Work done for the process shown in the figure is ............ $J$

  • A
    $1$
  • B
    $1.5$
  • C
    $4.5$
  • D
    $0.3$

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$P_A = 3 \times 10^{4} \text{ Pa}, V_A = 2 \times 10^{-3} \text{ m}^3, P_B = 8 \times 10^{4} \text{ Pa}, V_D = 5 \times 10^{-3} \text{ m}^3.$
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