- $\Delta S_{surr} = \frac{-q}{T}$ $\quad$ $\Delta S = \frac{q_{rev}}{T}$
- adiabatic wall
- q = 0
- $\Delta S_{surr}$ = 0
- adiabatic process
- $(P_1,T_1,V_1)(I) \rightarrow (P_2,T_2,V_2)(II)$
- from given information state I and state II
- $q_{rev}$, $\Delta S = \frac{q_{rev}}{T}$
![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-06-chapter-08-chemical-thermodynamics-l-8_8-yay-_qn2t7q-7.png)