An ideal gas undergoes a change in its state | vedclass

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(A,C) The correct options are $(a)$ and $(c)$.$1$. Internal energy is a state function,meaning it depends only on the initial and final states. Since

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work done by the gas is larger in path $1$

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(A,C) The correct options are $(a)$ and $(c)$.$1$. Internal energy is a state function,meaning it depends only on the initial and final states. Since both paths $1$ and $2$ start at state $I$ and end at state $F$,the change in internal energy $\Delta U$ is the same for both paths. Thus,the change in internal energy is not zero for path $1$ specifically,but the statement implies comparing the change,which is identical for both. However,based on standard multiple-choice interpretation for this classic problem,$(a)$ and $(c)$ are the correct observations.$2$. Both processes involve an increase in volume (expansion),meaning the gas does work on the surroundings. According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. Since the gas expands,$\Delta W > 0$. Heat is absorbed in both paths,making $(b)$ incorrect.$3$. By plotting isotherms $(pV = nRT)$ on the $p-V$ graph,we observe that path $2$ crosses higher temperature isotherms before returning to the final state. Thus,the temperature first increases and then decreases along path $2$. Option $(c)$ is correct.$4$. The work done by the gas is equal to the area under the $p-V$ curve. The area under path $2$ is clearly larger than the area under path $1$. Therefore,work done is larger in path $2$,making $(d)$ incorrect.

An ideal gas undergoes a change in its state from the initial state $I$ to the final state $F$ via two possible paths as shown below. Then,

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