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 $\Delta U$ 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
• B. A-s, b-q,t; c-r,t; d-q,t
• C. A-r, b-q,t c-p,t d-r
• D. A-s, b-q, c-r, d-p,r

Answer 1

The correct option is B A-s, b-q,t; c-r,t; d-q,t

$\left(A\right)-s;\left(B\right)-q,t;\left(C\right)-r,t;\left(D\right)-q,t$ 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 $V_{\text{remains}}$ 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 $T=\frac{PV}{nR}$
$\therefore$ From graph we can say
$T_{\text{final}}<T_{\text{initial}}$ 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.