# Work done by ideal gas

1. Jan 29, 2014

### theory.beta

Hi all,

I was wondering if I am having a definition problem on mechanical work.
Since dW = -PdV (as I was told in class), is it correct to say the pressure is fixed with W = -PV, since dW = d(PV) = -VdP - PdV = -PdV suggests dP = 0?

Thanks

S.

2. Jan 29, 2014

### Philip Wood

No, it is not correct. dW is misleading notation. It suggests that dW is an infinitesimal change in some quantity, W, which the system possesses (a so-called 'function of state'). This is not the case. [For the same reason, dQ is also misleading notation, suggesting, wrongly, that some quantity, Q, is possessed by the system.] Work and heat are both energy IN TRANSIT, rather than residing in the system.

PV is a function of state (because P and V are both functions of state), but it is a mistake to regard dW as a differential of pV (or -PV).

3. Jan 29, 2014

### theory.beta

Hi Philip,

Thank you for the reply. However, I still think P is fixed in general, leading W = PV. Let's look at an integral form of ideal gas law. In essence, we have dT = (PdV+VdP)/R; integrating this would give the ideal gas law. But the actual integrated solution as PV = nRT implies either dV or dP to be zero.
Another example would be the heat function in a system of fixed volume, where Q = W = E +PV (Landau textbook chapter 2, the W is a little confusing). He took the derivative of W into dP and dV separately.
Any thoughts? Thanks again.

S.

4. Jan 29, 2014

### Staff: Mentor

The differential work done by the gas on the surroundings is dW=PdV, not dW=-PdV. Even if the pressure isn't constant, the differential work is still PdV. This comes from dW=Fdl, where F is the force and dl is the differential displacement. Since F = PA, dW = PAdl. But Adl = dV. So dW = PdV. See my Blog on my PF personal page.

Chet

5. Jan 30, 2014

### Andrew Mason

The integral of dT is not PV unless you start from absolute 0.

$$nRΔT = nR\int_A^B dT = \int_A^B (Pdv + VdP) = \int_A^B d(PV) = P(B)V(B)-P(A)V(A) = ΔPV$$

P could be constant in which case nRΔT = PΔV (i.e. ∫VdP = 0). Or V could be constant, in which case nRΔT = VΔP. Or P and V could both change. One cannot determine $\int Pdv$ or $\int VdP$ separately for a given process without knowing how P or V varies during the process. But from the ideal gas law, PV = nRT, we know that d(PV) = nRdT

AM

6. Jan 30, 2014

### Philip Wood

Two of the best known ideal cases (but good approximations to real changes) are isothermal and adiabatic expansions. P varies in both of these.

When P is constant, then W = P ΔV, not PV.