Section 2.9: Quasi-Static Processes

Simpler (Ideal) Special Case The discussion so far has been general concerning 2 systems A & A' interacting with each other. Now, consider a Simpler (Ideal) Special Case Definition: Quasi-Static Process

Simpler (Ideal) Special Case The discussion so far has been general concerning 2 systems A & A' interacting with each other. Now, consider a Simpler (Ideal) Special Case Definition: Quasi-Static Process This is defined to be a general process by which system A interacts with system A', but the interaction is carried out so slowly that system A remains arbitrarily close to equilibrium at all stages of the process.

Simpler (Ideal) Special Case The discussion so far has been general concerning 2 systems A & A' interacting with each other. Now, consider a Simpler (Ideal) Special Case Definition: Quasi-Static Process This is defined to be a general process by which system A interacts with system A', but the interaction is carried out so slowly that system A remains arbitrarily close to equilibrium at all stages of the process. How slowly does this interaction have to take place? This depends on the system, but the process must be much slower than the time it takes system A to return to equilibrium if it is suddenly disturbed.

Quasi-Static Processes or Quasi-Equilibrium Processes These are defined to be sufficiently slow processes that any intermediate state can be considered an equilibrium state. The macroscopic parameters must be well defined for all intermediate states.

Advantages of Quasi-Static Processes The macrostate of a system that participates in such a process can be described with the same (small) number of macroscopic parameters as for a system in equilibrium (for a gas, this could be T & P). By contrast, for non-equilibrium Processes (e.g. turbulent flow of a gas), a huge number of macroscopic parameters is needed.

Quasi-Static Processes See figure for an Example A gas is confined to a container of volume V. At the top is a moveable piston. Sand is placed on the top of that to weigh it down. Several different types of Quasi-Static Processes can be carried out on this system. These processes & the terminology used to describe them are: Isochoric Process:  V = constant Isobaric Process:  P = constant Isothermal Process:  T = constant Adiabatic Processes: Q = 0

Quasi-Static Processes Let the external parameters of system A be: x1,x2,x3,…xn. For this case, the energy of the rth quantum state of the system may abstractly be written: Er = Er(x1,x2,…xn) When the values of one or more external parameters are changed, the energy Er also obviously changes. Let each external parameter change by an infinitesimal amount: xα xα + dxα How does Er change? From calculus, we know: dEr = ∑α(∂Er/∂xα)dxα

Xα,r ≡ The Generalized Force The exact differential of Er is: dEr = ∑α(∂Er/∂xα)dxα As we’ve already noted, if the external parameters change, some mechanical work must be done. For the system in it’s rth quantum state, that work may be written: đWr = - dEr = - ∑α(∂Er/∂xα)dxα ≡ ∑αXα,rdxα Here, Xα,r ≡ - (∂Er/∂xα) Xα,r ≡ The Generalized Force associated with the external parameter xα

<Xα> ≡ - (∂Ē/∂xα) The Macroscopic Work done when the system’s external parameters change is related to the change in it’s mean internal energy Ē by: đW = - dĒ ≡ ∑α<Xα>dxα Here, <Xα> ≡ - (∂Ē/∂xα) <Xα> ≡ Mean Generalized Force associated with external parameter xα

đW = <p>dV đW = <q>dU đW = <μ>dB đW = <P>dE Examples: External parameters & mean generalized forces: External Parameter xα Mean Generalized Force <Xα> Macroscopic Work đW = <Xα>dxα Position Coordinate x Mean Mechanical Force <Fx> đW = <Fx>dx Volume V Mean Pressure <p> đW = <p>dV Electric Potential U Mean Electric Charge <q> đW = <q>dU Magnetic Field B Mean Magnetic Moment <μ> đW = <μ>dB Electric Field E Mean Electric Dipole Moment <P> đW = <P>dE

Quasi-Static Work Done by Pressure Section 2.10: Quasi-Static Work Done by Pressure

Quasi-Static Work Done One of the most important examples is to have the system of interest be a gas & to look at the Quasi-Static Work Done by Pressure on the gas. See the figure. A gas is confined to a container of volume V, with a piston at the top. A weight is on the top of the piston, which is changed by adding small lead shot to it, as shown.

Quasi-Static Work Done by Pressure on the gas. Initially, the piston & the gas are in equilibrium. If the weight is increased, the piston will push down on the gas, increasing the pressure p & doing work ON the gas. If the weight is decreased, the gas will push up on it, decreasing the pressure p & doing work ON the piston.

F = pA đW = F ds A = Cross sectional area. V = As = gas volume. Elementary Physics: Differential work đW done by gas when piston undergoes differential vertical displacement ds is: đW = F ds F = Total vertical force on piston. Definition of (mean) pressure p: F = pA A = Cross sectional area. V = As = gas volume. So, đW = pAds = pdV

đW = pAds = pdV So, the work done by the gas as the volume changes from Vi to Vf is the integral of the pressure p as a function of V: Obviously, this is the area under the p(V) vs. V curve!

Figs. (a) & (b) are only 2 of the Many Possible Processes! There are Many Possible Paths in the P-V Plane to take the gas from initial state i to final state f. The work done is, in general, different for each. This is consistent with the fact that đW is an Inexact Differential. Figs. (a) & (b) are only 2 of the Many Possible Processes!

Figures (c), (d), (e), (f): 4 more of the Many Possible Processes!

Section 2.11: Brief Math Discussion: Exact & Inexact Differentials

đQ = dĒ + đW dĒ is an Exact Differential. We’ve seen that, for infinitesimal, quasi-static processes, the First Law of Thermodynamics is đQ = dĒ + đW dĒ is an Exact Differential. đQ, đW are Inexact Differentials. To understand what an Inexact Differential is, it helps to first briefly review what is meant by an Exact Differential

dF(x,y) ≡ A(x,y) dx + B(x,y) dy Exact Differentials Let F(x,y) = an arbitrary function of x & y. F(x,y) is a well behaved function satisfying all the math criteria for being an analytic function of x & y. It’s Exact Differential is: dF(x,y) ≡ A(x,y) dx + B(x,y) dy where A(x,y) ≡ (∂F/∂x)y & B(x,y) ≡ (∂F/∂y)x. Math Theorem If F(x,y) is an analytic function, then it’s 2nd cross partial derivatives MUST be equal: (∂2F/∂x∂y) ≡ (∂2F/∂y∂x)

(∂2F/∂x∂y) ≡ (∂2F/∂y∂x) (∂A/∂y)x ≡ (∂B/∂x)y or The Exact Differential of the function F(x,y) is dF(x,y) ≡ A(x,y) dx + B(x,y) dy , where A(x,y) ≡ (∂F/∂x)y & B(x,y) ≡ (∂F/∂y)x. As just discussed, if F(x,y) is an analytic function, then it must be true that: (∂A/∂y)x ≡ (∂B/∂x)y or (∂2F/∂x∂y) ≡ (∂2F/∂y∂x)

Independent of the path It can further be shown that, if F(x,y) is an analytic function, the integral of dF between any 2 arbitrary points 1 & 2 in the xy plane is Independent of the path between 1 & 2.

Arbitrary Analytic Function F(x,y): Also, it can be shown that, for an Arbitrary Analytic Function F(x,y): The integral of dF over an arbitrary closed path in the x-y plane is zero. = 0 All of this stuff is derived in Appendix A of the text book. I will provide you with a copy of the text book (or a photocopy). It is the property of exactness that makes, e.g. P, V. T, state variables, i.e. how you reach a particular state (the path) is unimportant. Therefore, a state defined by a set of state variables is unique. Work and heat are not state variables, i.e. a system may do work or exchange heat with its surroundings in many different ways. In other words, if a system starts of in one state, and it performs some work. The final state is not uniquely defined. For instance, if a gas does work adiabatically, it must have cooled (remember the adiabatic expansion demo I did in the first class – see Mocko web page). On the other hand, if the gas did work under isothermal conditions, its temperature does not change. Furthermore, heat must have flowed into the gas to compensate the cooling that would have occurred under adiabatic conditions. We will discuss heat next week. Oddly enough, there exists an infinite number of ways a system can exchange heat with its surroundings, whilst also doing work, that will take it from one unique state (P1,V1,T1) to another unique state (P2,V2,T2). Therefore, there exists a combination of dW and dQ that is exact, even though they themselves are not!!! This is the first law of thermodynamics which we will discuss in the next chapter.

3 Tests for an Exact Differential F(x,y) = arbitrary analytic function of x & y.

Gases: Quasi-Static Work Done by Pressure đW is clearly path dependent ∆Ē = đQ + đW does not depend on the path.

Summary: The Differential dF = Adx + Bdy is Exact if: or (∂2F/∂x∂y) ≡ (∂2F/∂y∂x)