Question

An ideal gas is taken through a cyclic thermodynamical process through four steps. The amounts of heat involved in these steps are $Q_{1}=5960 \,J , Q_{2}=-5585 \,J , Q_{3}=-2980\, J$ $Q_{4}=3645\, J$; respectively. The corresponding works involved are $W_{1}=2200\, J , W_{2}=-825\, J, W_{3}=-1100 \,J$ 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)

765 J

Answer 2

$\Delta Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}$
$=5960-5585-2980+3645=1040\, J$
$\Delta =W_{1}+W_{2}+W_{3}+W_{4}$
$=2200-825-1100+W_{4}=275+W_{4}$
For a cyclic process, $\Delta U=0$
ie, $U_{f}-U_{i}=0$
From first law of thermodynamics,
$\Delta Q =\Delta U+\Delta W$
$1040 =0+275+W_{4}$
$W_{4} =765 \,J$