Using van der Waals' equation, calculate the constant, a (atm Ltr2 mole -2) when two moles of a gas confined in a four litre flask exerts a pressure of 11 atm at a temperature of 300 K. The value of 'b' is 0.05 Litre mole-1.
Correct answer: A. 6.5
- A. 6.5
- B. 2.23
- C. 23.2
- D. 85
Explanation
Given:Pressure (P) = 11 atmVolume (V) = 4 LNumber of moles (n) = 2 molesTemperature (T) = 300 KIdeal gas constant (R) = 0.0821 L atm / (mol K)Van der Waals constant 'b' = 0.05 L/molVan der Waals Equation:The equation we'll use is:(P + (an^2 / V^2)) * (V - nb) = nRTSteps:Rearrange the equation to solve for 'a':(P + (an^2 / V^2)) = nRT / (V - nb)(an^2 / V^2) = (nRT / (V - nb)) - Pa = (V^2 / n^2) * ((nRT / (V - nb)) - P)Substitute the given values into the rearranged equation:a = ((4 L)^2 / (2 mol)^2) * (((2 mol * 0.0821 L atm / (mol K) * 300 K) / (4 L - (2 mol * 0.05 L/mol))) - 11 atm)Simplify the equation step by step:a = (16 L^2 / 4 mol^2) * (((49.26 L atm) / (4 L - 0.1 L)) - 11 atm)a = (4 L^2 / mol^2) * ((49.26 L atm / 3.9 L) - 11 atm)a = (4 L^2 / mol^2) * (12.63 atm - 11 atm)a = (4 L^2 / mol^2) * (1.63 atm)Calculate the final value of 'a':a = 6.52 atm L^2 / mol^2
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