How Does the Definition of Work Affect the First Law of Thermodynamics Equation?

[Nick tringali]

Homework Statement

I picked choice A but the Answer is choice B. The book literally states that when work is done on the system work is negative. I get that when it’s Adiabatic the equation simplifies to U=-W making it U=-(-W). Is this like a trick question? Why would the book tell me that when work is done on the system work is negative then also ask a question and state it’s actually positive. Hope this question makes sense.

Relevant Equations

Delta U= Q-W

 

[Steve4Physics]

Hi @Nick tringali.

You have to be very careful when thinking about the direction of energy-transfer when work is done (on something by something else).

Your textbook link says "... W is the work done by the system" [my underlining].
But the question is about "the work done on the gas" [my underlining; and of course 'the gas' is 'the system' here].

 

[kuruman]

Yes, one needs to be careful about the preposition and be careful about who does the work and on whom work is done. It doesn't help that the first law appears as $\Delta U=Q+W$ in some textbooks and as $\Delta U=Q-W$ in others. Both are correct depending on the definition of $W$ in the textbook. As usual, $\Delta U$ is the change in internal energy and $Q$ is the heat that enters the gas.

To Nick tringali: Can you tell in which equation $W$ stands for "work done on the gas" and why? If so, then you understand what is going on here.

 

[Nick tringali]

Yes, one needs to be careful about the preposition and be careful about who does the work and on whom work is done. It doesn't help that the first law appears as $\Delta U=Q+W$ in some textbooks and as $\Delta U=Q-W$ in others. Both are correct depending on the definition of $W$ in the textbook. As usual, $\Delta$ is the change in internal energy and $Q$ is the heat that enters the gas.

To Nick tringali: Can you tell in which equation $W$ stands for "work done on the gas" and why? If so, then you understand what is going on here.

I would say that U= Q+W equation is where W stands for work done on the gas. When work is done on the gas it increases the internal energy so +W being work done on the gas would make sense . Right?