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(D) For a cyclic process,the total change in internal energy is zero: $\Delta U_{ab} + \Delta U_{bc} + \Delta U_{ca} = 0$.Given $\Delta U_{ca} = -180\

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$130$

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(D) For a cyclic process,the total change in internal energy is zero: $\Delta U_{ab} + \Delta U_{bc} + \Delta U_{ca} = 0$.Given $\Delta U_{ca} = -180\, J$,we have $\Delta U_{ab} + \Delta U_{bc} = 180\, J$.From the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.For path $bc$,the process is isochoric (vertical line in $P-V$ diagram),so $\Delta W_{bc} = 0$. Thus,$\Delta U_{bc} = \Delta Q_{bc} = 60\, J$.Substituting this into the cyclic equation: $\Delta U_{ab} + 60 = 180 \implies \Delta U_{ab} = 120\, J$.Now,for path $ab$,$\Delta W_{ab} = \Delta Q_{ab} - \Delta U_{ab} = 250 - 120 = 130\, J$.The total work done along path $abc$ is $\Delta W_{abc} = \Delta W_{ab} + \Delta W_{bc} = 130 + 0 = 130\, J$.

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