Work done by a gas in thermodynamics A given mass of gas expands from state A to state B by three paths 1, 2, 3 as shown in the figure below.


If $W_1,$ $W_2$ and $W_3$ are respectively, be the work done by the gas along three paths, then
(A) $W_1 &gt; W_2 &gt; W_3$
(B) $W_1 &lt; W_2 &lt; W_3$
(C) $W_1 = W_2 = W_3$
(D) $W_1 &lt; W_2 : W_1 &lt; W_3$

The work done is $W = -P \,\mathrm dV$ that is the negative of the area under the graph; since Area 3 &gt; Area 2 &gt; Area 1, thus, $W_1 &gt; W_2 &gt; W_3.$
But the answer given is option (B). Why is this so?
Sign conventions can make 'work' confusing, but the question here asks for the work done by the gas. When a gas expands, it's doing work (it's losing energy), so it's actually $W = + P \, \mathrm dV$, inverting all the inequality signs. So option B) is correct.