Distinguish between. and non-thermal energy sources.
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1 Distinguish between System & Surroundings We also distinguish between thermal We also distinguish between thermal and non-thermal energy sources.
2 P Work The gas in the cylinder is the system How much work is performed on the gas in a cylinder (system) when compressing the gas? Generically: w=-f dist Expansion: dw=-p ex d 1 2 w P d ex
3 Expansion when P=0, w=-p d ex 2 work P d P (at constant pressure) ex ex 1 Compress Gas Δ = final - initial is negative Work is positive, work is done on system System gains energy Expand Gas = final - initial is positive Work is negative, work is done by system System loses energy
4 Reversible Processes A process effected by infinitesimal changes in a variable. Proceeds through h a sequence of equilibrium i states t One always remains on the surface of an equation of state Idealized process takes infinitely long to carry out The system remains in equilibrium throughout the process and can be reversed by an infinitesimal change in the variable. The work done by the system in a reversible expansion from A to B is the maximum work that the system can perform in changing from A to B
5 Reversible and Irreversible Work The path is a portion on of the eq. of state surface 1 The path is not Completely on the eq. of state surface 2 Reversible: system and surroundings in equilibrium P P P int ex Irreversible: system and surroundings not in equilibrium w P P P P P P ex
6 Reversible compression work with an ideal gas Reversible, so P ext = P sys = P For constant P we have already shown that w = -P ext Δ 2 nrt 2 w P d d 1 1 Isothermal: T= cons t P = nrt nrt ln ln nrt ln 2 2 w nrt d 2 A ibl di b ti (i l t d i t ) i A reversible adiabatic (insulated piston) compression requires is a bit more complex and will not be needed here.
7 Reversible adiabatic compression work with an ideal gas First law: du = dq + dw = dq - Pd (insulated gas conta ainer) P ext = P P = nrt 1 2 Adiabatic: dq = 0 So du + Pd = 0 2 nrt 2 w P d d 1 1 We might return to this case later, but for an ideal gas, we can derive an adiabatic equation of state: P P CP 5 where for an ideal gas, C 3 and w C T T 2 1
8 Heat Transactions du = dq + dw exp + dw e 0 0 At constant volume, dw exp =0 0, and for no additional work, such as electrical work, we have du=dq or U=q
9 Math for Heat Transactions U is a state function It depends only on state, t not on path to get there U = U final -U initial This means mathematically that du is an exact differential and U i f du For now, consider a system of constant composition. U can then be regarded d as a function of, T and P. Because there is an equation of state relating, T, and P, any two are sufficient to characterize U. So we could have U(P,), U(P,T) or U(,T).
10 Math for Heat Transactions So we could choose U(p,), U(pT) U(p,T) or U(,T). Exact differential review: F(x,y) F F df dy dx x y y Let us choose U = U(,T) When + d at cons t T, U changes to x U U' U d T
11 Math for Heat Transactions Or in general, U U U' U d dt T T For infinitesimal changes, U U du d dt T T Some terms are familiar: U Heat capacity at constant volume C U T TT T Internal pressure at constant temp
12 Heat Capacity U U du d dt T T () T C (T) If d = 0, then du = C (T)dT However, C (T) is approximately constant over small temperature changes and above room temperature so integrate both sides: U = C T Since constant volume: q =C T
13 Internal Energy (5) Many useful, general relationships are derived from manipulations of partial al derivatives, ves, but I will (mercifully) spare you more. Suffice it to say that U is best used for processes taking place at constant volume, with only P work: Then du = dq and U= U 2 U 1 = q The increase in internal energy of a system in a rigid container is thus equal to the heat q supplied to it. We would prefer a different state function for constant pressure processes, one that naturally accounts for P work on the surroundings: enthalpy.
14 Enthalpy Defined Enthalpy, H U + P dh du d P du Pd dp At Constant P, H = U + P U = q + w q = q P = U - w, and w = -P So q P = U + P H If, in addition, is constant, then U = H = q
15 Comparing H and U at constant P H = U + P A B C D 1. Reactions that do not involve gases 0 and H U 2. Reactions in which n gas = 0 0 and H H UU 3 Reactions in which n 0 3. Reactions in which n gas 0 0 and H U
16 ConcepTest #1 Which of the following reactions has the largest difference between H and U? A. NH 3 (g) + HCl (g) NH 4 Cl (s) B. CO (g) + Cl 2 (g) COCl 2 (g) C. ZnS (s) + 3/2 O 2 (g) ZnO (s) + SO 2 (g) D. ZnO (s) + CO (g) Zn (s) + CO 2 (g)
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