9.6 Work A change in the state of the system is usually accompanied by the transfer of heat and or work between the system and the surroundings. Work and heat are NOT properties of the system, but are only transfered between the system and the surroundings during a change in state. Can you make the statement “the system has 834 Joules of work”? Work can take on several different forms: hydrostatic or pressure-volume (PV) work done as the system expands or is compressed gravitational work associated with changing the position of a mass in a gravitational field electrical work associated with changing the position of a charge in an electrical field surface expansion work associated with expanding the system’s surface or tension work associated with stretching a system under tension

9.7 The work can be obtained by integrating over the differential change in the work: w =  dw While the sign convention for heat transfer is universally accepted, the sign convention for work varies. We will arbitrarily adopt a sign convention for work that is the same as the one adopted for heat: Work done by the system on the surroundings, i.e., work transferred from the system to the surroundings, is defined as negative: Consider the work done by a gas trapped in a cylinder by a massless frictionless piston (a fiction we will frequently use) of area A, as the gas expands against an opposing external pressure: P ext xixi xfxf x piston of area A

9.8 which in turn can be represented by the dot product of the external force vector and the distance vector: w =  F ext dx =  F ext dx cos (  ) = -  F ext dx Can you explain where the minus sign in the above equation originated (I’m looking for more than the cos of  is negative)? Since the external force is equal to the external pressure times the area of the piston on which it is acting: w = -  P ext A dx = -  P ext dV For an expansion, dV > 0 and the expansion work will be negative. Note that the formula just derived will be the starting point for all our hydrostatic or PV work calculations! What is the sign on the work transferred when the surroundings does work on the system and, for example compresses the system? When you stretch a rubber band, is the work done on the rubber band positive or negative?

9.9 Changes of state or processes can be viewed as occuring reversibly or irreversibly. An irreversible process: is a real or actual process occurs in a finite number of steps requires a finite amount of time is one in which the system and surroundings are not in equilibrium A reversible process: requires an infinite number of steps occurs in an infinite amount of time can be reversed on an infinitesimal scale by reversing the direction of the variables driving the change is one in which the system and surroundings are in equilibrium at each point in the process is an ideal or limiting process that can never actually occur

9.10 Expansions against a constant external pressure are irreversible. Consider the isothermal irreversible expansion of 1.00 mole of an ideal gas from 3.00 atm to 2.00 atm at 300 K against a constant opposing pressure of 1.00 atm: 1.00 atm 3.00 atm 2.00 atm ideal gas 1.00 mole 300 K irreversible thermostated at 300 K To calculate the work done by the gas during this expansion: w = -  P ext dV = - P ext  dV = - P ext [ V 2 - V 1 ] Why could we factor the pressure out of the integral? Since we don’t know the initial and final volumes, we have to calculate them using the ideal gas law: w = - P ext [ nRT/P 2 - nRT/P 1 ] = - n R T P ext [1/P 2 - 1/P 1 ] = - (1.00 mole)(8.314 J/mole K)(300 K)[1/2.00 atm - 1/3.00 atm] = J Does the negative sign for the work make sense?

9.11 It is instructive to look at this expansion on a plot of pressure versus volume or PV diagram: Note that the expansion work done by the gas: w = - P ext [ V 2 - V 1 ] is just opposite in sign and equal in magnitude to the yellow shaded area on PV diagram. How would you change the constant external pressure to get the maximum work out of the gas in a single stage (single step) expansion? Could you design a multi-step (e.g., a 2 step) expansion to get even more work out of the gas?

9.12 We can model a reversible process with a quasistatic process in which the system and surroundings are just enough out of equilibrium to allow the process to occur. Consider the same expansion as before, but this time we will add enough water on top of the piston to add 2.00 atm of pressure to the 1.00 atm of atmospheric pressure that is already present. The pressure on both sides of the piston is 3.00 atm and the piston is in mechanical equilibrium and not moving. Now a molecule of water evaporates and the total external pressure is slighlty (infinitesimally) less than the pressure of the gas and the piston moves up slighlty. As the water continues to evaporate the gas will expand until it reaches its final pressure of 2.00 atm: 1.00 atm 3.00 atm 2.00 atm ideal gas 1.00 mole 300 K reversible thermostated at 300 K 2.00 atm of water Note that the pressure of the gas (the system) is changing throughout the expansion and is equal to the external pressure at every point during the expansion.

9.13 To calculate the work done by the gas during this reversible expansion: w = -  P ext dV = -  P dV = -  nRT/V dV In this derivation: Why were we able to set P ext = P (the pressure of the gas or system)? Why did we not factor the pressure out of the integral, as we did in the calculation for the irreversible expansion? Why did we replace the pressure of the gas with nRT/V? Integrating the above expression gives: w = - nRT ln (V 2 /V 1 ) = - nRT ln (P 1 /P 2 ) = - (1.00 mole)(8.314 J/mole K)(300 K)ln (3.00 atm/2.00 atm) = x J Note that greater work is done in the reversible expansion than is done in the irreversible expansion between the same two states.

9.14 Examining this reversible expansion on a PV diagram: Once again the expansion work is the negative of the yellow shaded area, but in this reversible expansion the pressure opposing the expansion was always equal to the pressure of the gas and corresponded to points on the isotherm. Why does the reversible work represent the maximum possible work that can be achieved during the expansion (hint: a single step irreversible expansion against an external pressure of 3.50 atm or even 2.50 atm would result in a greater amount of work - why isn’t this possible)?

9.15 Is the work done during the cyclic change in state diagramed below positive or negative? initial and final state P V The amount of work done by the 1.00 mole of gas expanding isothermally at 300 K from 3.00 atm to 2.00 atm depended on whether the expansion was carried out irreversibly or reversibly: State functions are thermodynamic functions whose change during a change in state does not depend on the path taken between two states, but only depends on the difference in the values of these functions in the initial and final states. Is work a state function? You can drive to Whitehall from Butte via either Homestake or Pipestone passes. For this trip which of the following; mileage, elevation loss, and/or fuel consumption are state functions? (3.00 atm) (2.00 atm) w irreversible path = J w reversible path = x10 +3 J

moles of a gas that follows the van der Waals equation of state: P = n R T / (V - n b) - a n 2 / V 2 is isothermally and reversibly expanded from L to L at K. The van der Waals constants for this gas are a = L 2 atm/mole 2 and b = L/mole. Calculate the work in Joules done by this gas during this expansion.