1. Tardigrade
  2. Question
  3. Physics
  4. An ideal gas is taken through a cyclic thermodynamical process through four steps. The amounts of heat involved in these steps are Q1=5960 J , Q2=-5585 J , Q3=-2980 J Q4=3645 J; respectively. The corresponding works involved are W1=2200 J , W2=-825 J, W3=-1100 J and W4 respectively. The value of W4 is

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Q. 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

BHUBHU 2008Thermodynamics

Solution:

$\Delta Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}$
$=5960-5585-2980+3645=1040\, J$
$\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} $
$W_{4} =765 \,J$