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Pressure exerted by 1 mole of methane in a 0.25 litre container at 300 K using van der Waal's equation (given a = 2.253 atml2mol–2, b = 0.0428 litmol–1) is (in atm): [rounded up to two decimal places]
Correct answer is between '81.00,84.00'. Can you explain this answer?
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Pressure exerted by 1 mole of methane in a 0.25 litre container at 300...
For one mole , vanderwaal eq can be written as :
p = (RT/V-b) - (a/V^2)
p = (0.082×300 / 0.25-0.0428) - (2.253/0.25^2)
p = 118.8 - 36.048
p = 82.75atm
p = (RT/V-b) - (a/V^2)
p = (0.082×300 / 0.25-0.0428) - (2.253/0.25^2)
p = 118.8 - 36.048
p = 82.75atm
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Pressure exerted by 1 mole of methane in a 0.25 litre container at 300...
The van der Waals equation is given by:
(P + (an^2/V^2))(V - nb) = nRT
Where:
P = pressure
a = van der Waals constant
n = number of moles
V = volume
b = van der Waals constant
R = gas constant
T = temperature
In this case, we are given:
a = 2.253 atm L^2/mol^2
n = 1 mole
V = 0.25 L
R = 0.0821 L·atm/(mol·K)
T = 300 K
First, we need to calculate the value of b using the van der Waals constant b, which is given as 0.0427 L/mol:
b = nb
b = (1 mol)(0.0427 L/mol)
b = 0.0427 L
Next, we can substitute the given values into the van der Waals equation:
(P + (an^2/V^2))(V - nb) = nRT
(P + (2.253 atm L^2/mol^2)(1 mol^2)/(0.25 L^2))(0.25 L - (1 mol)(0.0427 L/mol)) = (1 mol)(0.0821 L·atm/(mol·K))(300 K)
Simplifying:
(P + 9.012 atm)(0.25 L - 0.0427 L) = 0.0821 L·atm/(mol·K) * 300 K
(P + 9.012 atm)(0.2073 L) = 24.63 L·atm/mol
0.2073P + 1.866 atm L = 24.63 L·atm/mol
0.2073P = 22.764 L·atm/mol
P = 22.764 L·atm/mol / 0.2073
P ≈ 109.92 atm
Therefore, the pressure exerted by 1 mole of methane in a 0.25 litre container at 300 K using the van der Waals equation is approximately 109.92 atm.
(P + (an^2/V^2))(V - nb) = nRT
Where:
P = pressure
a = van der Waals constant
n = number of moles
V = volume
b = van der Waals constant
R = gas constant
T = temperature
In this case, we are given:
a = 2.253 atm L^2/mol^2
n = 1 mole
V = 0.25 L
R = 0.0821 L·atm/(mol·K)
T = 300 K
First, we need to calculate the value of b using the van der Waals constant b, which is given as 0.0427 L/mol:
b = nb
b = (1 mol)(0.0427 L/mol)
b = 0.0427 L
Next, we can substitute the given values into the van der Waals equation:
(P + (an^2/V^2))(V - nb) = nRT
(P + (2.253 atm L^2/mol^2)(1 mol^2)/(0.25 L^2))(0.25 L - (1 mol)(0.0427 L/mol)) = (1 mol)(0.0821 L·atm/(mol·K))(300 K)
Simplifying:
(P + 9.012 atm)(0.25 L - 0.0427 L) = 0.0821 L·atm/(mol·K) * 300 K
(P + 9.012 atm)(0.2073 L) = 24.63 L·atm/mol
0.2073P + 1.866 atm L = 24.63 L·atm/mol
0.2073P = 22.764 L·atm/mol
P = 22.764 L·atm/mol / 0.2073
P ≈ 109.92 atm
Therefore, the pressure exerted by 1 mole of methane in a 0.25 litre container at 300 K using the van der Waals equation is approximately 109.92 atm.
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Pressure exerted by 1 mole of methane in a 0.25 litre container at 300 K using van der Waal's equation (given a = 2.253 atml2mol–2, b = 0.0428 litmol–1) is (in atm): [rounded up to two decimal places]Correct answer is between '81.00,84.00'. Can you explain this answer?
Question Description
Pressure exerted by 1 mole of methane in a 0.25 litre container at 300 K using van der Waal's equation (given a = 2.253 atml2mol–2, b = 0.0428 litmol–1) is (in atm): [rounded up to two decimal places]Correct answer is between '81.00,84.00'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Pressure exerted by 1 mole of methane in a 0.25 litre container at 300 K using van der Waal's equation (given a = 2.253 atml2mol–2, b = 0.0428 litmol–1) is (in atm): [rounded up to two decimal places]Correct answer is between '81.00,84.00'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Pressure exerted by 1 mole of methane in a 0.25 litre container at 300 K using van der Waal's equation (given a = 2.253 atml2mol–2, b = 0.0428 litmol–1) is (in atm): [rounded up to two decimal places]Correct answer is between '81.00,84.00'. Can you explain this answer?.
Pressure exerted by 1 mole of methane in a 0.25 litre container at 300 K using van der Waal's equation (given a = 2.253 atml2mol–2, b = 0.0428 litmol–1) is (in atm): [rounded up to two decimal places]Correct answer is between '81.00,84.00'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Pressure exerted by 1 mole of methane in a 0.25 litre container at 300 K using van der Waal's equation (given a = 2.253 atml2mol–2, b = 0.0428 litmol–1) is (in atm): [rounded up to two decimal places]Correct answer is between '81.00,84.00'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Pressure exerted by 1 mole of methane in a 0.25 litre container at 300 K using van der Waal's equation (given a = 2.253 atml2mol–2, b = 0.0428 litmol–1) is (in atm): [rounded up to two decimal places]Correct answer is between '81.00,84.00'. Can you explain this answer?.
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