### Transcription of Bonus * Bonus * Bonus Infinitesimal work done dw …

1 Definition of work using calculus: **Bonus** * **Bonus** * **Bonus** **Infinitesimal** work done dw by **Infinitesimal** change in volume of gas dV: A dL pext=F / A. Gas Movable Piston T1 T2. work = dw = F dL = (Apext) dL = pext (AdL). But, AdL = V2 - V1 = dV ( **Infinitesimal** volume change). dw = pext(V2 - V1) = pextdV But sign is arbitrary, so choose . dw = -pextdV (w < 0 is work done by gas, dV > 0). dw = -pextdV. (Note! p is the external pressure on the gas!). dw = - pextdV. Total work done in any change is the sum of little **Infinitesimal** increments for an **Infinitesimal** change dV. dw = - pextdV = w (work done by the system ).

2 Two Examples : ( 1 ) pressure = constant = pexternal, V changes vi vf vf vf w = -p vi extdV= . - pext vi dV = - pext ( vf - vi ) = -pextDV fi {Irreversible expansion if pext pgas That is if, pgas = nRT/V pexternal }. Example 2 : dV 0, but p const and T = const: pext= pgas = nRT (Called a reversible process.). V. dV. w = - nRT. V. vf vf dV dV. w = - vi nRT. V. = - nRT vi V. [Remembering that f(x) dx is the w = - nRT ln ( vf / vi ) area under f(x) in a plot of f(x) vs x, w = - pdV is the area under p in a plot of p vs V.]. P, V not const but PV = nRT = const (Isothermal change).

3 {Reversible isothermal expansion because pext= pgas }. Graphical representation of pext dV. P Expansion At constantPres- P1 sure pext=P2. no work P= nRT/V. nRT/V. Isothermal reversible P2 expansion shaded area = -w V. Vi = V1 Vf = V2. P pext= nRT/V PV=const is a hyperbola Compare the shaded constant area in the plot above P1 P = nRT/V. nRT/V. to the shaded area in the PV=. const plot for a reversible P2. isothermal expansion with shaded area = -w pext= pgas = nRT/V V. Vi = V1 Vf = V2. Work done is NOT independent of path : Change the State of a gas two different ways: Consider n moles of an **ideal** gas Initial condition: Ti = 300 K, Vi = 2 liter, pi = 2 atm.

4 Final condition: Tf = 300 K, Vf = 1 liter, pf = 4 atm. Path 1 consists of two steps: DV 0 for Step 1 : 2 atm, 2 l, 300K cool at 2 atm, 1 l, 150K this step const -p compress Step 2: Warm at constant V: 2 atm, 1 liter, 150 K DV=0 for 4 atm, 1 liter, 300 K. this step w = - pext ( Vf - Vi ) for the first step, pext = const = 2 atm w = - 2 atm ( 1 - 2 ) l = 2 l -atm w = 0 for 2nd step since V = const wtot = 2 l -atm Path 2 is a single step reversible isothermal compression: 2 atm, 2 l, 300K 4 atm, 1 l, 300K (T constant) pext=pgas=. nRT/V= p vf vf vf dV dV. w = - vi p dV = - nRT.

5 Vi V. = - nRT vi V. w = - nRT ln ( vf / vi ) = -nRT ln ( 1/2 ). Since nRT = const = PV = 4 l-atm . w = -4 l-atm ( ln 1/2 ) = ( .693 ) 4 l-atm = l-atm Compare to w for path 1: w = 2 l-atm w for two different paths between same initial and fianl states is NOT the same. Work is NOT a state Function! Heat : Just as work is a form of energy, heat is also a form of energy. Heat is energy which can flow between bodies that are in thermal contact. In general heat can be converted to work and work to heat -- can exchange the various energy forms. Heat is also NOT a state function.

6 The heat change occurring when a system changes state very definitely depends on the path. Can prove by doing experiments, or (for **ideal** gases) can use heat capacities to determine heat changes by different paths. The First Law of Thermodynamics I) Energy is a state function for any system : Ea Path a State 1 State 2. Eb Path b Ea and Eb are both for going from 1 2. If E not a state function then: DEa DEb Suppose DEa > DEb - now go from state 1 to state 2 along path a, then return to 1 along path b. Energy change = DE = DEa - DEb DE > 0. Have returned system to its original state and created energy.

7 Experimentally find no situation in which energy is created, therefore, DEa = DEb and energy is a state function. No one has made a perpetual motion machine of 1st kind. The First Law The energy increase of a system in going between two states equals the heat added to the system plus the work done on the system. DE = q+w (Here is where choice of sign for w is made). dE = dq + dw q > 0 for heat added to the system w > 0 for work done on the system (dV < 0). dw = -pextdV (w < 0 is work done by system, dV > 0). Totally empirical law. The result of observations in many, many experiments.

8 DE is a state function independent of the path. q and w are NOT state functions and do depend on the path used to effect the change between the two states of the system. Taking a system over different paths results in same DE but different q, w: qa , wa 1 2. qb , w b DE = E2 - E1. qc , wc Same for Paths a, b, c qa, qb, qc all different, wa, wb, wc all different, but qa + wa = qb + wb = qc + wc = DE = E2 E1. Measurements of DE. Suppose we want to measure DE for the following change : Initial State and system: O2 and N2 gas at 25 C and P(O2) = P(N2) = 1 atm. (1 mole each).

9 Final State : 2 moles NO at 25oC, 1 atm. (This is really a conversion of energy stored in the chemical bonds of O2 and N2 into stored chemical energy in the NO bond.). We know DE = q + w a) What is w? 1st let us carry the change above out at constant volume : N2 + O2 2NO. Then no mechanical work is done by the gases as they react to form NO because they are not coupled to the world --- no force moving through a distance --- nothing moves w = 0. DE = qv Change in energy for a chemical reaction carried out at constant volume is directly equal to the heat evolved or absorbed.

10 If qv > 0 then DE > 0 and energy or heat is absorbed by the system. This is called an endoergic reaction. If qv < 0 then DE < 0 and energy or heat is evolved by the system. This is called an exoergic reaction. Can we find or define a new state function which is equal to the heat evolved by a system undergoing a change at constant pressure rather than constant volume? is there a state function = qp? Yes! H E + pV will have this property Note E, p, V are state fcts. \ H must also be a state fct. Let us prove DH = qp : (for changes carried out at constant p). DE = q + w DH = DE + D (pV).