JEE Mains · Physics · STD 11 - 11. thermodynamics
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 \(\Delta {Q_A}\, = \,\Delta {Q_B}\,;\,\Delta {U_A} = \Delta {U_B}\)
- 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 & Solution
Correct Answer
(B) \(\Delta {Q_A}\, > \,\Delta {Q_B}\,;\,\Delta {U_A} = \Delta {U_B}\)
Step-by-step Solution
Detailed explanation
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 \(\Delta Q=\Delta U+W\)…
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