JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Chemical Thermodynamics

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

Δ S

as change in entropy and

w

as work done]

Answer: (A), (C)

Solution

(A) Entropy (S) is a state function, so its change depends only on initial and final states, not on the path. Therefore,

Δ S_{X \to Z} = Δ S_{X \to Y} + Δ S_{Y \to Z}

is correct.

(B) Work (w) is a path function, not a state function. The work done along the direct path X→Z is generally different from the sum

w_{X \to Y} + w_{Y \to Z}

. This is incorrect.

(C) From the diagram, Y→Z is an isochoric (constant volume) process, so

w_{Y \to Z} = 0

. Therefore,

w_{X \to Y \to Z} = w_{X \to Y} + 0 = w_{X \to Y}

. This is correct.

(D) Since entropy is a state function and Z ≠ Y,

Δ S_{X \to Y \to Z} \neq = Δ S_{X \to Y}

. This is incorrect.

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · One or more correct options