An ideal gas goes from state A to state B via three different processes as indicated in the P - V diagram:

If Q1, Q2, Q3 indicate the heat absorbed by the gas along the three processes and \DeltaU1, \DeltaU2, \DeltaU3 indicate the change in internal energy along the three processes respectively, then:

  • Option 1)

    Q3 > Q2 > Q1 and \DeltaU1 = \DeltaU2 = \DeltaU3

  • Option 2)

    Q1 = Q2 = Q3 and \DeltaU1 > \DeltaU2 > \DeltaU3

  • Option 3)

    Q3 > Q2 > Q1 and \DeltaU1 > \DeltaU2 > \DeltaU3

     

  • Option 4)

    Q1 > Q2 > Q3 and \DeltaU= \DeltaU2 = \DeltaU3

 

Answers (1)
V Vakul

As we learnt in

Property of internal energy -

U (internal energy) is a state function.

- wherein

i.e. It depends only on initial and final state.

 

 

 

Since change in internal energy is a state function hence

\Delta U_{1}=\Delta U_{2}=\Delta U_{3}

work done is W_{3}< W_{2}< W_{1}

Since Q=\Delta U+W

\therefore Q_{1}> Q_{2}> Q_{3}

Correct option is 3.


Option 1)

Q3 > Q2 > Q1 and \DeltaU1 = \DeltaU2 = \DeltaU3

This is incorrect option

Option 2)

Q1 = Q2 = Q3 and \DeltaU1 > \DeltaU2 > \DeltaU3

This is incorrect option

Option 3)

Q3 > Q2 > Q1 and \DeltaU1 > \DeltaU2 > \DeltaU3

 

This is incorrect option

Option 4)

Q1 > Q2 > Q3 and \DeltaU= \DeltaU2 = \DeltaU3

This is correct option

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