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A gas expands from volume $V_{1}$ to volume $V_{2}$ according to three different processes 1, 2 and 3. Change in internal energy of the gas in the processes are $Δ U_{1}, Δ U_{2}$ and $Δ U_{3}$ respectively. If same quantity of heat is supplied in three cases, then

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ΔU1<ΔU2<ΔU3

ΔU1>ΔU2>ΔU3

ΔU2<ΔU3>ΔU1

ΔU1>ΔU3<ΔU2

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detailed solution

Correct option is A

By first law of thermodynamics, ΔQ=ΔU+ΔW⇒ΔU=∆Q−ΔWNow ΔW1>ΔW2>ΔW3∴ ΔU1<ΔU2<ΔU3

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A gas expands from volume V1 to volume V2 according to three different processes 1, 2 and 3. Change in internal energy of the gas in the processes are ΔU1,ΔU2 and ΔU3 respectively. If same quantity of heat is supplied in three cases, then

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Ready to Test Your Skills?

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A gas expands from volume $V_{1}$ to volume $V_{2}$ according to three different processes 1, 2 and 3. Change in internal energy of the gas in the processes are $Δ U_{1}, Δ U_{2}$ and $Δ U_{3}$ respectively. If same quantity of heat is supplied in three cases, then