Discussion Forum



1)

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?


A) $\triangle S_{X \rightarrow Z}=\triangle S_{X \rightarrow Y}+\triangle S_{Y \rightarrow Z}$

B) $ W_{X \rightarrow Z }= W_{X \rightarrow Y}+W _{Y \rightarrow Z}$

C) $ W_{X \rightarrow Z \rightarrow Z}= W_{X \rightarrow Y}$

D) $\triangle S_{X \rightarrow Y\rightarrow Z}=\triangle S_{X \rightarrow Y}$

Answer:

Option A,C

Explanation:

(a) Entropy is a state function, change tn entropy in a cyclic process is zero'
Therefore.

$\triangle S_{X \rightarrow Y}+\triangle S_{Y \rightarrow Z}+\triangle S_{Z \rightarrow X}$=0

$\Rightarrow$  $-\triangle S_{Z \rightarrow X}=\triangle S_{X \rightarrow Y}+\triangle S_{Y \rightarrow Z}$

=$\triangle S _{X \rightarrow Z}$

Analysis of options (b) and (c)

Work is a non-stable function, it does depend on the Path followed.

$W_{Y\rightarrow Z}=0$ as $\triangle V=0$ Therefore $W_{X\rightarrow Y \rightarrow Z}=W_{X\rightarrow Y}$

Also work is the area under the curve on p-V diagram.

12112021995_m4.PNG

 As shown

 $ W_{X \rightarrow Y}+W_{Y\rightarrow Z}=W_{X\rightarrow Y}= W_{X\rightarrow Y \rightarrow Z}=W_{X\rightarrow Y} $

but not equal to $W_{X \rightarrow Z}$