physics-class-11-unit-12-chapter-05-entropy-and-t-s-diagram-l-9-9-qv1pgs-mhhy

### <Success style="font-size:25px"> Entropy And T S Diagram L-9 </Success>
### <primary style="font-size:24px"> Second Law of Thermodynamics </primary>
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- <span style="color:blue">Kelvin-Planck Statement:</span> No (cyclic) process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat to work.

- No heat engine with efficiency one!

- <span style="color:blue">Clausius Statement:</span> No (cyclic) process is possible whose sole result is the transfer of heat from a colder object to a hotter object.

- No refrigerator with   $\varphi \rightarrow \infty \Rightarrow W=0$

- These two are equivalent statements!

- Carnot Engine:

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<footer style = "font-size: 18px"> $\rightarrow$ Entropy and T-S Diagram $\rightarrow$ <span style = "color:blue"> Second Law of Thermodynamics </span> $\rightarrow$ Second Law of Thermodynamics $\rightarrow$ Second Law of Thermodynamics </footer>

### <Success style="font-size:25px"> Entropy And T S Diagram L-9 </Success>
### <primary style="font-size:24px"> Second Law of Thermodynamics </primary>
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- A reversible engine: No dissipation

- Working substance: Chosen to be Ideal Gas (one mole)

- Works in a complete cycle between a hot reservoir $T_1$ and a cold reservoir $T_2$

- Efficiency should be maximum: But is not unity

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<footer style = "font-size: 18px">Entropy and T-S Diagram $\rightarrow$ Second Law of Thermodynamics $\rightarrow$ <span style = "color:blue"> Second Law of Thermodynamics </span> $\rightarrow$ Second Law of Thermodynamics $\rightarrow$ Second Law of Thermodynamics </footer>

### <Success style="font-size:25px"> Entropy And T S Diagram L-9 </Success>
### <primary style="font-size:24px"> Second Law of Thermodynamics </primary>
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- The Carnot Engine: (One mole of ideal Gas)

- Four processes:

-  1. Isothermal expansion $P_1, V_1, T_1 \rightarrow P_2, V_2, T_1$

-  2. Adiabatic Expansion $\quad P_2, V_2, T_1 \rightarrow P_3, V_3, T_2$

-  3. Isothermal compression $P_3, V_3, T_2 \rightarrow P_4, V_4, T_2$

-  4. Adiabatic Compression $\quad P_4, V_4, T_2 \rightarrow P_1, V_1, T_1$

- These processes can be executed in any order ensuring the initial and final states are the same. We shall calculate the work done And the heat absorbed. The change over a cycle in internal energy is zero.

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<footer style = "font-size: 18px">Second Law of Thermodynamics $\rightarrow$ Second Law of Thermodynamics $\rightarrow$ <span style = "color:blue"> Second Law of Thermodynamics </span> $\rightarrow$ Second Law of Thermodynamics $\rightarrow$ Entropy </footer>

### <Success style="font-size:25px"> Entropy And T S Diagram L-9 </Success>
### <primary style="font-size:24px"> Second Law of Thermodynamics </primary>
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- $\eta=1-\frac{T_2}{T_1}$

- Carnot theorem:

- Given two heat reservoirs a Carnot engine has the maximum efficiency.

- Efficiency of all reversible engines working between two given reservoirs is the reservoirs is the same: regardless of the working Substance employed or the Operation details.

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### <Success style="font-size:25px"> Entropy And T S Diagram L-9 </Success>
### <primary style="font-size:24px"> Entropy </primary>
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- $\eta_C> \eta_I \Rightarrow W'> W$

- <span style="color:blue">Entropy: An extensive Thermodynamic variable</span> $S=S(U, N, V)$

- Given an equilibrium state: has a definitely value of entropy; it is a state function.

- In the limit of absolute zero temperature entropy vanishes

- Let us consider that a small amount of heat is transferred to a System which is at a temperature T (Reversible process)

- $\Delta S_f=\frac{\Delta Q}{ T}  \text{general}:  \Delta S=S_f-S_i=\int_{T_1}^T \frac{\Delta Q}{T}$

- In an adiabatic process the entropy Change is zero

- Irreversible Process: $S_f-S_i>\int_T^T \frac{\Delta Q}{T}$
 
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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-11-unit-12-chapter-05-entropy-and-t-s-diagram-l-9_9-qv1pgs-mhhy-6.jpg)

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<footer style = "font-size: 18px">Second Law of Thermodynamics $\rightarrow$ Second Law of Thermodynamics $\rightarrow$ <span style = "color:blue"> Entropy </span> $\rightarrow$ Entropy $\rightarrow$ T-S Diagram and Carnot Engine </footer>

### <Success style="font-size:25px"> Entropy And T S Diagram L-9 </Success>
### <primary style="font-size:24px"> Entropy </primary>
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- Entropy is somehow related to "randomness", we lose the Information about the system when entropy increases.

- $S=0$

- $\delta Q=dU+PdV$

- $\frac{\delta Q}{T}=dS$

- Consider an Isolated system: (System + Resorvoir)

- In a thermodynamic process: $\Delta S \geq 0$  Second law

- $\Delta S \geq 0 \Rightarrow \Delta S_{system}+\Delta S_{resorvoir}$

- Entropy always increases (Entropy of the universe in increasing)

- $\Delta S_{total}=0 \longrightarrow$ Reversible process

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<footer style = "font-size: 18px">Second Law of Thermodynamics $\rightarrow$ Entropy $\rightarrow$ <span style = "color:blue"> Entropy </span> $\rightarrow$ T-S Diagram and Carnot Engine $\rightarrow$ T-S Diagram and Carnot Engine </footer>

### <Success style="font-size:25px"> Entropy And T S Diagram L-9 </Success>
### <primary style="font-size:24px"> T-S Diagram and Carnot Engine </primary>
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- PV Diagram: Area under the curve provides the work done on/by the system

- TS diagram: Area under the curve net heat exchanged.

- Adiabatic Process: Isentropic Process

- Isothermal Process: Entropy changes

- Change of entropy (of the system and the reservoir) is zero in a reversible process.

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-11-unit-12-chapter-05-entropy-and-t-s-diagram-l-9_9-qv1pgs-mhhy-8.jpg)
![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-11-unit-12-chapter-05-entropy-and-t-s-diagram-l-9_9-qv1pgs-mhhy-9.jpg)

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<footer style = "font-size: 18px">Entropy $\rightarrow$ Entropy $\rightarrow$ <span style = "color:blue"> T-S Diagram and Carnot Engine </span> $\rightarrow$ T-S Diagram and Carnot Engine $\rightarrow$ T-S Diagram and Carnot Engine </footer>

### <Success style="font-size:25px"> Entropy And T S Diagram L-9 </Success>
### <primary style="font-size:24px"> T-S Diagram and Carnot Engine </primary>
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-  Step I: $T=T_1$ fixed, $S$ increases

-  Step II: $T$ decreases, $S$ fixed

-  Step III: $T=T_2$ fixed, $S$ decreases

-  Step IV: $T$ decreases, $S$ fixed

- We are focusing on the system: In any process, the net change in Entropy of the system and Reservoir zero.

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-11-unit-12-chapter-05-entropy-and-t-s-diagram-l-9_9-qv1pgs-mhhy-8.jpg)
![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-11-unit-12-chapter-05-entropy-and-t-s-diagram-l-9_9-qv1pgs-mhhy-9.jpg)

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<footer style = "font-size: 18px">Entropy $\rightarrow$ T-S Diagram and Carnot Engine $\rightarrow$ <span style = "color:blue"> T-S Diagram and Carnot Engine </span> $\rightarrow$ T-S Diagram and Carnot Engine $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Entropy And T S Diagram L-9 </Success>
### <primary style="font-size:24px"> T-S Diagram and Carnot Engine </primary>
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- Steps I: Heat absorbed $Q_1$ and temperature $\mathrm{T}_1: \quad S_1=\frac{Q_1}{T_1}$

- Steps II and IV: Adiabatic Processes: $\quad \Delta S=0$

- Stept III. Heat released $Q_2$ and temperature $\mathrm{T}_2: \quad S_2=-\frac{Q_2}{T_2}$

- Net change in Entropy of the system in a reversible closed loop is the :

- $\Delta S_{total}=\frac{Q_1}{T_1}-\frac{Q_2}{T_2}=0 \Rightarrow \frac{Q_1}{Q_2}=\frac{T_1}{T_2} $

- $\eta=1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1}$

- $\Delta S_{total}^{reservoir}=0$

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-11-unit-12-chapter-05-entropy-and-t-s-diagram-l-9_9-qv1pgs-mhhy-10.jpg)
![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-11-unit-12-chapter-05-entropy-and-t-s-diagram-l-9_9-qv1pgs-mhhy-11.jpg)

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<footer style = "font-size: 18px">T-S Diagram and Carnot Engine $\rightarrow$ T-S Diagram and Carnot Engine $\rightarrow$ <span style = "color:blue"> T-S Diagram and Carnot Engine </span> $\rightarrow$ Problem $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Entropy And T S Diagram L-9 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- $\eta =1-\frac{Q_2}{Q_1}= 1-\frac{C_P{T_1-T_2}}{C_V{T_3-T_2}}$

- $\eta=1-\frac{C_P\left(T_1-T_2\right)}{C_V\left(T_3-T_2\right)}=1-\gamma \frac{\left(T_1-T_2\right)}{T_3-T_2} $

- $\eta=1-\gamma \frac{P_1\left(V_1-V_2\right)}{V_2\left(P_2-P_1\right)} $

- $=1-\gamma\frac{(V_1 / V_2-1)}{(P_2 / V_1-1)}$

- $PV=RT$

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<footer style = "font-size: 18px">T-S Diagram and Carnot Engine $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ Thank You </footer>

### <Success style="font-size:25px"> Entropy And T S Diagram L-9 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- Isochoric slope $\quad \frac{T}{C_v}$

- Isobaric Curve slope $\quad T / C_p$

- $\left(\frac{\partial T}{\partial S}\right)_v>\left(\frac{\partial T}{\partial S}\right)_p$

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<footer style = "font-size: 18px">Problem $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Thank You $\rightarrow$  </footer>