Question

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

The value of $W_4$ ( in $J$) is:

(write the absolute value of an answer)

Answer 1

Internal energy of the system depends on initial and final states. So in a cyclic thermodynamic process, internal energy will not change, i.e; $\Delta U=0$

From first law of thermodynamics, $Q=W+U$

Ideal gas is taken through 4 steps ,

$\Delta Q=\Delta W+\Delta U$

But $\Delta U=0$, therefore,

$\Delta Q=\Delta W$

Change in heat energy after 4 steps

$\Delta Q=Q_1{+Q}_2{+Q}_3{+Q}_4$

Total work done after 4 steps

$\Delta W=W_1{+W}_2{+W}_3{+W}_4$

$⟹\Delta Q=\Delta W$

$⟹W_4=Q_1{+Q}_2{+Q}_3{+Q}_4-(W_1+W_2+W_3)$

$⟹W_4=(+5960-5585-2980+3645)-(+2200-825+1100)$

$⟹W_4=-1435J$