In the P-V diagram shown,the gas does 5 , J o | vedclass

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(C) For a complete cyclic process $abca$,the total change in internal energy is zero,i.e.,$\Delta U_{total} = \Delta U_{ab} + \Delta U_{bc} + \Delta U

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$4$

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(C) For a complete cyclic process $abca$,the total change in internal energy is zero,i.e.,$\Delta U_{total} = \Delta U_{ab} + \Delta U_{bc} + \Delta U_{ca} = 0$.For the isothermal process $ab$,the change in internal energy $\Delta U_{ab} = 0$.For the adiabatic process $bc$,the work done $W_{bc} = 4 \, J$. According to the first law of thermodynamics,$\Delta Q = \Delta U + W$. Since $\Delta Q = 0$ for an adiabatic process,$\Delta U_{bc} = -W_{bc} = -4 \, J$.Substituting these values into the cyclic equation: $0 + (-4 \, J) + \Delta U_{ca} = 0$.Therefore,$\Delta U_{ca} = 4 \, J$.Since the change in internal energy for the path $c$ to $a$ is $\Delta U_{ca} = -\Delta U_{ac}$,we are looking for the change in internal energy from $c$ to $a$,which is $\Delta U_{ca} = 4 \, J$.

In the $P-V$ diagram shown,the gas does $5 \, J$ of work in the isothermal process $ab$ and $4 \, J$ of work in the adiabatic process $bc$. What will be the change in internal energy of the gas in the straight path $c$ to $a$ (in $, J$)?

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