1. Lecture 45 Entropy Clausius theorem Entropy as a state function
Second law statement using entropy
Calculating entropy increase

2. Clausius theorem Suppose a system absorbs heat 𝛿𝑄 at temperature 𝑇.
Since the value of 𝛿𝑄 𝑇 does not depend on the details of how the heat is transferred, we can assume it is from a Carnot engine, which in turn absorbs heat 𝛿 𝑄 0 from a heat reservoir with constant temperature 𝑇 0 .
For Carnot cycle
𝛿𝑄 𝑇 = 𝛿 𝑄 0 𝑇 0 → 𝛿 𝑄 0 = 𝑇 0 𝛿𝑄 𝑇
Therefore in one cycle, the total heat absorbed from the reservoir is
𝑄 0 = 𝑇 𝛿𝑄 𝑇
Since after a cycle, the system and the Carnot engine as a whole return to its initial status, the difference of the internal energy is zero.
𝑄 0 =Δ 𝐸 𝑖𝑛𝑡 +𝑊+ 𝑊 0 =𝑊+ 𝑊 0 = 𝑊 𝑡𝑜𝑡𝑎𝑙

3. According to the Kelvin-Planck statement of Second Law of thermodynamics, we cannot drain heat from one reservoir and convert them entirely into work without making any other changes, so 𝑊 𝑡𝑜𝑡𝑎𝑙 ≤０ Therefore， 𝛿𝑄 𝑇 ≤0 Which is called Clausius Inequality. If the system is reversible, then reverse its path and do the experiment again we can get − 𝛿𝑄 𝑇 ≤0 Thus for reversible case 𝛿𝑄 𝑇 =0

4. The Carnot Cycle
𝑄 ℎ =𝑛𝑅 𝑇 ℎ ln 𝑉 2 𝑉 1 , 𝑄 𝑐 =𝑛𝑅 𝑇 𝑐 ln 𝑉 4 𝑉 3 𝑝 1 𝑉 1 𝛾 = 𝑝 4 𝑉 4 𝛾 → 𝑝 1 𝑉 1 =𝑛𝑅 𝑇 1 , 𝑝 4 𝑉 4 =𝑛𝑅 𝑇 4 𝑇 1 𝑇 4 = 𝑝 1 𝑉 1 𝑝 4 𝑉 4 = 𝑉 4 𝛾−1 𝑉 1 𝛾−1 , 𝑇 2 𝑇 3 = 𝑉 3 𝛾−1 𝑉 4 𝛾−1 , 𝑉 4 𝑉 1 = 𝑉 3 𝑉 2 → 𝑉 2 𝑉 1 = 𝑉 3 𝑉 4 𝑄 ℎ 𝑇 ℎ = 𝑄 𝑐 𝑇 𝑐 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 This implies that there is a function of state which changes over the Carnot cycle when heat is added and subtracted. Clausius called this state function entropy.

5. Entropy
For reversible processes 𝛿𝑄=𝑑 𝐸 𝑖𝑛𝑡 +𝛿𝑊 = 𝐶 𝑉 𝑑𝑇+𝑝𝑑𝑉= 𝑛𝑐 𝑉 𝑑𝑇+𝑛𝑅𝑇 𝑑𝑉 𝑉 In order to calculus the integral, we should know how 𝑇 and 𝑉 change during the process Rewrite the equation as 𝛿𝑄 𝑇 =𝑛 𝑐 𝑉 𝑑𝑇 𝑇 +𝑛𝑅 𝑑𝑉 𝑉 =𝑑 𝑛 𝑐 𝑉 ln𝑇+𝑛𝑅ln𝑉 Define 𝑑𝑆= 𝛿𝑄 𝑇 Which is a total differentiation. We call the function 𝑆 as entropy.

6. Entropy is a function of state
𝑑𝑆= 𝛿𝑄 𝑇
The change of entropy is
Δ𝑆= 𝑖 𝑓 𝑑𝑆 = 𝑆 𝑓 − 𝑆 𝑖
= 𝑖 𝑓 𝑑 𝑛 𝑐 𝑉 ln𝑇+𝑛𝑅ln𝑉 =𝑛 𝑐 𝑉 ln 𝑇 𝑓 𝑇 𝑖 +𝑛𝑅ln 𝑉 𝑓 𝑉 𝑖
Entropy was discovered through mathematics rather than through laboratory results.
It is a mathematical construct and has no easy physical analogy.
In adiabatic process, the entropy does not change. It changes only when the heat is added and subtracted.

7. Principle of entropy increase
For an isolated system, it spontaneously changes from state 1 to state 2.
The process can be reversible or irreversible.
We can always find a reversible process to reverse the state from 2 to 1.
Then we get
1 2 𝛿𝑄 𝑇 𝛿𝑄 𝑇 = 1 2 𝛿𝑄 𝑇 +( 𝑆 1 − 𝑆 2 )≤0
Δ𝑆= 𝑆 2 − 𝑆 1 ≥ 1 2 𝛿𝑄 𝑇
For the right part respective to an isolated system 𝛿𝑄=0, finally we get
Δ𝑆≥0

8. General statement of the second law of thermodynamics
The entropy of an isolated system never decreases. It either stays constant (reversible processes) or increases (irreversible processes).
Since all real processes are irreversible, we can equally well state the second law as:
The total entropy of any system plus that of its environment increases as a result of any natural process:
Δ𝑆=Δ 𝑆 𝑠𝑦𝑠𝑡 +Δ 𝑆 𝑒𝑛𝑣 >0

9. "Time's Arrow"
For the reverse of any of the processes in the last few Examples, the entropy would decrease; and we never observe them.
For example,
We never observe heat flowing spontaneously from a cold object to a hot object.
Nor do we ever observe a gas spontaneously compressing itself into a smaller volume.
Nor do we see thermal energy transform into kinetic energy of a rock so the rock rises spontaneously from the ground.
Any of these processes would be consistent with the first law of thermodynamics (conservation of energy). But they are not consistent with the second law of thermodynamics, and this is why we need the second law.
Hence entropy has been called time's arrow, for it can tell us in which direction time is going.

10. Entropy increases in free expansion (not a reversible process)
For free expansion is not a quasistatic process, we can not calculus the integral ∫𝑑𝑠 directly.
However, we can find a quasistatic process to connect the initial and final states of free expansion.
Then we get
Δ𝑆=𝑛 𝑐 𝑉 ln 𝑇 𝑓 𝑇 𝑖 +𝑛𝑅ln 𝑉 𝑓 𝑉 𝑖 =𝑛𝑅ln 𝑉 𝑓 𝑉 𝑖 >0
The inverse process of free expansion will get Δ𝑆<0. But you “never” see a free shrinkage process.