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A 8.3 L cylinder contains 128 . 0 g O2 gas at 27°C. What mass of O2 gas must be released to reduce the pressure to 6.0 bar ?
- a)64.0 g
- b)32.0 g
- c)16.0 g
- d)8.0 g
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A 8.3 L cylinder contains 128 .0 g O2 gas at 27°C. What mass of O...
By applying ideal gas eqn. PV = nRT
Further solving it we get an eq.
w = PVM/RT
Putting values, we get 64g.
Hence, the correct answer is Option A.
Further solving it we get an eq.
w = PVM/RT
Putting values, we get 64g.
Hence, the correct answer is Option A.
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A 8.3 L cylinder contains 128 .0 g O2 gas at 27°C. What mass of O...
To find the pressure of the gas, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
First, we need to convert the mass of O2 gas to moles. The molar mass of O2 is 32.0 g/mol, so we can calculate the number of moles using the formula:
moles = mass / molar mass
moles = 128.0 g / 32.0 g/mol = 4.0 mol
Next, we need to convert the temperature from Celsius to Kelvin. To do this, we add 273.15 to the Celsius temperature:
T = 27°C + 273.15 = 300.15 K
Now, we can substitute the values into the ideal gas law equation:
PV = nRT
P(8.3 L) = (4.0 mol)(0.0821 L·atm/mol·K)(300.15 K)
P(8.3 L) = 98.472 L·atm·K/mol
P = 98.472 L·atm·K/mol / 8.3 L
P = 11.87 atm
Therefore, the pressure of the O2 gas in the cylinder is 11.87 atm.
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
First, we need to convert the mass of O2 gas to moles. The molar mass of O2 is 32.0 g/mol, so we can calculate the number of moles using the formula:
moles = mass / molar mass
moles = 128.0 g / 32.0 g/mol = 4.0 mol
Next, we need to convert the temperature from Celsius to Kelvin. To do this, we add 273.15 to the Celsius temperature:
T = 27°C + 273.15 = 300.15 K
Now, we can substitute the values into the ideal gas law equation:
PV = nRT
P(8.3 L) = (4.0 mol)(0.0821 L·atm/mol·K)(300.15 K)
P(8.3 L) = 98.472 L·atm·K/mol
P = 98.472 L·atm·K/mol / 8.3 L
P = 11.87 atm
Therefore, the pressure of the O2 gas in the cylinder is 11.87 atm.
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A 8.3 L cylinder contains 128 .0 g O2 gas at 27°C. What mass of O2 gas must be released to reduce the pressure to 6.0 bar ?a)64.0 gb)32.0 gc)16.0 gd)8.0 gCorrect answer is option 'A'. Can you explain this answer? for Class 11 2026 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A 8.3 L cylinder contains 128 .0 g O2 gas at 27°C. What mass of O2 gas must be released to reduce the pressure to 6.0 bar ?a)64.0 gb)32.0 gc)16.0 gd)8.0 gCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 8.3 L cylinder contains 128 .0 g O2 gas at 27°C. What mass of O2 gas must be released to reduce the pressure to 6.0 bar ?a)64.0 gb)32.0 gc)16.0 gd)8.0 gCorrect answer is option 'A'. Can you explain this answer?.
A 8.3 L cylinder contains 128 .0 g O2 gas at 27°C. What mass of O2 gas must be released to reduce the pressure to 6.0 bar ?a)64.0 gb)32.0 gc)16.0 gd)8.0 gCorrect answer is option 'A'. Can you explain this answer? for Class 11 2026 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A 8.3 L cylinder contains 128 .0 g O2 gas at 27°C. What mass of O2 gas must be released to reduce the pressure to 6.0 bar ?a)64.0 gb)32.0 gc)16.0 gd)8.0 gCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 8.3 L cylinder contains 128 .0 g O2 gas at 27°C. What mass of O2 gas must be released to reduce the pressure to 6.0 bar ?a)64.0 gb)32.0 gc)16.0 gd)8.0 gCorrect answer is option 'A'. Can you explain this answer?.
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