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Question on First law What is the meaning of thermodynamic, Sign convention"work done on or by" a gas?

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I have had this doubt since forever.

1)TheThe question first of all asks work for done BY gasby or work done ONon the gas which already means two different things. Isn't this like askingHow does that make sense? If I do work one$mgh$ on a block by hand in raising is mgh. Whereasit, the work done by gravitythe block on me is -mgh? Two of them$-mgh$. They have the same magnitude and opposite sign so.

In physics we use the sign convention of ∆W=P∆V for work done BY gas. So if it should be both +expansion it is Positive and vice versa. However in chemistry we use ∆W=-P∆V for work done by gas and expansion is negative. I understand that overall the sign convention give same answer for ∆U and ∆Q as one has Q=U+W whereas other one is U=Q+W. But doesn't this mean work done according to your sign convention will change?

  1. In physics we use the sign convention of ∆W=P∆V for work done BY gas. So if it expansion it is Positive and vice versa. However in chemistry we use ∆W=-P∆V for work done by gas and expansion is negative. I understand that overall the sign convention give same answer for ∆U and ∆Q as one has Q=U+W whereas other one is U=Q+W. But dissent this mean work done according to your sign convention will change?

CorrectThe correct answer is given assupposed to be D).

Question on First law of thermodynamic, Sign convention

enter image description here

I have had this doubt since forever.

1)The question first of all asks work done BY gas or work done ON gas which already means two different things. Isn't this like asking work one on block by hand in raising is mgh. Whereas work done by gravity is -mgh? Two of them have same magnitude and opposite sign so it should be both + and -.

  1. In physics we use the sign convention of ∆W=P∆V for work done BY gas. So if it expansion it is Positive and vice versa. However in chemistry we use ∆W=-P∆V for work done by gas and expansion is negative. I understand that overall the sign convention give same answer for ∆U and ∆Q as one has Q=U+W whereas other one is U=Q+W. But dissent this mean work done according to your sign convention will change?

Correct answer is given as D).

What is the meaning of "work done on or by" a gas?

enter image description here

The question asks work for done by or on the gas. How does that make sense? If I do work $mgh$ on a block by raising it, the work done by the block on me is $-mgh$. They have the same magnitude and opposite sign.

In physics we use the sign convention of ∆W=P∆V for work done BY gas. So if it expansion it is Positive and vice versa. However in chemistry we use ∆W=-P∆V for work done by gas and expansion is negative. I understand that overall the sign convention give same answer for ∆U and ∆Q as one has Q=U+W whereas other one is U=Q+W. But doesn't this mean work done according to your sign convention will change?

The correct answer is supposed to be D.

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Qmechanic
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Question on First law of thermodynamic, Sign convention.

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Question on First law of thermodynamic, Sign convention.

enter image description here

I have had this doubt since forever.

1)The question first of all asks work done BY gas or work done ON gas which already means two different things. Isn't this like asking work one on block by hand in raising is mgh. Whereas work done by gravity is -mgh? Two of them have same magnitude and opposite sign so it should be both + and -.

  1. In physics we use the sign convention of ∆W=P∆V for work done BY gas. So if it expansion it is Positive and vice versa. However in chemistry we use ∆W=-P∆V for work done by gas and expansion is negative. I understand that overall the sign convention give same answer for ∆U and ∆Q as one has Q=U+W whereas other one is U=Q+W. But dissent this mean work done according to your sign convention will change?

Correct answer is given as D).