Chapter 2 : fundamental ideas

1. Chapter 2 : fundamental ideas Energy is a fundamental quantity in thermodynamics. The First Law: energy can neither be created nor destroyed. In Classical Thermodynamics energy is divided into two parts: heat and work: ΔU = q + w . Heat and work are transfers of energy to/from the system Open systems exchange both matter and energy with the surroundings. Closed systems can exchange energy with the surroundings, but not matter. CHEM 471: Physical Chemistry Isolated systems have no exchange of any kind —neither matter nor energy — with the surroundings.

2. Chapter 2: work against a constant force Work = external force × displacement. Both force and displacement are vectors. w = (xFext, x + yFext, y + zFext, z) w = Fext . r Fextis the force we apply to the system to displace it against some internal force F applied by the surroundings. Example: The acceleration due to gravity near the earth’s surface doesn’t change much with height. What is the work done in carrying a 20 kg television up 3 flights of dorm stairs (about 10 m)? First: let’s define upwards as positive. So the distance is positive, the gravitational force F is negative, and the external force is positive. CHEM 471: Physical Chemistry F = mg = 20 kg × –9.80665 m/s2 = –196.13 N w = Fext.r = –F.r = –(–196.13 N) × 10 m = 1961.3 J Sign convention Work done on the system (in this case, the TV) is positive. Work done by the system (if the TV falls downstairs) is negative.

3. Chapter 2: example: work against a varying force Example: Hooke’s law says the force exerted by a spring is proportional to the displacement from its equilibrium distance. Fx = –k(x – x0) As we stretch or compress the spring, the force changes. We therefore have to replace w = Fext . r by dw = Fext . dr Since force and displacement are both along the x direction, we get dw = Fext, xdx. Integrating... CHEM 471: Physical Chemistry

4. Chapter 2: example: work against a varying force So if we stretch the spring from length x1 to x2 Changing variables: x′ = x – x0 and so dx′ = dx CHEM 471: Physical Chemistry

5. Chapter 2: practical example: MFM of DNA Single stranded DNA Simplified diagram of a molecular force microscope (MFM). Exercise: Between B and C, the force needed to stretch the DNA is 65 pN. What work is needed to stretch the DNA by 0.3 nm (i.e. break 1 b.p. of DNA)? Double stranded DNA CHEM 471: Physical Chemistry Answer: 1 nm = 10–9m, and 1 pN = 10–12 N. w = 65 × 10–12 N × 0.3 × 10–9 m = 1.95 × 10–20 J

6. Chapter 2: practical example: MFM of bacteriorhodopsin Exercise: If the force required to extract the first pair of alpha helices in the example is given by (Fext/pN) = 20(x/nm)+ (x/nm)2, up to an extension of 10 nm, what is the work needed to extract the helices? Answer: First of all, let's convert to SI. 1 nm = 10–9m, and 1 pN = 10–12 N. (Fext/pN) × (1012 pN/N) = 20(x/nm) × (109nm/m) + (x/nm)2 × (109nm/m)2 ⇒ 1012Fext = 2 ×1010 x + 1018x2 ⇒Fext = 2 ×10–2 x + 106x2 CHEM 471: Physical Chemistry = (10–18 + 3.333 × 10–19)J = 1.333 × 10–18J

7. Chapter 2: practical example: MFM of titin Titin is the protein that maintains muscle equilibrium length Reversible Unfolding of Individual Titin Immunoglobulin Domains by AFM Matthias Rief, Mathias Gautel, Filipp Oesterhelt, Julio M. Fernandez, Hermann E. Gaub Science 16 May 1997: Vol. 276. no. 5315, pp. 1109 - 1111 CHEM 471: Physical Chemistry Work done in unfolding the Igg-like domain is the area under the curve!

8. Introduction: for Friday… Read pages 19-23! Useful problems: 3, 4, 19 CHEM 471: Physical Chemistry