MCQ
Following figure shows two processes $A$ and $B$ for a gas. If $\Delta Q_A$ and $\Delta Q_B$ are the amount of heat absorbed by the system in two cases, and $\Delta U_A$ and $\Delta U_B$ are changes in internal energies, respectively, then
  • A
    $\Delta {Q_A}\, = \,\Delta {Q_B}\,;\,\Delta {U_A} = \Delta {U_B}$
  • $\Delta {Q_A}\, > \,\Delta {Q_B}\,;\,\Delta {U_A} = \Delta {U_B}$
  • C
    $\Delta {Q_A}\, < \,\Delta {Q_B}\,;\,\Delta {U_A} < \Delta {U_B}$
  • D
    $\Delta {Q_A}\, > \,\Delta {Q_B}\,;\,\Delta {U_A} > \Delta {U_B}$

Answer

Correct option: B.
$\Delta {Q_A}\, > \,\Delta {Q_B}\,;\,\Delta {U_A} = \Delta {U_B}$
b
Initial and final states for both the processes are same,

$\therefore \Delta \mathrm{U}_{\mathrm{A}}=\Delta \mathrm{U}_{\mathrm{B}}$

Work done during process $A$ is greater than in process $B$. Because area is more

By First law of thermodynamics

$\Delta Q=\Delta U+W$

$\Rightarrow \Delta \mathrm{Q}_{\mathrm{A}}>\Delta \mathrm{Q}_{\mathrm{B}}$

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