For an ideal gas, consider only `P-V` work in going from an initial state `X` to the final state `Z`. The final state `Z` can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct? [Take `DeltaS` as change in entropy and `w` as work done]

Options

A. `DeltaS_(XtoZ)=DeltaS_(XtoY)+DeltaS_(YtoZ)`
B. `w_(XtoZ)=w_(XtoY)+w_(YtoZ)`
C. `w_(XtoYtoZ)=w_(XtoY)`
D. `DeltaS_(XtoYtoZ)=DeltaS_(XtoY)`

Answer

A, C

Explanation

`DeltaS` is a state function. So, it does not depend on path and only depends on initial and final stages.
`DeltaS_(XtoZ)=Delta_(XtoY)+DeltaS_(YtoZ)`
and `DeltaS_(YtoZ)` is not zero
`:. DeltaS_(XtoYtoZ)!=DeltaS_(XtoY)`
Work is not a state function and depends on path,
`:. w_(XtoZ)=w_(XtoY)+w_(YtoZ)`
`w_(XtoZ)=PdV` (`P` constant)
`w_(YtoZ)=0` (`V` is constant)
`w_(XtoZ)=2.303nRT"log"(V_(2))/(V_(1))`
`w_(XtoYtoZ)=w_(XtoY)+w_(YtoZ)`
As `w_(YtoZ)=0`, hence `w_(XtoYtoZ)=w_(XtoY)`