An ideal gas is made to undergo cyclic process shown in figure below. Let Delta W depict work done,

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11.Thermodynamics

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An ideal gas is made to undergo the cyclic process shown in the figure below. Let $\Delta W$ depict the work done, $\Delta U$ be the change in internal energy of the gas and $Q$ be the heat added to the gas. Sign of each of these three quantities for the whole cycle will be (0 refers to no change)

$-, 0,-$

$+, 0,+$

$0,0,0$

$+,+,+$

(KVPY-2018)

$(a)$ Given cyclic process is

Area under compression process $C A$ is more than area under expansion process AB. So, net work done is negative.

i.e. $\Delta W < 0$

Also, in a cyclic process, change in internal energy is zero.

i.e. $\Delta U=0$

Now, by using first law of thermodynamics, we have

$\Delta Q=0+\Delta W$

we see that, $\quad \Delta Q < 0$

Standard 11

Physics

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(JEE MAIN-2024)

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(NEET-2024)

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(JEE MAIN-2021)

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