Question

Figure displays a close cycle for a gas? The change in internal energy along path ca is $-160J$. The energy transferred to the gas as heat is $200J$ along path $ab$, and $40J$along path $bc$. How much work is done by the gas along path $abc$

Answer 1

Regarding part (a), it is important to recognize that the problem is asking for the
total work done during the two-step "path":$a\to b$ followed by $b\to c.$ During the latter part of this "path"
there is no volume change and consequentlt no work done. Thus the answer to part (b) is also
the answer to part (a). Since $△U$ for process $c\to a$ is $-160J,$ then $U_c-U_a=160J$. THerefore, using the
First Law of Thermodynamic, we have
$160=U_c-U_b+U_b-U_a$
$=Q_{b\to c}-W_{b\to c}+Q_{a\to b}-W_{a\to b}$
$=40-0+200-W_{a\to b}$
Therefore, $W_{a\to b}{W}_{a\to b\to c}=W_{a\to b}=80J$.