MCQ
$\Delta\text{S}(\text{A}\rightarrow\text{C})=50\text{ev},$$\Delta\text{S}(\text{C}\rightarrow\text{D})=30\text{ev},$
$\Delta\text{S}(\text{B}\rightarrow\text{D})=20\text{ev}$
$\text{A}\rightarrow\text{B}$
$\downarrow\ \ \ \ \ \uparrow$
$\text{C}\rightarrow\text{D}$
The entropy change for $A \rightarrow B$ would be :
  • A
    $100ev$
  • $60ev$
  • C
    $-\ 60ev$
  • D
    $-\ 100ev$

Answer

Correct option: B.
$60ev$
$\Delta\text{S}(\text{A}\rightarrow\text{B})=\Delta\text{S}(\text{A}\rightarrow\text{C})$
$+\Delta\text{S}(\text{C}\rightarrow\text{D})+\Delta\text{S}(\text{D}\rightarrow\text{B})$
$=50+30-\Delta\text{S}(\text{D}\rightarrow\text{B})$
$=50+30-20=60\text{ev}$

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