An ideal gas undergoes a cyclic transformatio | vedclass

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(D) Internal energy $(U)$ is a state function,which means its value depends only on the state of the system and not on the path taken. In any cyclic p

Vedclass · Accepted Answer

$\Delta U (	ext{Case-}I) = \Delta U (	ext{Case-}II) = \Delta U (	ext{Case-}III)$

Vedclass · Answer

(D) Internal energy $(U)$ is a state function,which means its value depends only on the state of the system and not on the path taken. In any cyclic process,the system returns to its initial state. Therefore,the change in internal energy $(\Delta U)$ for a complete cycle is always zero. $\Delta U_{	ext{cycle}} = U_{	ext{final}} - U_{	ext{initial}} = 0$. Since all three cases represent cyclic processes starting and ending at point $A$,the change in internal energy for each case is equal to zero. Thus,$\Delta U (	ext{Case-}I) = \Delta U (	ext{Case-}II) = \Delta U (	ext{Case-}III) = 0$.

An ideal gas undergoes a cyclic transformation starting from the point $A$ and coming back to the same point by tracing the path $A$ $\rightarrow B$ $\rightarrow C$ $\rightarrow D$ $\rightarrow A$ as shown in the three cases above. Choose the correct option regarding $\Delta U$.

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