Question

Fo an ideal gas, consider only P-V work in going from 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.

[Take $△$ S as change in entropy and W as work done].
Which of the following choice(s) is (are) correct?
(2012)

• A. $△S_{X\to Z}=△S_{X\to Y}+△S_{Y\to Z}$
• B. $W_{X\to Z}=W_{X\to Y}+W_{Y\to Z}$
• C. $W_{X\to Z△Z}=W_{X\to Y}$
• D. $△S_{X\to Y\to Z}=△S_{X\to Y}$

Answer 1

Answer: (A), (C)

The correct options are
A $△S_{X\to Z}=△S_{X\to Y}+△S_{Y\to Z}$
C $W_{X\to Z△Z}=W_{X\to Y}$

Entropy is state function, change in entropy in a cyclic process is zero.
Therefore, $△S_{X\to Y}+△S_{Y\to Z}+△S_{Z\to X}=0$
$\Rightarrow -△S_{Z\to X}=△S_{X\to Y}+△S_{Y\to Z}$
$=△S_{X\to Z}$

Analysis of option (b) and (c)
Work is a non-stable function, it does depends on the pathe followed. $W_{Y\to Z}=0as△V=0.$
Therefore, $W_{X\to Y\to Z}=W_{X\to Y}.$ Also, work is the area under the curve on p-V diagram.

As shown above $W_{X\to Y}+W_{Y\to Z}=W_{X\to Y}=W_{X\to Y\to Z}$ but not equal to $W_{X\to Z}.$