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Thermodynamics
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Section:
Chemistry
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© examsnet.com
Question : 34 of 42
Marks:
+1
,
-0
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 ΔS as change in entropy and W as work done)
[JEE Adv 2012 P1]
Δ
S
X
→
Z
\Delta S_{X \rightarrow Z}
Δ
S
X
→
Z
​
=
Δ
S
X
→
Y
+
Δ
S
Y
→
Z
\Delta S_{X \rightarrow Y} + \Delta S_{Y \rightarrow Z}
Δ
S
X
→
Y
​
+
Δ
S
Y
→
Z
​
W
X
→
Z
W_{X \rightarrow Z}
W
X
→
Z
​
=
W
X
→
Y
+
W
Y
→
Z
W_{X \rightarrow Y} + W_{Y \rightarrow Z}
W
X
→
Y
​
+
W
Y
→
Z
​
W
X
→
Y
→
Z
W_{X \rightarrow Y \rightarrow Z}
W
X
→
Y
→
Z
​
=
W
X
→
Y
W_{X \rightarrow Y}
W
X
→
Y
​
Δ
S
X
→
Y
→
Z
\Delta S_{X \rightarrow Y \rightarrow Z}
Δ
S
X
→
Y
→
Z
​
=
Δ
S
X
→
Y
\Delta S_{X \rightarrow Y}
Δ
S
X
→
Y
​
Validate
Solution:
As entropy is a state function and is additive
Δ
S
X
→
Z
\Delta S_{X \rightarrow Z}
Δ
S
X
→
Z
​
=
Δ
S
X
→
Y
+
Δ
S
Y
→
Z
\Delta S_{X \rightarrow Y} + \Delta S_{Y \rightarrow Z}
Δ
S
X
→
Y
​
+
Δ
S
Y
→
Z
​
On moving from Y to Z, the work done is zero as the volume is kept constant (isochoric process), so
W
X
→
Y
→
Z
W_{X \rightarrow Y \rightarrow Z}
W
X
→
Y
→
Z
​
=
W
X
→
Y
W_{X \rightarrow Y}
W
X
→
Y
​
© examsnet.com
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