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Reversible adiabatic expansion using Van der Waals' equation |
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| May6-12, 10:20 AM | #1 |
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Reversible adiabatic expansion using Van der Waals' equation
1. The problem statement, all variables and given/known data
A real gas obeys Van der Waals‟ equation, which for one mole of gas is (p + A/V2)(V-B) = RT and its internal energy is given by U = CvT - A/V where the molar heat capacity at constant volume, Cv , is independent of the temperature and pressure. Show that the relation between the pressure p and the volume V of the Van der Waals‟ gas during a reversible adiabatic expansion can be written as (p + A/V2)(V-B)[itex]\gamma[/itex] = const. and find the expression for the parameter [itex]\gamma[/itex] in terms of Cv and R . 2. Relevant equations (p + A/V2)(V-B) = RT U = CvT - A/V Q= U + W 3. The attempt at a solution There is already a given solution and method for this equation. I worked through this much: 0 = U + W 0 = dU + PdV dU = (dU/dT)dT - (dU/dV)dV = CvdT + (A/V2)dV 0 = CvdT + (P + A/V2)dV = CvdT + RT/(V-B) ∫R/(V-B) dV = -∫Cv(dT/T) Rln(V-B) + Cvln(T) = const. ln(V-B)R + ln(T)CV = const ln[(V-B)R(T)CV] = const. (V-B)R(T)CV = const. I got stuck here and checked the method. My process was right, but according to it, the next line of work is: (V-B)R(RT)CV = const. I don't understand where this mystery R comes from. I've tried rearranging the ideal gas equation, and the first given equation to no avail. Could someone please explain how I get this R in the process? Thanks! |
| May6-12, 04:48 PM | #2 |
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R is the universal molargas constant and is part of the definition of a van der walal gas, just as it is for an ideal gas.
I haven't worked this out completely, but I would: let p1 = p + A/v2 v1 = v - B then p1v1 = RT. By your du = -dw and dT = (1/R)(p1dv1 + v1dp1) you can eliminate T, integrate by parts to get p1v1γ = constant. I think. |
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| adiabatic, pressure, relation, van der waals, volume |
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