3 More applications of derivatives

Transcription

1 3 More alications of derivatives 3.1 Eact & ineact differentials in thermodynamics So far we have been discussing total or eact differentials ( ( u u du = d + dy, (1 but we could imagine a more general situation If the differential is eact, M = ( u y du = M(, yd + N(, ydy. (2 ( y and N =. By the identity of mied artial derivatives, we have ( ( M 2 ( u N = = y E: Ideal gas V = R (for 1 mole, take V = V (,, so ( ( dv = d + d = R R d d (4 2 Now the work done in changing the volume of a gas is dw = dv = Rd R u (3 d. (5 Let s calculate the total change in volume and work done in changing the system between two oints A and C in, sace, along aths AC or ABC. 1. Path AC: d d = & 1 = 1 so d = d (6 1 = ( 1 (7 so (8 dv = R dw d R 2[ 1 + ( 1]d = R 2( 1 1d (9 = R ( 1 1d (10 1

2 ( 2, 2 C (, ( 1, 1 A B Figure 1: Path in, lane for thermodynamic rocess. Now we can calculate the change in volume and the work done in the rocess: V 2 V 1 = W 1 W 2 = AC AC dv = R( = R ( dv = R( ln = R ln 2 ( Path ABC: Note along AB d = 0, while along BC d = 0. V 2 V 1 = = W 2 W 1 = ABC ( ( d + 2 R R 1 d ( 1 d d = 2 1 ( 2 ( d + d 1 d = R ( ( R d = R 1 ln 2 + R( 2 1 V 1 Note the change in V is indeendent of the ath the volume is characteristic of a oint (, in equilibrium but the work done in the rocess is not! What s the difference? In the system with, as indeendent variables, dv is an eact differential, while dw is not. How can you see the difference? Go back and eamine the forms we had for dv and dw in (4 and (5. In the case of dv, we had dv = Md + Nd, with M = R M = R 2 2 and N = R 2 (14 N = R 2, (15

3 which is indeed eact, whereas dw = M d + N d M = R ; N = R (16 M = 0 N = R, (17 is not. his is a demonstration (we won t use the word roof that for a for a thermodynamic rocess involving changes in the lane, the volume of the system is a state variable, i.e. (for 1 mole of gas it simly deends on what and are; if you have secified,, you know the volume of the system. he change in volume between two oints will therefore always be V 2 V 1 indeendent of which ath is chosen. he work done in the same rocess is however not indeendent of the ath of integration. 3.2 Maima/minima roblems with constraints (Lagrange multiliers I m going to make a mathematical detour before coming back to thermodynamics, in order to give you some tools you need to solve homework roblems. In hysics we often need to find the etrema of a function of several variables subject to a constraint of some kind. In math, a simle eamle would be the distance function in 3-sace, d = 2 + y 2 + z 2. Of course the minimum of this function over all 3-sace is just 0, achieved at = y = z = 0. But suose we were to look for the minimum of d over the constrained set of oints defined by the lane 2y 2z = 3? he usual way to roceed is often the simlest, if it works: eress y = y(, then set dy/d = 0 and solve. But sometimes one can t solve for y( elicitly, so one can try the method of imlicit differentials (see Boas ch. 4, or use an elegant technique called the method of Lagrange multiliers (Joseh Lagrange ( was a French mathematical hysicist. he idea is, if you can eress the roblem in terms of the minimization of a function f(, y,..., together with a constraint g(, y,... = 0, to imagine you are solving the unconstrained roblem of finding the minimum of F (, y,... ; λ = f(, y,... + λg(, y,.... hen you can simly treat λ as an additional variable, and minimize with resect to it as well. In the rocess of solving the roblem, you eliminate λ from the solution. E: Coming back to our roblem with the lane, let s first make our life a bit easier by recalling that if we minimize d = 2 + y 2 + z 2, the square root will also 3

4 be a minimum. So define and set all the derivatives equal to zero: F = 2 + y 2 + z 2 + λ( 2y 2z 3 (18 z λ = 2 + λ = 0 (19 = 2y 2λ = 0 (20 = 2z 2λ = 0 (21 = 2y 2z 3 = 0. (22 Note the equation / λ = 0 is always just the constraint equation itself. Now we have a roblem with 4 equations and 4 unknowns, and it can be solved. he usual idea is to eliminate the constraint as quickly as ossible. he first eqn. tells us that λ = 2, so 2y + 4 = 0 (23 2z + 4 = 0 (24 2y 2z 3 = 0, (25 which we can solve to find (, y, z = (1/3, 2/3, 2/ Differentiation of integrals Leibnitz rule for differentiating integrals: d d v( u( f(, tdt = f(, v dv f(, udu d d + v( u( ( f dt. (26 E : 2 e t I = dt (27 t di e 2 e 2 d = e t dt (28 = 2 (e22 e 2 (29 4

5 3.4 Laws of thermodynamics Let s come back to the idea of eact and ineact differentials in thermodynamics. Here s another eamle of an ineact differential: dq = c }{{}}{{} d + Λ d, (30 }{{} heat absorbed heat caacity const. latent heat. (31 We showed already that the work done is also not an eact differential, however the combination of the two is: du dq dw (32 [ ( ] [ ( ] = c d + Λ d (33 his is called the internal energy, and this equation is sometimes referred to as the 1st law of thermodynamics, eressing energy conservation, i.e. the change in internal energy of a gas in a thermodynamic cycle goes either into heating the system (dq is the infinitesimal heat absorbed by the system is the or into doing work (done by the system. Another eact differential is ds = dq/. Remark 1: sometimes you will see the notation dq and dw for infinitesimal heat absorbed and work done. his is just a more careful notation to remind you that they are ineact differentials. Remark 2: since the work done in a thermodynamic rocess deends on the ath, it is really nonsense what I wrote above W 2 W 1. You are to think of this as the work erformed by the system in the rocess of going 1 2, but W 2 and W 1 have no indeendent meaning, since they are not characteristic of a macrostate. 3.5 Legendre transformation If for a function f(, y we have the differential df = d + qdy, where and q are equal to f/ and f/, resectively, we might ask the question, how do we make a simle change of variables to a new function g(, q with q one of the the indeendent variables? We simly make use of the fact that d(f qy = df dq y q dy = d + q dy dq y q dy = d y dq. (34 his function is by definition associated with an eact differential dg. You elored this on the HW for various thermodynamic quantities. 5