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

• 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

  $\Delta \mathrm{U}_{1}=\Delta \mathrm{U}_{2}>\Delta \mathrm{U}_{3}$
• Correct answer
  D

  $\Delta \mathrm{U}_{1}>\Delta \mathrm{U}_{2}>\Delta \mathrm{U}_{3}$

Solution

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

Hence, the correct answer is option: 'D'.