The three processes in a thermodynamic cycle | vedclass

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(D) The total work done in a complete thermodynamic cycle is the sum of the work done in each individual process:$W_{	ext{total}} = W_{12} + W_{23} +

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$30$

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(D) The total work done in a complete thermodynamic cycle is the sum of the work done in each individual process:$W_{	ext{total}} = W_{12} + W_{23} + W_{31}$Given that the total work $W_{	ext{total}} = 10 \, J$.In an isochoric process $(2 ightarrow 3)$,the volume remains constant,so the work done $W_{23} = 0 \, J$.The work done in the adiabatic process $(3 ightarrow 1)$ is given as $W_{31} = -20 \, J$.Substituting these values into the total work equation:$10 \, J = W_{12} + 0 \, J + (-20 \, J)$$10 \, J = W_{12} - 20 \, J$$W_{12} = 30 \, J$For an isothermal process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta U + W$. Therefore,for the isothermal process $(1 ightarrow 2)$:$\Delta Q_{12} = 0 + W_{12} = 30 \, J$.Thus,the heat added to the system in the isothermal process is $30 \, J$.

List-$I$	List-$II$
$(A)$ Isothermal	$(I)$ $\Delta W = 0$
$(B)$ Adiabatic	$(II)$ $\Delta Q = 0$
$(C)$ Isobaric	$(III)$ $\Delta U \neq 0$
$(D)$ Isochoric	$(IV)$ $\Delta U = 0$

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Match List-$I$ with List-$II$.
List-$I$ List-$II$
$(A)$ Isothermal $(I)$ $\Delta W = 0$
$(B)$ Adiabatic $(II)$ $\Delta Q = 0$
$(C)$ Isobaric $(III)$ $\Delta U \neq 0$
$(D)$ Isochoric $(IV)$ $\Delta U = 0$

Choose the correct answer from the options given below:

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The $P-V$ diagram of a certain process (Carnot cycle) is as shown. The process is also represented as:

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