Work done by expanding gas?

[OneDay]

Quick question

 

-expanding gas cool = -q and is the work negative too?

-compressed gas warms = +q and the work is positive?

 

Thanks for the help!

[AndyDude]

This may be right but I believe work and heat are two exclusive properties; expanding or compressing gas is limited to work only. Heat as positive or negative is determined whether the system gains (+) or loses (-) heat.

 

Regarding work, there are two conventions used interchangeably:

 

1) Work is defined as -PdV. In that case if the volume expands, work is negative and the internal energy is reduced. U = Q + W.

 

2) Work can also be defined as simply PdV. In this case if the volume expands, the work done is positive and the internal energy is reduced. U = Q - W. (Kaplan method)

[srindogg@hotmail.com]

To clarify:

 

The chemistry defn: Gas expansion is -ve work and Q + W = delta E

The engineering defn: Gas expansion is +ve work and Q - W = delta E

 

The MCAT always uses the chemistry definition. From above, you can see that both defns are equivalent

[GatewayMD]

Lol where did you learn engineering....there's no difference in defintions or conventions between the two

 

This is all you need to know,

 

When the piston expands work is done BY the system W = +ve

When the piston compresses work in done ON the system W=-ve

 

Change in internal energy (U not E) = Q - W

 

now the Q depends on your system: open, closed, or isolated if evergy is allowed to cross the boundary of the system then to preserve the internal energy (delta U = 0) heat will have to be added to the system therefore positive Q

 

essentially the relationship between work and heat depends on whether or not you are trying to preseve internal energy or not...phrase your question better and you will get a better answer, right now you have basically 3 different ones and it's just going to screw you up worse

 

oh and be glad you'll never need to use this as a physician

[srindogg@hotmail.com]

Haha, actually there is a difference in conventions... I studied engineering at UBC and am a thermodynamics T.A.

 

For the MCAT, the CHEMISTRY defn is used.

[Sergem]

Yea, srindogg is right. It's also in the Kaplan text.

[GatewayMD]

Either they're nuts at UBC, or you are trying to **** this guy up

 

From Thermodynamics: An Engineering Approach 6th Ed, Cenegel, Boles

 

"The generally accepted formal sign convention for heat and work interactions is as follows: heat transfer to a system and work done by a system are positive, heat transfer from a system and work done on a system are negative"

 

Which is exactly what I said...

 

And I'd like to know what hacker kaplan text you have because i'm looking at mine too and I don't see any mention of two conventions

 

Also what I was trying to get to but epic failed

 

"...We will use an intuitive approach in this book as it eliminates the need to adopt a formal sign convention"

 

Essentially if you look at where the energy is going and what the question is asking you can figure everything out without having to MEMORIZE any silly conventions, but a good engineering thermodynamics T.A. at UBC would know that right....

[candiolp]

WOW, people get saucy on this forum!! I think its hilarious!

[srindogg@hotmail.com]

Congratulations GatewayMD, you are half way there.

 

(1) You are absolutely right about the ENGINEERING convention in the Cengel and Boles textbook. On page 200 of the book: Q - W = delta U

 

(2) Look at any chemistry book or in the "hacker" Kaplan book. For the sake of consistency, lets look at the equation sheet provided by prep101 available at the following link.

 

http://www.prep101.com/mcat/ES_MCATChemistry.pdf

 

On page 2, on the right side, the equation for First law according to CHEMISTRY convention is: delta E = Q + W

 

Both defns are equivalent as explained in the earlier posts, its just a difference in convention.