[Dev]

Hi.

My teacher told us that the value of PV gives an idea of the energy a system has.
i.e PV is value of a certain type of internal energy.

So i can say that dU = dQ + d(PV)
                         
dU = dQ + VdP + PdV

but the famous equation of the First law is actually dU = dQ + dW
where, dW = PdV

1) Where is the VdP term?

(Pext = External Pressure
Pint = Internal Pressure)


Whenever we do use dW = PdV, we use external pressure.
That must mean that earlier PV is actually PextV.
For Reversible process - P(internal) = P(external)
For Irreversible Process - We use P(external)

2) Why Pext?

This can  somewhat explain why VdP term is not mentioned beause dPext = 0.
But in reversible process, external pressure changes.
So in that case, we can't ignore that term

[TyPie]

It's dU=dQ-Pdv 
I think you're just repeating terms with the VdP, so your internal energy would be miscalculated. I'm not sure about external vs internal pressure, but it probably has to do with the sign of your q and work.  I've heard that it's used both ways, in which some books use +q while other books use negative q.  If you look at the units of PV, then you will probably see a lot of L*atm.  This can be converted to joules, which is a unit for work too.   
http://en.wikipedia.org/wiki/First_law_of_thermodynamics

I hope I helped.

[Yggdrasil]

Hi.

My teacher told us that the value of PV gives an idea of the energy a system has.
i.e PV is value of a certain type of internal energy.

For an ideal gas, its internal energy is directly proportional to its temperature:

U = n C_v T

By the ideal gas law, the quantity nT is directly proportional to PV

PV = nRT

Therefore, the internal energy of an ideal gas is directly proportional to PV:

U = (C_v/R) PV

[curiouscat]

So i can say that dU = dQ + d(PV)

AFAIK you cannot. That expression is just wrong.

I might be wrong; thermo is tricky.But I think that expression you wrote is wrong.