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Two gases occupy two containers A and B the gas in A, of volume 0.10m3, exerts a pressure of 1.40 MPa and that in B of volume 0.15m3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)
- a)0.70
- b)0.98
- c)1.40
- d)210
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Two gases occupy two containers A and B the gas in A, of volume 0.10m3...
Given:
- Volume of gas in container A = 0.10 m^3
- Pressure of gas in container A = 1.40 MPa
- Volume of gas in container B = 0.15 m^3
- Pressure of gas in container B = 0.7 MPa
To find:
- Final pressure in the container
Assumption:
- The gases are ideal gases.
- The temperature remains constant.
Solution:
When the two containers are united by a tube and the gases are allowed to intermingle, they will reach a state of equilibrium. At this state, the pressure will be equal throughout the system.
Step 1: Calculate the number of moles of each gas:
We can use the ideal gas equation to calculate the number of moles of each gas:
PV = nRT
For container A:
nA = (PA * VA) / (RT)
nA = (1.40 MPa * 0.10 m^3) / (8.314 J/(mol·K) * T) (1)
For container B:
nB = (PB * VB) / (RT)
nB = (0.70 MPa * 0.15 m^3) / (8.314 J/(mol·K) * T) (2)
Step 2: Apply the law of conservation of moles:
Since the two gases are allowed to mix, the total number of moles of gas remains constant. Therefore:
nA + nB = nA' + nB' (3)
where nA' and nB' are the final number of moles of gas in containers A and B respectively.
Step 3: Calculate the final pressure:
At equilibrium, the final pressure will be the same in both containers. Therefore:
PA' = PB'
Substituting equations (1) and (2) into equation (3), we get:
(1.40 MPa * 0.10 m^3) / (8.314 J/(mol·K) * T) + (0.70 MPa * 0.15 m^3) / (8.314 J/(mol·K) * T) = (PA' * VA) / (RT) + (PB' * VB) / (RT)
Simplifying the equation:
0.14 / T + 0.105 / T = PA' * 0.10 / (8.314 * T) + PB' * 0.15 / (8.314 * T)
Dividing both sides of the equation by 0.10:
1.4 / T + 1.05 / T = PA' / (8.314 * T) + PB' * 0.15 / (8.314 * T)
Combining like terms:
2.45 / T = (PA' + 0.15 * PB') / (8.314 * T)
Cross-multiplying:
2.45 * 8.314 * T = PA' + 0.15 * PB'
Rearranging the equation:
PA' = 2.45 * 8.314 * T - 0.15 * PB' (4)
Since PA'
- Volume of gas in container A = 0.10 m^3
- Pressure of gas in container A = 1.40 MPa
- Volume of gas in container B = 0.15 m^3
- Pressure of gas in container B = 0.7 MPa
To find:
- Final pressure in the container
Assumption:
- The gases are ideal gases.
- The temperature remains constant.
Solution:
When the two containers are united by a tube and the gases are allowed to intermingle, they will reach a state of equilibrium. At this state, the pressure will be equal throughout the system.
Step 1: Calculate the number of moles of each gas:
We can use the ideal gas equation to calculate the number of moles of each gas:
PV = nRT
For container A:
nA = (PA * VA) / (RT)
nA = (1.40 MPa * 0.10 m^3) / (8.314 J/(mol·K) * T) (1)
For container B:
nB = (PB * VB) / (RT)
nB = (0.70 MPa * 0.15 m^3) / (8.314 J/(mol·K) * T) (2)
Step 2: Apply the law of conservation of moles:
Since the two gases are allowed to mix, the total number of moles of gas remains constant. Therefore:
nA + nB = nA' + nB' (3)
where nA' and nB' are the final number of moles of gas in containers A and B respectively.
Step 3: Calculate the final pressure:
At equilibrium, the final pressure will be the same in both containers. Therefore:
PA' = PB'
Substituting equations (1) and (2) into equation (3), we get:
(1.40 MPa * 0.10 m^3) / (8.314 J/(mol·K) * T) + (0.70 MPa * 0.15 m^3) / (8.314 J/(mol·K) * T) = (PA' * VA) / (RT) + (PB' * VB) / (RT)
Simplifying the equation:
0.14 / T + 0.105 / T = PA' * 0.10 / (8.314 * T) + PB' * 0.15 / (8.314 * T)
Dividing both sides of the equation by 0.10:
1.4 / T + 1.05 / T = PA' / (8.314 * T) + PB' * 0.15 / (8.314 * T)
Combining like terms:
2.45 / T = (PA' + 0.15 * PB') / (8.314 * T)
Cross-multiplying:
2.45 * 8.314 * T = PA' + 0.15 * PB'
Rearranging the equation:
PA' = 2.45 * 8.314 * T - 0.15 * PB' (4)
Since PA'
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Community Answer
Two gases occupy two containers A and B the gas in A, of volume 0.10m3...
We know that
PAVA = nA RT, PBVB = nBRT and Pf (VA + VB) = (nA + nB) RT
Pf (VA + VB) = PAVA + PBVB
PAVA = nA RT, PBVB = nBRT and Pf (VA + VB) = (nA + nB) RT
Pf (VA + VB) = PAVA + PBVB
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Two gases occupy two containers A and B the gas in A, of volume 0.10m3, exerts a pressure of 1.40 MPa and that in B of volume 0.15m3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)a)0.70b)0.98c)1.40d)210Correct answer is option 'B'. Can you explain this answer?
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Two gases occupy two containers A and B the gas in A, of volume 0.10m3, exerts a pressure of 1.40 MPa and that in B of volume 0.15m3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)a)0.70b)0.98c)1.40d)210Correct 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 gases occupy two containers A and B the gas in A, of volume 0.10m3, exerts a pressure of 1.40 MPa and that in B of volume 0.15m3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)a)0.70b)0.98c)1.40d)210Correct 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 gases occupy two containers A and B the gas in A, of volume 0.10m3, exerts a pressure of 1.40 MPa and that in B of volume 0.15m3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)a)0.70b)0.98c)1.40d)210Correct answer is option 'B'. Can you explain this answer?.
Two gases occupy two containers A and B the gas in A, of volume 0.10m3, exerts a pressure of 1.40 MPa and that in B of volume 0.15m3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)a)0.70b)0.98c)1.40d)210Correct 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 gases occupy two containers A and B the gas in A, of volume 0.10m3, exerts a pressure of 1.40 MPa and that in B of volume 0.15m3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)a)0.70b)0.98c)1.40d)210Correct 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 gases occupy two containers A and B the gas in A, of volume 0.10m3, exerts a pressure of 1.40 MPa and that in B of volume 0.15m3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)a)0.70b)0.98c)1.40d)210Correct answer is option 'B'. Can you explain this answer?.
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