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 $\Delta U_1, \Delta U_2, \Delta U_3 $ indicate the change in internal energy along the three processes respectively, then

• A. $Q_1>Q_2>Q_3 $ and $\Delta U_1=\Delta U_2=\Delta U_3 $
• B. $Q_3>Q_2>Q_1 $ and $ \Delta U_1=\Delta U_2=\Delta U_3 $
• C. $Q_1=Q_2=Q_3 $ and $\Delta U_1>\Delta U_2>\Delta U_3 $
• D. $Q_3>Q_2>Q_1 $ and $ \Delta U_1>\Delta U_2>\Delta U_3 $

Answer 1

Answer: (A)

Change in internal energy is path independent and depends only on the initial and final states.
As the initial and final states in the three processes are same. Therefore
$\Delta U_{1}=\Delta U_{2}=\Delta U_{3}$
Workdone, $W =$ Area under $P-V$ graph
As area under curve $1 >$ area under curve $2 >$ area under curve $3$
$\therefore W_{1} >W_{2} >W_{3}$
According to first law of thermodynamics,
$Q=W+\Delta U$
As $W_{1} >W_{2} >W_{3}$ and $\Delta U_{1}=\Delta U_{2}=\Delta U_{3}$
$\therefore Q_{1} >Q_{2} >Q_{3}$