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JEE Main 20205 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
5 September
Shift
Morning Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2020 · 5 September · Morning Shift · Physics · Thermodynamics

Three different processes that can occur in an ideal monoatomic gas...

Three different processes that can occur in an ideal monoatomic gas are shown in the PP vs VV diagram. The paths are labelled as A→B,A→CA \to B, A \to C and A→DA \to D. The change in internal energies during these process are taken as EAB,EAC{E}_{AB}, {E}_{AC} and EAD{E}_{AD} and the work done as WAB,WAC{W}_{AB}, {W}_{AC} and WAD{W}_{AD}. The correct relation between these parameters are:

Three different processes that can occur in an ideal monoatomic gas are shown in the P vs V diagram. The paths are

Options

  1. AEAB=EAC0,WAC=0,WAD<0{E}_{AB} = {E}_{AC} 0, {W}_{AC} = 0, {W}_{AD} < 0
  2. BEAB=EAC=EAD,WAB>0,WAC=0,WAD<0{E}_{AB} = {E}_{AC} = {E}_{AD}, {W}_{AB} > 0, {W}_{AC} = 0, {W}_{AD} < 0
  3. CEAB0,WAC>WAD{E}_{AB} 0, {W}_{AC} > {W}_{AD}
  4. DEAB>EAC>EAD,WAB<WAC<WAD{E}_{AB} > {E}_{AC} > {E}_{AD}, {W}_{AB} < {W}_{AC} < {W}_{AD}

Correct answer

B. EAB=EAC=EAD,WAB>0,WAC=0,WAD<0{E}_{AB} = {E}_{AC} = {E}_{AD}, {W}_{AB} > 0, {W}_{AC} = 0, {W}_{AD} < 0

Step-by-step solution

EAB=EAC=EAD{E}_{AB} = {E}_{AC} = {E}_{AD}

dU=nfR2(Tf−Ti)dU = \dfrac{nfR}{2}\left({T}_{f} - {T}_{i}\right)

WAB>0(+)asV↑{W}_{AB} > 0 ( + ) \text{as}V ↑

WAC=0asV=constant{W}_{AC} = 0\text{as}V = \text{constant}

WAD<0(−)asV↓{W}_{AD} < 0 ( - ) \text{as}V ↓

Three different processes that can occur in an ideal monoatomic gas are shown in the P vs V diagram. The paths are

\DeltaT\DeltaT is same for EAB=EAC=EAD{E}_{AB} = {E}_{AC} = {E}_{AD}
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