Internal Energy + Work Question

[rooksai]

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Book equation: Delta E= q + w

For an adiabatic process, no heat is transferred between the system and the environment and all energy is transferred as work.

Book says that since heat is theoretically not transferred since q = 0, Delta E= -w.

I was wondering how the book got Delta E = -w for an adiabatic process, since
q = 0, wouldn't equation be:

Delta E = w?

Additionally, I am confused about the:
w = -p Delta v, because if you set up the equation. If there is no heat transfer and heat does equal -w, then would Delta E = -(-p delta v) which would be positive, or is that the negative sign convention already set it up as -p delta v?

 

[N1st]

Book equation: Delta E= q + w

For an adiabatic process, no heat is transferred between the system and the environment and all energy is transferred as work.

Book says that since heat is theoretically not transferred since q = 0, Delta E= -w.

I was wondering how the book got Delta E = -w for an adiabatic process, since
q = 0, wouldn't equation be:

Delta E = w?

Additionally, I am confused about the:
w = -p Delta v, because if you set up the equation. If there is no heat transfer and heat does equal -w, then would Delta E = -(-p delta v) which would be positive, or is that the negative sign convention already set it up as -p delta v?

There are two possible sign conventions for work.

The first law of thermodynamics is ΔU = q - w, given that you define w as positive when work is done by the system, or equivalently ΔU = q + w, if you define w as positive when it is done on the system.

I prefer this second option, because q is heat into the system, and w is work done on the system, when both are positive.

So yes, if q = 0, then ΔU = w by the second sign convention.