# Thermodynamical Work

1. Mar 16, 2012

### emailanmol

I seem to have a conceptual doubt.

In thermodynamics first law
Q + W= U
W is the work done by the surrounding on gas and has value integral of [-P(external)dV].
Where dV is change in volume of gas

Now in some other conventions , we define the 1st law as Q-W = U
Where W is work done by gas on surrounding.

My question is.

How Is work done by surrounding on gas = -(Work done by gas on surrounding)

According to me work by gas on surrounding is integral of [P(internal)dV]

So these two are equivalent only for reversible processes where Pext is nearly equal to Pint.

Also whenever a gas compresses due to external pressure, the surrounding does work on gas and gas gains energy.But while getting compressed the gas does work done surrounding too and loses some of its energy.(Am i right?)
So isnt the net work done on gas a difference between the two i.e isn't it w=-(Pext -Pint)dV

Where am I going wrong?
It will be really kind of you to be descriptive and to help me identifying the mistakes in my analogy, and the required corrections.
Thanks

2. Mar 16, 2012

### DrDu

That's entirely true. You can only replace P(ext) by -P(int) if they are in equilibrium .
To the second part of your question: According to Newton actio=reactio always and you have to use one of the two (usually actio) to calculate work.

3. Mar 16, 2012

### emailanmol

Hey thanks.Now its clear