Thermodynamics and Thermochemistry 1 Question 42

45. For an ideal gas, consider only PV 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 ΔS as change in entropy and W as work done]. Which of the following choice(s) is (are) correct? (2012)

(a) ΔSXZ=ΔSXY+ΔSYZ

(b) WXZ=WXY+WYZ

(c) WXYZ=WXY

(d) ΔSXYZ=ΔSXY

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Solution:

  1. (a) Entropy is a state function, change in entropy in a cyclic process is zero.

 Therefore, ΔSXY+ΔSYZ+ΔSZX=0 ΔSZX=ΔSXY+ΔSYZ =ΔSXZ

Analysis of options (b) and (c)

Work is a non-stable function, it does depends on the path followed. WYZ=0 as ΔV=0. Therefore, WXYZ=WXY. Also, work is the area under the curve on pV diagram.

As shown above WXY+WYZ=WXY=WXYZ but not equal to WXZ.