Q146 / 180 · COMEDK 2023 (Memory Based) · Physics

An ideal gas goes from state $A$ to state $B$ via three different processes as indicated in the $p-V$ diagram. If $Q_1, Q_2$ and $Q_3$ indicate the heat absorbed by the three processes and $\Delta U_1, \Delta U_2$ and $\Delta U_3$ indicate the change in internal energy along the three processes respectively, then Figure

  • Correct answer
    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 $\Delta U_1>\Delta U_2>\Delta U_3$

Solution

For all processes 1,2 and 3 Change in internal energy, i.e. $\begin{gathered} \Delta U=U_B-U_A \\ \therefore \Delta U_1=\Delta U_2=\Delta U_3 \Rightarrow Q=\Delta U+\Delta W \end{gathered}$ Now, $\Delta W=$ work done by the gas, i.e. area under $p-V$ curve. As, area $(1)>$ area $(2)>$ area $(3)$ $\begin{array}{ll} \therefore & \Delta W_1>\Delta W_2>\Delta W_3 \\ \therefore & Q_1>Q_2>Q_3 \end{array}$

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

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