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

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(C) For a cyclic process $abca$,the total change in internal energy is zero,so $\Delta U_{ab} + \Delta U_{bc} + \Delta U_{ca} = 0$.$1$. For the isothe

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(C) For a cyclic process $abca$,the total change in internal energy is zero,so $\Delta U_{ab} + \Delta U_{bc} + \Delta U_{ca} = 0$.$1$. For the isothermal process $ab$,the temperature remains constant,so the change in internal energy $\Delta U_{ab} = 0$.$2$. For the adiabatic process $bc$,the first law of thermodynamics is $\Delta Q = \Delta U + \Delta W$. Since $\Delta Q = 0$,we have $\Delta U_{bc} = -\Delta W_{bc}$. Given $\Delta W_{bc} = 4 \, J$,we get $\Delta U_{bc} = -4 \, J$.$3$. Substituting these into the cyclic equation: $0 + (-4 \, J) + \Delta U_{ca} = 0$,which gives $\Delta U_{ca} = 4 \, J$.Therefore,the change in internal energy for the path $c$ to $a$ is $4 \, J$.

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 $a$?

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