According to the first law of thermodynamics, ΔU = Q + W, where ΔU Is the increase in internal energy of the system, Q is the heat transferred to the system and W is the external work done by the system.Which of the following is NOT a correct expression?
Correct answer: D. When work is done by the system: ΔU = Q - W
- A. At constant temperature: Q = -W
- B. When no work is done: ΔU = Q
- C. In gaseous system: ΔU = Q + P Δ V
- D. When work is done by the system: ΔU = Q - W
Explanation
There are two sign conventions:1. ∆U= Q+W: Here, work done BYthe gas/system is taken POSITIVE. 2. ∆U= Q-W: Here, work done ON the gas/system is taken NEGATIVE. The equation provided in the question follows the first convention, according to which D is incorrect.
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About First Law of Thermodynamics
The first law relates heat supplied, work done and the change in internal energy through energy conservation. Problems use sign conventions and apply the law to isothermal, adiabatic, isobaric and isochoric processes. Internal energy is a state function, while heat and work depend on the path followed.
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