The three processes in a thermodynamic cycle | vedclass

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Question

(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} +

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(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 $30 \, J$.

Column-$I$	Column-$II$
$(a)$ Adiabatic	$(i)$ $\Delta Q = \Delta U$
$(b)$ Isothermal	$(ii)$ $\Delta Q = \Delta W$
	$(iii)$ $\Delta U = -\Delta W$

Similar Questions

In Column-$I$ processes and in Column-$II$ the first law of thermodynamics are given. Match them appropriately:

Column-$I$ Column-$II$

$(a)$ Adiabatic $(i)$ $\Delta Q = \Delta U$

$(b)$ Isothermal $(ii)$ $\Delta Q = \Delta W$

$(iii)$ $\Delta U = -\Delta W$

What are the differences between mechanics and thermodynamics?

$A$ cyclic process on an ideal monatomic gas is shown in the figure. The correct statement is:

One mole of an ideal gas goes from an initial state $A$ to a final state $B$ via two processes: It first undergoes isothermal expansion from volume $V$ to $3V$ and then its volume is reduced from $3V$ to $V$ at constant pressure. The correct $P-V$ diagram representing the two processes is:

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