Question

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

• A. 1315 J
• B. 275 J
• C. 765 J
• D. 675 J

Answer 1

Answer: (C)

$\Delta Q=Q_1+Q_2+Q_3+Q_4$
$=5960-5585-2980+3645=1040J$
$\Delta =W_1+W_2+W_3+W_4$
$=2200-825-1100+W_4=275+W_4$
For a cyclic process, $\Delta U=0$
ie, $U_f-U_i=0$
From first law of thermodynamics,
$\Delta Q=\Delta U+\Delta W$
$1040=0+275+W_4$
$W_4=765J$