Question

An ideal gas is taken through a cyclic thermodynamic process involving four steps. The amounts of heat involved in these steps are $Q_1=5960J,Q_2=-5585J,Q_3=-2980J$ and $Q_4=3645J$, respectively. The corresponding amounts of work done are $W_1=2200J,W_2=-825J$ and $W_3=-1100J$ and $W_4=respectively$.
Find the value of $W_4$.

Answer 1

Since $W_2$ and $W_3$ are negative, it means that the work is done on the gas. Hence $Q_2$ and $Q_3$ are negative which implies that heat is evolved in processes $2$ and $3$. Since $Q_1$ and $Q_4$ are positive, heat is absorbed by the gas in processes $1$ and $4$. As $(Q_1+Q_4)$ is greater than $(Q_2+Q_3)$, the gas absorbs a net amount of heat energy in a complete cycle, which is given by
$△Q=Q_1+Q_2+Q_3+Q_4$
$=5960-5585-2980+3645=1040joule$
The net work done by the gas is
$△W=W_1+W_2+W_3+W_4$
$=2200-825-1100+W_4=(275+W_4)joule$
Since the process is cyclic, the change in internal energy $△U=0$. From the first law of thermodynamics, we have
$△W=△Q-△U=△Q$
or $275+W_4=1040$ or $W_4=1040-275=765J$.