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A physical chemist happened to know that the ๐‘ value for a gas that followed the Van der Waals equation was ๐‘ = 0.05 L / m o l .ย  But he didn't know the ๐‘Ž value.ย  He got 1.00 mole of the gas at a pressure of 1.00 atm at 10.0 ยฐC and found that it had a volume of 23.0 L.ย  Based on this single measurement, what is the probable value of ๐‘Ž ?ย  Note that ( ๐‘ + ๐‘Ž ๐‘› 2 ๐‘‰ 2 ) ( ๐‘‰ โˆ’ ๐‘› ๐‘ ) = ๐‘› ๐‘… ๐‘‡ is the Van der Waals equation of state, ๐‘… = 0.08206 L a t m m o l โˆ’ 1 K โˆ’ 1

Options
A.6.6
B.-510
C.-6.6
D.none of the others.
E.510
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Step-by-Step Analysis
The problem gives b = 0.05 L/mol for a gas described by the Van der Waals equation. Weโ€™re told: n = 1 mole, P = 1.00 atm, T = 10.0 ยฐC (which is 283.15 K), V = 23.0 L. The Van der Waals equation is (P + a n^2 / V^2)(V โˆ’ nb) = nRT. With n = 1, this becomes (P + a / V^2)(V โˆ’ b) = RT. First, compute V โˆ’ b: V โˆ’ nb = 23.0 โˆ’ 0.05 = 22.95 L. Next, compute P + a / V^2: V^2 = 23.0^2 = 529 L^2, so P + a / V^2 ......Login to view full explanation

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