * Tf = 3RTf.
- For the monoatomic ideal gas:
The final internal energy of the monoatomic ideal gas is (3/2) * 2 * R * Tf = 3RTf.
Therefore, the total energy of the system after mixing is 6RTf.
4. Equating Total Energies:
According to the principle of energy conservation, the total energy before mixing is equal to the total energy after mixing.
3RT0 + 15RT0 = 6RTf
18RT0 = 6RTf
Dividing both sides by 6R:
3T0 = Tf
Therefore, the final temperature of the mixture is T0/3.
Answer:
The correct option is (a) 2.5 T0.
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A box containing 2 moles of a diatomic ideal gas at temperature T0 is ...
2.5
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A box containing 2 moles of a diatomic ideal gas at temperature T0 is connected to another identical box containing 2 moles of a monoatomic ideal gas at temperature 5T0. There are no thermal losses and the heat capacity of the boxes and vibrational degree of freedom are negligible. Find the final temp, of the mixturea)2.5 T0b)2.0 T0c)2.66 T0d)1.5 T0Correct answer is option 'A'. Can you explain this answer?
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A box containing 2 moles of a diatomic ideal gas at temperature T0 is connected to another identical box containing 2 moles of a monoatomic ideal gas at temperature 5T0. There are no thermal losses and the heat capacity of the boxes and vibrational degree of freedom are negligible. Find the final temp, of the mixturea)2.5 T0b)2.0 T0c)2.66 T0d)1.5 T0Correct answer is option 'A'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A box containing 2 moles of a diatomic ideal gas at temperature T0 is connected to another identical box containing 2 moles of a monoatomic ideal gas at temperature 5T0. There are no thermal losses and the heat capacity of the boxes and vibrational degree of freedom are negligible. Find the final temp, of the mixturea)2.5 T0b)2.0 T0c)2.66 T0d)1.5 T0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A box containing 2 moles of a diatomic ideal gas at temperature T0 is connected to another identical box containing 2 moles of a monoatomic ideal gas at temperature 5T0. There are no thermal losses and the heat capacity of the boxes and vibrational degree of freedom are negligible. Find the final temp, of the mixturea)2.5 T0b)2.0 T0c)2.66 T0d)1.5 T0Correct answer is option 'A'. Can you explain this answer?.
A box containing 2 moles of a diatomic ideal gas at temperature T0 is connected to another identical box containing 2 moles of a monoatomic ideal gas at temperature 5T0. There are no thermal losses and the heat capacity of the boxes and vibrational degree of freedom are negligible. Find the final temp, of the mixturea)2.5 T0b)2.0 T0c)2.66 T0d)1.5 T0Correct answer is option 'A'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A box containing 2 moles of a diatomic ideal gas at temperature T0 is connected to another identical box containing 2 moles of a monoatomic ideal gas at temperature 5T0. There are no thermal losses and the heat capacity of the boxes and vibrational degree of freedom are negligible. Find the final temp, of the mixturea)2.5 T0b)2.0 T0c)2.66 T0d)1.5 T0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A box containing 2 moles of a diatomic ideal gas at temperature T0 is connected to another identical box containing 2 moles of a monoatomic ideal gas at temperature 5T0. There are no thermal losses and the heat capacity of the boxes and vibrational degree of freedom are negligible. Find the final temp, of the mixturea)2.5 T0b)2.0 T0c)2.66 T0d)1.5 T0Correct answer is option 'A'. Can you explain this answer?.
Solutions for A box containing 2 moles of a diatomic ideal gas at temperature T0 is connected to another identical box containing 2 moles of a monoatomic ideal gas at temperature 5T0. There are no thermal losses and the heat capacity of the boxes and vibrational degree of freedom are negligible. Find the final temp, of the mixturea)2.5 T0b)2.0 T0c)2.66 T0d)1.5 T0Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Physics.
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