21.1 Carnot Cycle In 1824 Sadi Carnot, a French engineer published a treatise in which he abstracted the essential features of heat engines. We will analyze a four step reversible Carnot cycle applied to an ideal gas: Step 1: The ideal gas in thermal equilibrium with a high temperature thermal reservoir at T H expands reversibly and isothermally from V 1 to V 2. During this expansion q H of heat is transferred from the high temperature reservoir to the gas to keep the gas isothermal during the expansion. The reversible work done by the ideal gas during this isothermal expansion is: w 1 = - n R T H ln (V 2 / V 1 ) = - q H Why in this step is the reversible work the negative of the heat transfered?

21.2 Sadi Carnot was born in 1796 the son of Lazare Carnot, a noted French mathematician, engineer, and general. He began his university studies at age 16 at the Ecole Polytechnique in Paris. After two years Sadi left the Ecole to work in the Corps of Engineers. Unchallenged in the Corps of Engineers he spent some time analyzing heat engines, about which little was known of their theory of operation. In 1824 he published his only work Reflexions sur la puissance motrice du feu (Reflections on the Motive Power of Fire), which was mostly ingnored by his contemporaries. This analysis set upper limits on the efficiency of steam engines and paved the way for the development of the 2nd Law of Thermodynamics. He died in 1832 at the age of 46, the victim of a cholera epidemic. The figure is taken from a ChemTeam WEB page maintained by John L. Park at dbhs.wvusd.k12.ca.us/Gallery/GalleryMenu.html

21.3 Step 2: The ideal gas expands reversibly and adiabatically from T H to T L. The reversible work done by the ideal gas during this adiabatic expansion is: w 2 = n C V (T L - T H ) Could you derive this expression for the work done in the adiabatic step? Step 3: The ideal gas in thermal equilibrium with a low temperature thermal reservoir at T L is compressed reversibly and isothermally from V 3 to V 4. During this expansion q L of heat is transferred from the ideal gas to the low temperature reservoir to keep the gas isothermal during the compression. The reversible work done by the ideal gas during the isothermal expansion is: w 3 = - n R T L ln (V 4 / V 3 ) = - q L Step 4: The ideal gas is reversibly and adiabatically compressed back to its initial state to complete the cycle. The reversible work done by the ideal gas during this adiabatic compression is: w 4 = n C V (T H - T L ) The net work done in one complete cycle is the sum of the work done in each of the cycle steps: w net = w 1 + w 2 + w 3 + w 4 = - n R T H ln (V 2 / V 1 ) + n C V (T L - T H ) - n R T L ln (V 4 / V 3 ) + n C V (T H - T L ) The work done in the adiabatic steps cancels to give: w net = - n R T H ln (V 2 / V 1 ) - n R T L ln (V 4 / V 3 )

21.4 We can equate relations that we derived earlier for the adiabatic steps (see the section on adiabatic processes): C V ln (T H / T L ) = R ln (V 3 / V 2 ) = R ln (V 4 / V 1 ) to simplify the equation for the net work even more: w net = - n R T H ln (V 2 / V 1 ) - n R T L ln (V 1 / V 2 ) = - n R ln (V 2 / V 1 ) (T H - T L ) = - n R T H ln (V 2 / V 1 ) (T H - T L ) / T H = + w 1 (T H - T L ) / T H = - q H (T H - T L ) / T H Could you justify the steps in the above derivation? Remembering the definition of efficiency of a heat engine we have for the efficiency of our reversible cyclic Carnot engine: efficiency = - w net / q H = (T H - T L ) / T H Note that the efficiency of the reversible heat engine does not depend on the nature of the working fluid, but only depends on the absolute temperatures between which the engine is operating. While we used an ideal gas as the working fluid to derive this result, the result is general and applies to any reversible heat engine and in fact represents the maximum efficiency possible for any engine. Which is more efficient, your refrigerator or your freezer? Which has the greater efficiency, an engine operating between 80 o C and 90 o C or an engine operating between 90 o C and 100 o C? Could you sketch how a single reversible Carnot cycle would appear on a TS diagram, a plot of temperature versus entropy?

21.5 A temperature scale is defined by at a minimum two temperatures and a interpolatable scale between the values of these two temperatures. The temperatures that define the scale are known as fixed points in temperature. They are associated with highly reproducible systems and are determined by international agreement. The efficiency of a heat engine will be a maximum, if the low temperature reservoir is at 0 degrees on whatever absolute temperature scale we choose. The results of the Carnot cycle therefore define the thermodynamic scale of temperature in which the lowest fixed point in temperature is absolute zero. All that remains to complete the definition of the thermodynamic temperature scale is to choose one other fixed point in temperature. If we choose the other fixed point to be the triple point of water at which air free ice, liquid water, and water vapor are all in equilibrium at a temperature of K, then the thermodynamic and Kelvin scales of temperature agree with each other.

21.6 The efficiency of a single reversible Carnot cycle can also be written as: efficiency = - w net / q H = + q net / q H = (q H + q L ) / q H Equating this expression for efficiency with the one we developed previously: (q H + q L ) / q H = (T H - T L ) / T H gives after rearrangement: q H / T H + q L / T L = 0

21.7 By extending this result for the single reversible Carnot cycle to any general reversible cyclic process an analytic expression for entropy can be developed: If we slice the general reversible cycle with a series of adiabats and isotherms the general cyclic process can be represented as a sum of single reversible Carnot cycles:  ( q H / T H + q L / T L ) i = 0 Why is this sum equal to zero? The sum is over all of the indvidual single reversible Carnot cycles that are at least partly in the general reversible cycle. Notice that two adjacent single reversible Carnot cycles will share an edge (see the colored arrows in the above sketch) and that in the Carnot analysis of each of these adjacent cycles that edge will be traversed in opposite directions and the corresponding terms from the two Carnot cycles will cancel in the sum. The only terms which will not cancel are for those edges that are outside the general cycle (see the edge labeled with the green arrow). isotherms adiabats general reversible cycle

21.8 As we let the number of adiabats and isotherms slicing the general reversible cyclic process approach infinity: 1) the temperatures of the high, T H, and low, T L, thermal reservoirs and the heats withdrawn, q H, and dumped, q L, into them, respectively, approach each other for each individual single reversible Carnot cycle and we can drop the subscripts distinguishing between them and: 2)the sum over all the single reversible Carnot cycles is replaced by an integral, which just represents an infinite sum over differential quantities: lim  (q H / T H + q L / T L ) -->  dq rev / T = 0 T H --> T L --> T q H + q L --> dq The rev subscript on dq rev is a reminder that these results apply to a reversible process. Why does q H + q L approach dq rev, the differential difference in heat transfer for the single reversible Carnot cycle under consideration?

21.9 Remembering that if the cyclic integral of some function is equal to zero, then that function is a state function, we define an analytical expression for the differential change in the state function entropy as: dS  dq rev / T