Question

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

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

Answer 1

Answer: (C)

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