Question

An ideal gas undergoes the cyclic process $abca$ as shown in the figure. The internal energy change of the gas along the path ca is $-180\, J$. The gas absorb $250\, J$ of heat along the path $ab$ and also absorbs $60\, J$ of heat along the path $bc$ Find the work done by the gas along the path $abc$

• A. $130 \, J$
• B. $140 \,J$
• C. $120 \,J$
• D. $100 \,J$

Answer 1

Answer: (A)

Change in internal energy for a cyclic process is zero. Thus,
$\Delta U_{abc}+\Delta U_{ca}=0\Delta U_{abc}=-\Delta U_{ca}=180\,J$
$ΔQ=ΔW+ΔU$
$250+60=ΔW+180$
$ΔW=130\,J$