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Question: Answered & Verified by Expert


In the given figure, 1 represents isobaric, 2 represents isothermal and 3 represents adiabatic processes of an ideal gas. If $\Delta \mathrm{U}_{1}, \Delta \mathrm{U}_{2}, \Delta \mathrm{U}_{3}$ be the changes in internal energy in these processes respectively, then
PhysicsThermodynamicsJEE Main
Options:
  • A $\Delta \mathrm{U}_{1} < \Delta \mathrm{U}_{2} < \Delta \mathrm{U}_{3}$
  • B $\Delta \mathrm{U}_{1}>\Delta \mathrm{U}_{3} < \Delta \mathrm{U}_{2}$
  • C $\Delta \mathrm{U}_{1}=\Delta \mathrm{U}_{2}>\Delta \mathrm{U}_{3}$
  • D $\Delta \mathrm{U}_{1}>\Delta \mathrm{U}_{2}>\Delta \mathrm{U}_{3}$
Solution:
1239 Upvotes Verified Answer
The correct answer is: $\Delta \mathrm{U}_{1}>\Delta \mathrm{U}_{2}>\Delta \mathrm{U}_{3}$


$\Delta \mathrm{T}_{1}=\mathrm{T}_{0}$
$\Delta \mathrm{T}_{2}=0$
$\Delta \mathrm{T}_{3}=-\mathrm{ve}$
$\Rightarrow \Delta \mathrm{U}_{1}>\Delta \mathrm{U}_{2}>\Delta \mathrm{U}_{3}$

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