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CHEE2325 Thermodynamics Lecture 3 Summary

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Summary of lecture 3 for CHEE2325, thermodynamics, at the University of Newcastle.

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CHEE2325 – Lecture 3


Defining processes with 1st law of thermodynamics:
- Consider a hot coffee cooling down. This satisfies the 1 st law of
thermodynamics as the energy lost by the coffee is gained by the
surroundings.
- Consider the reverse process, whereby the coffee was heated up by the
surroundings. The energy gained by the coffee is lost by the surroundings, so
it satisfies the 1st law of thermodynamics. But this process would never occur,
so there must be another law that emphasises the direction of a process.



Entropy:
- Consider 2 blocks in contact with each other at different temperatures as
such:




- The entropy of a system, denoted by S, is defined by:
δ Qrev
dS= where:
T
dS=¿ incremental change in entropy (J K-1)
δ Q rev=¿ incremental, reversible addition of heat (J)
T =¿ absolute temperature at the boundary (K)

δ Qrev
∆ S=∫
T
- Entropy is a state function, meaning that ∆ S between 2 specified states is
constant, regardless of the path, reversible or irreversible.
- Entropy, S, is an extensive property. However as specific or molar entropy, ^S,
is intensive.
- While energy can be transferred by heat or work, entropy can only be
transferred by heat. Work is entropy free and does not contribute to entropy
transfer.



Isothermal heat transfer processes:

, - Consider a reversible isothermal process, i.e., T =T 0.
δ Qrev 1 Q
∆ S=∫ = ∫ δ Q rev =
T T0 T0




2nd law of thermodynamics:
- For a closed system and any process (reversible or irreversible):
2
δQ
S2−S 1=∫ + S gen
1 T
- S gen is called the entropy generation. This is typically a result of a gradient
existing in the system, which is then dissipated to approach equilibrium.
- The second law of thermodynamics states that S gen ≥ 0.
- If S gen is:
¿ 0 – irreversible process,
¿ 0 – reversible process,
¿ 0 – impossible.
- Rearranging the above equation, we get:
2
δQ
∆ S−∫ =S gen
1 T

- If the system is isolated (adiabatic),
∆ S=S gen
- Because S gen ≥ 0, ∆ Sisolated ≥ 0.
- Consider a system and its surroundings. They constitute an isolated system
since both can be enclosed by a sufficiently large arbitrary boundary across
which there is no mass or energy exchange. Therefore, the system and its
surroundings can be viewed as 2 subsystems of an isolated system.
Therefore:
∆ Suniverse =∆ S system +∆ S surroundings=S gen
- This means that ∆ Suniverse ≥ 0. This is another statement of the 2nd law of
thermodynamics.



Overall entropy balance:

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