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Question

A sample of gas goes from state A to state B in four different manners, as shown by the graphs. Let W be the work done by the gas and ΔU be change in internal energy along the path AB. Correctly match the graphs with the statements provided.

A
A-s, b-q,t; c-p, d-r
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B
A-s, b-q,t; c-r,t; d-q,t
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C
A-r, b-q,t c-p,t d-r
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D
A-s, b-q, c-r, d-p,r
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Solution

The correct option is B A-s, b-q,t; c-r,t; d-q,t
(A)s;(B)q,t;(C)r,t;(D)q,t in (A) , V is on vertical axis.


As V is icreasing, W is positive.


V is decreasing, W is negative,
As magnitude of negative work in part - II is greater than positive work in part - I, net work during the process is negative. Using PV = nRT and as Vremains same for initial and final points of the process, it is obvious that final temp. is greater than initial temperature as pressure has increased. Therefore dU is positive.
As T=PVnR
From graph we can say
Tfinal<Tinitial in part B,C & D
Similar arguments can be applied to other graphs.
Why this question?

Importance in JEE: Classic graphical question on how different thermodynamic variables inter-dependably change in a process.

Caution: P is not necessarily on y-axis.



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