Question

In a cyclic process shown in the figure an ideal gas is adiabatically taken from $A$ to $B$, the work done on the gas during the process $A\to B$ is $80J$, when the gas is taken from $B$ to $A$ the heat absorbed by the gas is $40J$. What will be work done by the gas in process $B\to A$?

Answer 1

$△Q_{AB}=0,△W_{AB}=-80J$
$△Q_{BA}=40J,△W_{BA}=?$
From the first law of themodynamics
$(\sum Q{)}_{cycle}=(\sum W{)}_{cycle}$
$△Q_{AB}+△Q_{BA}=△W_{AB}+△W_{BA}$
$0+40=-80+△W_{BA}$
$△W_{BA}=120J$.