Verified Solution • 2025–26 Session moderate

QPage 62: Q13: HC Verma: Concepts of Physics - Volume 2 Laws of Thermodynamics Page 62: Q13 Solution: Refer to figure. Let and be the change in internal energy in…

⚠️ Common Exam Traps & Conceptual Pitfalls

Remember the first law is $Delta U=Q-W$ (heat added minus work done by the system). For a cyclic process $Delta U_{ ext{total}}=0$, so $Q_{ ext{total}}=W_{ ext{total}}$, giving $Q-W=0$. Students often mistakenly write $Q+W=0$ or $Delta U_1-Delta U_2=0$, confusing the net change with the difference between individual steps.

Refer to figure. Let ΔU1\Delta U_1 and ΔU2\Delta U_2 be the change in internal energy in processes A and B respectively, ΔQ\Delta Q be the net heat given to the system in process A + B and ΔW\Delta W be the net work done by the system in the process A + B.

Multiple Choice Options

  • A ΔU1+ΔU2=0\Delta U_1 + \Delta U_2 = 0
  • B ΔU1−ΔU2=0\Delta U_1 - \Delta U_2 = 0
  • C ΔQ−ΔW=0\Delta Q - \Delta W = 0
  • D ΔQ+ΔW=0\Delta Q + \Delta W = 0

Step-by-Step Verified Solution

**Answer:** (a), (c). **Step-by-Step Explanation:** _U_ + _U_ = 0 (a) because the system returns to initial state. _Q - _W = 0 (c) for a cyclic process.

All Questions in this Chapter

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