4a - Ludwig Boltzmann (1844-1906). Sadi Carnot (1796-1832)...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Sadi Carnot (1796-1832) Ludwig Boltzmann (1844-1906). The first law is not enough!!! 2 1 T T > 12 TT = Both directions satisfy the first law (energy conservation). However, only one is observed and the other is not !!! Heat flows from high to low temp, but never the other way around! A gas expands to fill a container and never spontaneously contracts to its original volume! We need a second law in order to determine which processes will take place spontaneously, and which won’t!!!
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 The definition of Entropy The first law of thermodynamics (for any process, either reversible or irreversible): dU D Dq w =+ The first law of thermodynamics for a reversible process: r rev v rev e Dq D DP d V q w dU =− The definition of entropy: rev d Dq ST = The first law of thermodynamics for a reversible process: dU d TdS PV •The entropy is a state function •The entropy is an extensive and additive quantity. Properties of the entropy: dS T dU V T Pd isindependent of the path f rev fi i dq SS S T ∆= − = 12 1 2 S + 1 S 2 S
Background image of page 2
3 Entropy change in an isothermal expansion of a monoatomic ideal gas 1 33 0 22 ( is consant in an isothermal process!) ln Points to note (1) The enrtopy change is linear in ( is an extensive quantity). (2) Th f i V V fi P dS dU dV TT U nRT dU nRdT dU T dS nRdV V S nR dV V Sn R VV nS =+ =⇒ = = ⇒= = ⎡⎤ ⇒∆ = ⎣⎦ e entropy increases with volume. (3) - ln in a reversible isothermic expansion (see chapter 3) . rev rev f i rev qw n R T V V Sq T == Entropy change due to mixing of two ideal gases The initial and of each of the gases are assumed equal. Hence, the final and also has teh same value. Interactions between gas particle are negligable (ideal gases). Hence, U and T B A mix TP PT S S S = ∆+ () remains constant. Hence, the entropy change of each gas is the same as in a reversible and isothermal expansion from the initial to final volume: / ln ln / A A AA A A B B n V n n V n V RT P RR R RT P n + + = ⎢⎥ [] [] [] {} 1 ln ln ln where is teh mole fraction of gas A. ln ln ln ln ln 0 A A A A aaA A B BB B B B B A B m B ix n nn n Vn n V R SR n n nx x x xnn n x n x x + = −+ ++ =− + ⇒∆ + + > A V B V
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
4 Entropy change due to heating at constant pressure Under constant pressure: D 11 ln In the case of a monoatomic ideal gas: 55 5 ln ln ln 22 2 Points to n ff ii rev P rev P TT f PP P i f i Dq dH C dT q dT dS C T SC d T C d T C T TP V n R V SR R R V n R V = = == ⎡⎤ ⇒∆ = = ⎢⎥ ⎣⎦ ∆= = = ∫∫ ote: (1) We assume that is independent of temperature. (2) The entropy increases with temperature. (3) ln regardless of whether the heating from to is done in a reversible or irre P Pf i if C SC TT T T versible manner, since the entropy is a state function. constant pressure ( )and temperature ( ): Consider a liquid gas phase transition process, which takes place under 00 Points to note: (1) Similar expressions ap vap vap vap re e vv a p v p b x a b H qH S PT S P T T H ∆∆ = =⇒ = > = >⇒ ply for other phase trnsitions: Melting (solid liquid): Sublimation (solid gas): (2) solid liquid, liquid gas and solid
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 15

4a - Ludwig Boltzmann (1844-1906). Sadi Carnot (1796-1832)...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online