Question

A sample of an ideal gas is taken through the cyclic process $abca$ as shown in the figure. The change in the internal energy of the gas along the path ca is $-180J$, The gas absorbs $250J$ of heat along the path ab and $60J$ along the path be. The work done (in joule) by the gas along the path $abc$ is :

Answer 1

Answer: 130

$\Delta U_{ac}=-\left(\Delta U_{ca}\right)=-\left(-180\right)=180J$
$Q=250+60=310J$
Now $Q=\Delta U+W$
or $310=180+W$
or $W=130J$