Gibbs’ Paradox.

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Presentation transcript:

Gibbs’ Paradox

Entropy of an Ideal Gas (Monatomic ideal gas) The Sackur-Tetrode equation: (Monatomic ideal gas) an average volume per molecule an average energy per molecule In general, for a gas of polyatomic molecules: f  3 (monatomic), 5 (diatomic), 6 (polyatomic)

Two cylinders (V = 1 liter each) are connected by a valve Two cylinders (V = 1 liter each) are connected by a valve. In one of the cylinders – Hydrogen (H2) at P = 105 Pa, T = 200C , in another one – Helium (He) at P = 3·105 Pa, T=1000C. Find the entropy change after mixing and equilibrating. Problem For each gas: The temperature after mixing: H2 : He :

Consider two different ideal gases (N1, N2) kept in two separate volumes (V1,V2) at the same temperature. To calculate the increase of entropy in the mixing process, we can treat each gas as a separate system. In the mixing process, U/N remains the same (T will be the same after mixing). The parameter that changes is V/N: Entropy of Mixing if N1=N2=1/2N , V1=V2=1/2V The total entropy of the system is greater after mixing – thus, mixing is irreversible.

- applies only if two gases are different ! Gibbs “Paradox” - applies only if two gases are different ! If two mixing gases are of the same kind (indistinguishable molecules): Stotal = 0 because U/N and V/N available for each molecule remain the same after mixing. Quantum-mechanical indistinguishability is important! (even though this equation applies only in the low density limit, which is “classical” in the sense that the distinction between fermions and bosons disappear.

at T1=T2, S=0, as it should be (Gibbs paradox) Problem Two identical perfect gases with the same pressure P and the same number of particles N, but with different temperatures T1 and T2, are confined in two vessels, of volume V1 and V2 , which are then connected. find the change in entropy after the system has reached equilibrium. - prove it! at T1=T2, S=0, as it should be (Gibbs paradox)

An Ideal Gas: from S(N,V,U) - to U(N,V,T) (fN degrees of freedom)  - the “energy” equation of state - in agreement with the equipartition theorem, the total energy should be ½kBT times the number of degrees of freedom. The heat capacity for a monatomic ideal gas:

The “Pressure” Equation of State for an Ideal Gas The “energy” equation of state (U  T): Ideal gas: (fN degrees of freedom) The “pressure” equation of state (P  T): - we have finally derived the equation of state of an ideal gas from first principles!