An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram- If Q 1- Q 2- Q 3 indicate the heat absorbed by the gas along the three processes and - U 1- - U 2- - U 3 indicate the change in internal energy along the three processes respectively- thenp -A- Q 1- Q 2- Q 3 and - U 1- U 2- U 3B- Q 3- Q 2- Q 1 and - U 1- U 2- U 3C- Q 3- Q 2- Q 1 and - U 1- U 2- U 3D- Q 1- Q 2- Q 3 and - U 1- U 2- U 3

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Byju's Answer

Standard XII

Physics

Equipartition Theorem

An ideal gas ...

Question

An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram. If $Q_{1}, Q_{2}, Q_{3}$ indicate the heat absorbed by the gas along the three processes and $Δ U_{1}, Δ U_{2}, Δ U_{3}$ indicate the change in internal energy along the three processes respectively, then

Q_{1}

Q_{2}

Q_{3}

and

Δ U_{1} = Δ U_{2} = Δ U_{3}

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Q_{3}

Q_{2}

Q_{1}

and

Δ U_{1} = Δ U_{2} = Δ U_{3}

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Q_{1} = Q_{2} = Q_{3}

and

Δ U_{1}

Δ U_{2}

Δ U_{3}

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Q_{3}

Q_{2}

Q_{1}

and

Δ U_{1} > Δ U_{2} > Δ U_{3}

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Solution

The correct option is A $Q_{1}$ > $Q_{2}$ > $Q_{3}$ and $Δ U_{1} = Δ U_{2} = Δ U_{3}$

Initial and final condition is same for all process
$Δ U_{1} = Δ U_{2} = Δ U_{3}$
$Δ Q = Δ U + Δ W$
Work done $Δ W_{1}$ > $Δ W_{2}$ > $Δ W_{3}$ (Area of P.V. graph)
So $Q_{1}$ . $Q_{2}$ > $Q_{3}$ .

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