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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Laws of Thermodynamics

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Laws of Thermodynamics

A system transitions from an initial state p_1, V_1 to a final...

A system transitions from an initial state p1,V1p_1, V_1 to a final state p2,V2p_2, V_2 via two distinct methods. If ΔQ\Delta Q and ΔW\Delta W respectively denote the heat supplied to the system and the work done by the system, which of the following quantities must be identical for both methods?

Options

  1. AΔQ−ΔW\Delta Q - \Delta W
  2. BΔQ\Delta Q
  3. CΔQ+ΔW\Delta Q + \Delta W
  4. DΔW\Delta W

Correct answer

A. ΔQ−ΔW\Delta Q - \Delta W

Step-by-step solution

According to the first law of thermodynamics, the heat supplied to the system is given by ΔQ=ΔU+ΔW\Delta Q = \Delta U + \Delta W, where ΔU\Delta U is the change in internal energy and ΔW\Delta W is the work done by the system.

Rearranging the equation gives ΔU=ΔQ−ΔW\Delta U = \Delta Q - \Delta W.

Internal energy is a state function, meaning its change ΔU\Delta U depends only on the initial and final states of the system and is independent of the path taken.

Since the initial state p1,V1p_1, V_1 and the final state p2,V2p_2, V_2 are the same for both methods, the change in internal energy ΔU\Delta U must be identical for both methods.

Therefore, the quantity ΔQ−ΔW\Delta Q - \Delta W must be identical for both methods.
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