Cv dT = 0 - P dV = - (n R T / V) dV Adiabatic Processes An adiabatic process is one in which no heat is transferred between the system and the surroundings. Consider the adiabatic reversible expansion of 1.00 mole of a monatomic ideal gas initially at 300 K from 3.00 atm to 1.00 atm: 1.00 atm 3.00 atm ideal gas 1.00 mole adiabatic reversible 2.00 atm of water adiabatic wall 300 K What do you think happened to the temperature during this adiabatic expansion; did it stay the same, increase, decrease? To answer this question lets calculate the final temperature. Since the gas is ideal we can write: dE = Cv dT and then using the 1st law we have: Cv dT = dE = dq + dw using the conditions on the problem that this is a reversible adiabatic expansion of an ideal gas gives: Cv dT = 0 - P dV = - (n R T / V) dV Could you justify the developments in this last expression? 14.1

Cv Ti Tf (dT / T) = - n R Vi  Vf (dV / V) Separating variables and integrating yields: Cv Ti Tf (dT / T) = - n R Vi  Vf (dV / V) Cv ln (Tf / Ti) = - n R ln (Vf / Vi) = - n R ln [(Tf / Pf) / (Ti / Pi)] It is common to express these relations in terms of the heat capacity ratio, g, the ratio of the constant pressure heat capacity to the constant volume heat capacity: g = Cp / Cv ln (Tf / Ti) = ln [(Tf / Pf) / (Ti / Pi)]- n R / Cv = ln [(Tf / Pf) / (Ti / Pi)]+ (Cv - Cp) / Cv = ln [(Tf / Pf) / (Ti / Pi)]+ (1 - g ) After some rearrangement: Tf / Ti = ( Pf / Pi ) + (1 - 1/g ) Could you derive this result? Similarly expressions can be developed relating inital and final temperatures and volumes and inital and final pressures and volumes: Tf / Ti = ( Vf / Vi ) + (1 - g ) Pf Vfg = Pi Vig Could you derive these equations? 14.2

g = Cp / Cv = (n 5/2 R) / (n 3/2 R) = 5/3 The final temperature in the adiabatic reversible expansion of this ideal gas can now be calculated? g = Cp / Cv = (n 5/2 R) / (n 3/2 R) = 5/3 Tf = Ti ( Pf / Pi ) + (1 - 1/g ) = (300 K) (1.00 atm / 3.00 atm)(1 - 1/ 5/3) = 193 K Temperatures in adiabatic expansions of gases always decrease (there is one exception). Why does the temperature fall? We could now calculate the change in internal energy for this expansion: DE = 300 K  193 K CV dT = 300 K  193 K n (3/2 R) dT = (1.00 mole) 3/2 (8.314 J/mol K) (193 K - 300 K) = - 1.33 x 10+3 J Why did the internal energy decrease during this expansion? What was the work done in this expansion? Can you calculate the enthalpy change for this expansion? How would you expect the temperature to change in an adiabatic compression? Can you think of any examples of this? 14.3

Cv dT = 0 - Pext dV = - Pext dV Suppose the same expansion is carried out irreversibly by letting 1.00 mole of a monatomic ideal gas initially at 300 K expand from 3.00 atm to 1.00 atm against a constant external pressure of 1.00 atm: 1.00 atm 3.00 atm ideal gas 1.00 mole adiabatic irreversible adiabatic wall 300 K Once again we need to calculate the final temperature and we begin as in the previous example: Cv dT = dE = dq + dw but now, since the expansion is irreversible dw is developed differently: Cv dT = 0 - Pext dV = - Pext dV integrating both sides between the initial and final states gives: Cv ( Tf - Ti ) = - Pext (Vf - Vi) = - Pext ( n R Tf / Pf - n R Ti / Pi ) = - n R Pext ( Tf / Pf - Ti / Pi ) 14.4

Cv ( Tf - Ti ) = - n R Pext ( Tf / Pf - Ti / Pi ) Substituting into this equation allows the final temperature to be calculated: Cv ( Tf - Ti ) = - n R Pext ( Tf / Pf - Ti / Pi ) n 3/2 R ( Tf - Ti ) = - n R Pext ( Tf / Pf - Ti / Pi ) 3/2 (Tf - 300 K) = - (1.00 atm) (Tf / 1.00 atm - 300 K / 3.00 atm) Tf = 220 K Why is the temperature drop less in the irreversible expansion than in the similar reversible expansion considered just previously? Calculate the work done, internal energy change, and enthalpy change in this irreversible expansion. 14.5

Note the frost coating the pipe cooled by the expanding gas. Most expansions of real gases from a higher to a lower pressure are to some degree adiabatic (primarily because gases are poor thermal conductors) and as a result gases generally cool during these expansions. This effect often referred to as Joule cooling is the basis of most refrigeration involving the mechanical compression and subsequent expansion of some working fluid. The picture below was sent to me by a former “P-Chem” student, Blane Thingelstad, who was working for Amerada Hess: As Blane described the picture, “This is an offshore well in the Gulf of Mexico, it was recently completed in a shallow sand and had some high pressures. Because the sand is shallow the temperature is not very high. The point where the pipe turns to ice the gas undergoes a 1100 psi pressure drop.” Note the frost coating the pipe cooled by the expanding gas. 14.6

Coupled Adiabatic Expansion Bonus Problem This is a bonus problem worth 20 points. The problem is due in class one week from the date on which is assigned and will be graded on the answer only. Note this is a particularly challenging problem! Consider the piston cylinder arrangement diagramed below: 10.00 atm & 25.0 oC 0.20 atm & 25.0 oC The cylinder walls and piston are adiabatic walls and no heat can pass through them. The piston is massless and frictionless. 1.000 mol of monatomic ideal gas is trapped on each side of the piston. Initially the gas on the left side of the piston is at 25.0 oC and 10.00 atm, while the gas on the right side is at 25.0 oC and 0.2000 atm. Initially the piston is held in place by stops. When the stops holding the piston are removed the gas trapped on the left expands adiabatically, while the gas trapped on the right is adiabatically compressed. In the above drawings the positions of the piston are not correct and are only drawn where they are for visual clarity. Calculate the final temperatures, pressures, and volumes for the gas on both sides of the piston.

When a warm air parcel rises, it expands as it encounters lower pressure in the upper atmosphere. Since air is a poor conductor, this expansion approximates an adiabatic expansion and the warm air parcel cools as it rises: earth’s surface Ps & Ts = pressure & temperature at the earth’s surface P & T = pressure & temperature at height, h, above the earth’s surface If we make the reasonable assumptions that the expansion is reversible and that air can be treated as an ideal gas, then we can start to analyze this situation with the equation we developed earlier for the reversible adiabatic expansion of an ideal gas: T / Ts = ( P / Ps ) + (1 - 1/g ) = ( P / Ps ) + (1 - Cv / Cp ) = ( P / Ps ) + R / Cp We can differentiate this expression to get the rate of change of temperature with pressure: dT / dP = Ts ( R / Cp ) ( 1 / Ps ) ( P / Ps ) + (R / Cp ) - 1 = Ts ( R / Cp ) ( 1 / Ps ) ( P / Ps ) + (R / Cp ) ( P / Ps ) - 1 = Ts ( R / Cp ) ( 1 / Ps ) ( P / Ps ) + (R / Cp ) ( Ps / P ) = ( R / P Cp ) [ Ts ( P / Ps ) + (R / Cp ) ] = ( R / P Cp ) T (see the starting expression) 14.7

dP / dh = - r g = - ( P Mavg / R T ) g Since air is a fluid we can differentiate the hydrostatic equation: P = - r g h where r is the density of the fluid and g is gravitational acceleration, to obtain an expression for how the pressure of an ideal gas varies with height: dP / dh = - r g = - ( P Mavg / R T ) g Remember that Mavg is the average molecular weight of the ideal gas. We can multiply the rate of change of temperature with pressure and the rate of change of pressure with height to obtain the adiabatic lapse rate which describes how temperature varies with height in the atmosphere: dT / dh = (dT / dP) (dP / dh) = (R T / Cp P) ( - P M g / R T ) = - Mavg g / Cp The rate at which temperature falls with altitude in dry air is given by this expression as: dT/dh = - (0.0288 kg / mole) (9.808x10-3 km/sec2) (1 oC / 1 K) (29.1 J / mole K) [(1 kg m2 / sec2) / 1 J] (1 km / 103 m)2 = - 9.7 oC / km Is the adiabatic lapse rate in humid air greater, less, or the same as in dry air? This is the rate at which temperature falls in the troposphere. In the stratosphere exothermic chemical reactions initiated by the absorption of ultraviolet radiation cause the temperature to rise. The elevation in the atmosphere where this temperature reversal occurs marks the boundary between the troposphere and stratosphere and occurs roughly 10 km above the earth’s surface. 14.8

These temperature changes define the terminology used to describe the structure of the atmosphere: This figure is taken from the website developed by John P. Stimac, Geology/Geography faculty member at Eastern Illinois University for his course ESC 1400G, Weather and Climate http://www.ux1.eiu.edu/~cfjps/1400/atmos_struct.html .

The radiosonde (balloon) temperature data versus altitude shown below was taken on October 1st, 1994, at Eureka in the NWT as part of the Atmospheric Environment Canada's Beaufort and Artic Storms Experiment to better predict the weather in the Beaufort Sea and surrounding Artic. 14.9