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Using van der Waals’ equation, calculate the constant ‘a’ when two moles of a gas confined in a four litre flask exert a pressure of 11.0 atm at a temperature of 300 K. The value of ‘b’ is 0.05 L mol–1:
Correct answer is '6.46'. Can you explain this answer?
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**Van der Waals Equation**
The Van der Waals equation is used to account for the non-ideal behavior of gases. It is given by:
\[ \left(P + \frac{an^2}{V^2} \right) \left(\frac{V}{n} - b \right) = RT \]
Where:
- P is the pressure of the gas
- V is the volume of the gas
- n is the number of moles of the gas
- T is the temperature in Kelvin
- R is the ideal gas constant
- a and b are the Van der Waals constants
**Given Data**
- P = 11.0 atm
- V = 4 L
- n = 2 mol
- T = 300 K
- b = 0.05 L mol^(-1)
We are asked to calculate the value of constant a.
**Solving for a**
We can rearrange the Van der Waals equation to solve for constant a:
\[ \frac{a}{V^2} = \frac{P}{n^2} - \frac{RT}{V} + \frac{b}{V} \]
Substituting the given values:
\[ \frac{a}{(4 \, \text{L})^2} = \frac{(11.0 \, \text{atm})}{(2 \, \text{mol})^2} - \frac{(0.0821 \, \text{L atm mol}^{-1} \, \text{K}^{-1})(300 \, \text{K})}{4 \, \text{L}} + \frac{(0.05 \, \text{L mol}^{-1})}{4 \, \text{L}} \]
Simplifying:
\[ \frac{a}{16 \, \text{L}^2} = \frac{11.0 \, \text{atm}}{4 \, \text{mol}^2} - \frac{0.0821 \, \text{L atm mol}^{-1} \, \text{K}^{-1} \times 300 \, \text{K}}{4 \, \text{L}} + \frac{0.05 \, \text{L mol}^{-1}}{4 \, \text{L}} \]
\[ a = 16 \, \text{L}^2 \left( \frac{11.0 \, \text{atm}}{4 \, \text{mol}^2} - \frac{0.0821 \, \text{L atm mol}^{-1} \, \text{K}^{-1} \times 300 \, \text{K}}{4 \, \text{L}} + \frac{0.05 \, \text{L mol}^{-1}}{4 \, \text{L}} \right) \]
Evaluating the expression:
\[ a = 16 \, \text{L}^2 \left( \frac{11.0 \, \text{atm}}{16 \, \text{mol}^2} - \frac{0.0821 \, \text{L atm mol}^{-1} \, \text
The Van der Waals equation is used to account for the non-ideal behavior of gases. It is given by:
\[ \left(P + \frac{an^2}{V^2} \right) \left(\frac{V}{n} - b \right) = RT \]
Where:
- P is the pressure of the gas
- V is the volume of the gas
- n is the number of moles of the gas
- T is the temperature in Kelvin
- R is the ideal gas constant
- a and b are the Van der Waals constants
**Given Data**
- P = 11.0 atm
- V = 4 L
- n = 2 mol
- T = 300 K
- b = 0.05 L mol^(-1)
We are asked to calculate the value of constant a.
**Solving for a**
We can rearrange the Van der Waals equation to solve for constant a:
\[ \frac{a}{V^2} = \frac{P}{n^2} - \frac{RT}{V} + \frac{b}{V} \]
Substituting the given values:
\[ \frac{a}{(4 \, \text{L})^2} = \frac{(11.0 \, \text{atm})}{(2 \, \text{mol})^2} - \frac{(0.0821 \, \text{L atm mol}^{-1} \, \text{K}^{-1})(300 \, \text{K})}{4 \, \text{L}} + \frac{(0.05 \, \text{L mol}^{-1})}{4 \, \text{L}} \]
Simplifying:
\[ \frac{a}{16 \, \text{L}^2} = \frac{11.0 \, \text{atm}}{4 \, \text{mol}^2} - \frac{0.0821 \, \text{L atm mol}^{-1} \, \text{K}^{-1} \times 300 \, \text{K}}{4 \, \text{L}} + \frac{0.05 \, \text{L mol}^{-1}}{4 \, \text{L}} \]
\[ a = 16 \, \text{L}^2 \left( \frac{11.0 \, \text{atm}}{4 \, \text{mol}^2} - \frac{0.0821 \, \text{L atm mol}^{-1} \, \text{K}^{-1} \times 300 \, \text{K}}{4 \, \text{L}} + \frac{0.05 \, \text{L mol}^{-1}}{4 \, \text{L}} \right) \]
Evaluating the expression:
\[ a = 16 \, \text{L}^2 \left( \frac{11.0 \, \text{atm}}{16 \, \text{mol}^2} - \frac{0.0821 \, \text{L atm mol}^{-1} \, \text
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Using van der Waals’ equation, calculate the constant ‘a’ when two moles of a gas confined in a four litre flask exert a pressure of 11.0 atm at a temperature of 300 K. The value of ‘b’ is 0.05 L mol–1:Correct answer is '6.46'. Can you explain this answer?
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Using van der Waals’ equation, calculate the constant ‘a’ when two moles of a gas confined in a four litre flask exert a pressure of 11.0 atm at a temperature of 300 K. The value of ‘b’ is 0.05 L mol–1:Correct answer is '6.46'. 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 Using van der Waals’ equation, calculate the constant ‘a’ when two moles of a gas confined in a four litre flask exert a pressure of 11.0 atm at a temperature of 300 K. The value of ‘b’ is 0.05 L mol–1:Correct answer is '6.46'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Using van der Waals’ equation, calculate the constant ‘a’ when two moles of a gas confined in a four litre flask exert a pressure of 11.0 atm at a temperature of 300 K. The value of ‘b’ is 0.05 L mol–1:Correct answer is '6.46'. Can you explain this answer?.
Using van der Waals’ equation, calculate the constant ‘a’ when two moles of a gas confined in a four litre flask exert a pressure of 11.0 atm at a temperature of 300 K. The value of ‘b’ is 0.05 L mol–1:Correct answer is '6.46'. 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 Using van der Waals’ equation, calculate the constant ‘a’ when two moles of a gas confined in a four litre flask exert a pressure of 11.0 atm at a temperature of 300 K. The value of ‘b’ is 0.05 L mol–1:Correct answer is '6.46'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Using van der Waals’ equation, calculate the constant ‘a’ when two moles of a gas confined in a four litre flask exert a pressure of 11.0 atm at a temperature of 300 K. The value of ‘b’ is 0.05 L mol–1:Correct answer is '6.46'. Can you explain this answer?.
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