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A thermodynamic system is taken through the cycle $P Q R S P$ process. The net work done by the system is
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$-20 J$
Work done by the system \(=\) Area of shaded portion on \(\mathrm{P}-\mathrm{V}\) diagram
\(=(300-100) 10^{-} 6 \times(200-10) \times 10^3=20 \mathrm{~J}\)
and direction of process is anticlockwise so work done will be negative i.e.
\(\Delta \mathrm{W}=-20 \mathrm{~J}\)
\(=(300-100) 10^{-} 6 \times(200-10) \times 10^3=20 \mathrm{~J}\)
and direction of process is anticlockwise so work done will be negative i.e.
\(\Delta \mathrm{W}=-20 \mathrm{~J}\)
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