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A bulb of unknown volume 'V' contains an ideal gas at 2 atm pressure. It was connectedto another evacuated bulb of volume 0.5 litrethrough a stopcock. When the stopcock wasopened the pressure in each bulb became0.5atm. Then Vis 1) 17 ml 2) 1.7 L 3) 0.17 L 4) 0.34?
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A bulb of unknown volume 'V' contains an ideal gas at 2 atm pressure. ...
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A bulb of unknown volume 'V' contains an ideal gas at 2 atm pressure. ...
Given:
- Initial pressure in the bulb = 2 atm
- Final pressure in both bulbs = 0.5 atm
- Volume of the second bulb (evacuated bulb) = 0.5 litre
To Find:
The volume of the unknown bulb (V)
Solution:
To solve this problem, we can use the combined gas law, which states that the product of pressure and volume is directly proportional to the temperature of the gas.
Step 1: Write the combined gas law equation:
(P1 * V1) / T1 = (P2 * V2) / T2
Here, P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, and T1 and T2 are the initial and final temperatures.
Step 2: Identify the given values:
P1 = 2 atm
P2 = 0.5 atm
V2 = 0.5 litre
Step 3: Convert the volume to the same unit:
To use the combined gas law equation, we need to make sure that the volumes are in the same unit. Let's convert the volume of the second bulb from liters to milliliters:
V2 = 0.5 litre * 1000 ml/litre = 500 ml
Step 4: Substitute the values into the equation:
(2 atm * V1) / T1 = (0.5 atm * 500 ml) / T2
Step 5: Simplify the equation:
2 atm * V1 = 0.5 atm * 500 ml
Step 6: Cancel out the units:
2 * V1 = 0.5 * 500
Step 7: Solve for V1:
V1 = (0.5 * 500) / 2
V1 = 250 ml
Step 8: Convert the volume to the desired unit:
The volume of the unknown bulb (V) is 250 ml. Let's convert it to liters:
V = 250 ml / 1000 ml/litre = 0.25 litre
Answer:
Therefore, the volume of the unknown bulb (V) is 0.25 litre.
- Initial pressure in the bulb = 2 atm
- Final pressure in both bulbs = 0.5 atm
- Volume of the second bulb (evacuated bulb) = 0.5 litre
To Find:
The volume of the unknown bulb (V)
Solution:
To solve this problem, we can use the combined gas law, which states that the product of pressure and volume is directly proportional to the temperature of the gas.
Step 1: Write the combined gas law equation:
(P1 * V1) / T1 = (P2 * V2) / T2
Here, P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, and T1 and T2 are the initial and final temperatures.
Step 2: Identify the given values:
P1 = 2 atm
P2 = 0.5 atm
V2 = 0.5 litre
Step 3: Convert the volume to the same unit:
To use the combined gas law equation, we need to make sure that the volumes are in the same unit. Let's convert the volume of the second bulb from liters to milliliters:
V2 = 0.5 litre * 1000 ml/litre = 500 ml
Step 4: Substitute the values into the equation:
(2 atm * V1) / T1 = (0.5 atm * 500 ml) / T2
Step 5: Simplify the equation:
2 atm * V1 = 0.5 atm * 500 ml
Step 6: Cancel out the units:
2 * V1 = 0.5 * 500
Step 7: Solve for V1:
V1 = (0.5 * 500) / 2
V1 = 250 ml
Step 8: Convert the volume to the desired unit:
The volume of the unknown bulb (V) is 250 ml. Let's convert it to liters:
V = 250 ml / 1000 ml/litre = 0.25 litre
Answer:
Therefore, the volume of the unknown bulb (V) is 0.25 litre.
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A bulb of unknown volume 'V' contains an ideal gas at 2 atm pressure. It was connectedto another evacuated bulb of volume 0.5 litrethrough a stopcock. When the stopcock wasopened the pressure in each bulb became0.5atm. Then Vis 1) 17 ml 2) 1.7 L 3) 0.17 L 4) 0.34?
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A bulb of unknown volume 'V' contains an ideal gas at 2 atm pressure. It was connectedto another evacuated bulb of volume 0.5 litrethrough a stopcock. When the stopcock wasopened the pressure in each bulb became0.5atm. Then Vis 1) 17 ml 2) 1.7 L 3) 0.17 L 4) 0.34? 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 A bulb of unknown volume 'V' contains an ideal gas at 2 atm pressure. It was connectedto another evacuated bulb of volume 0.5 litrethrough a stopcock. When the stopcock wasopened the pressure in each bulb became0.5atm. Then Vis 1) 17 ml 2) 1.7 L 3) 0.17 L 4) 0.34? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bulb of unknown volume 'V' contains an ideal gas at 2 atm pressure. It was connectedto another evacuated bulb of volume 0.5 litrethrough a stopcock. When the stopcock wasopened the pressure in each bulb became0.5atm. Then Vis 1) 17 ml 2) 1.7 L 3) 0.17 L 4) 0.34?.
A bulb of unknown volume 'V' contains an ideal gas at 2 atm pressure. It was connectedto another evacuated bulb of volume 0.5 litrethrough a stopcock. When the stopcock wasopened the pressure in each bulb became0.5atm. Then Vis 1) 17 ml 2) 1.7 L 3) 0.17 L 4) 0.34? 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 A bulb of unknown volume 'V' contains an ideal gas at 2 atm pressure. It was connectedto another evacuated bulb of volume 0.5 litrethrough a stopcock. When the stopcock wasopened the pressure in each bulb became0.5atm. Then Vis 1) 17 ml 2) 1.7 L 3) 0.17 L 4) 0.34? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bulb of unknown volume 'V' contains an ideal gas at 2 atm pressure. It was connectedto another evacuated bulb of volume 0.5 litrethrough a stopcock. When the stopcock wasopened the pressure in each bulb became0.5atm. Then Vis 1) 17 ml 2) 1.7 L 3) 0.17 L 4) 0.34?.
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