1. H. Saibi January 20 th, 2016

2.  The internal Energy of an Ideal Gas  Work and the PV Diagram for a Gas  Heat capacities of Gases  Heat capacities of Solids  The Quasi-Static Adiabatic Compression of a Gas.

3.  The translational kinetic energy of the molecules in an ideal gas is related t the absolute temperature T by equation: Where n is the number of moles of gas and R is the universal gas constant. If the internal energy of a gas is just this translational kinetic energy, then E int =K, and: ©2008 by W.H. Freeman and Company Figure 1: Free expansion of a gas. When the stopcock on the gas is opened, the gas expands rapidly into the evacuated chamber. Because no work is done on the gas and the whole system is thermally insulated, the initial and final internal energies of the gas are equal. (1)

4. ©2008 by W.H. Freeman and Company  Quasi-Static Processes The magnitude of the force F exerted by the gas on the piston is PA, where A is the area of the piston and P is the gas pressure. As the piston moves a small distance dx, the work done by the gas on the piston is: Where dV=Adx is the increase in the volume of the gas. During the expansion the piston exerts a force of magnitude PA on the gas, but opposite in direction to the force of the gas on the piston. Thus, work done by the piston on the gas is just the negative of the work done by the gas: Note that for an expansion, dV is positive, the gas does work on the piston, so dW on gas is negative, and for a compression, dV is negative, work is done on the gas, so dW on gas is positive. The work done on the gas during an expansion or a compression from a volume of V i to a volume of V f is: To calculate this work, we need to know how the pressure varies during the expansion or compression. The various possibilities can be illustrated most easily using a PV diagram. Figure 2: Gas confined in a thermally insulated cylinder with a movable piston. If the piston moves a distance dx, the volume of the gas changes by dV=Adx. The work done by the gas is PAdx=PdV, where P is the pressure. (2) (3) (4) Work done on a Gas

5. ©2008 by W.H. Freeman and Company  For a volume change of  V (  V is negative for a compression), we have  Because pressures are often given in atmospheres and volumes are often given in liters, it is convenient to have a conversion factor between liter-atmospheres and joules: Figure 3: Each point on a PV diagram, such as (Po, Vo), represents a particular state of the gas. The horizontal line represents states with a constant pressure Po. The shaded area, Po  V , represents the work done on the gas as it is compressed an amount  V . (5)

6. The work done on the gas along the constant-volume (vertical) part of path A is zero; along the constant pressure (horizontal) part of the path A, it is We can calculate the work done on the gas along path C by using P=nRT/V. Hence, the work done on the gas as it is compressed from V i to V f along path C is: Because T is constant for an isothermal process, we can factor it from the integral. We then have Figure 4: Three paths on PV diagrams connecting an initial state (Pi, Vi) and a final state (Pf, Vf). The corresponding shaded area indicates the work done on the gas along each path. (6) Work done on Gas during Isothermal Compression

7.  Calculating Work done by an ideal gas during a constrained Quasi-Static Process o Picture: The increment of work done by a gas is equal to the pressure times the increment of volume. That is,. It follows that. The constraint dictates how to evaluate this integral.  SOLVE  1- If the volume V is constant, then dV equal zero and  2- If the pressure P is constant, then  3- If the temperature T is constant, then and  4- If no energy is transferred to or from the gas via heat, then (Quasi-Static Adiabatic Compression of a gas)  Check: if the volume increases then the W by must be positive, and vice versa.

8.  Writing Q v for the amount of heat transfer into the gas at constant volume, we have  Because W=0, we have from the first law of thermodynamics,  Taking the limit as  T approaches zero, we obtain  The heat capacity at constant volume is the rate of change of the internal energy with temperature. Because E int and T are both state functions, Equations 7a and 7b hold for any process.  Now let us calculate the difference C p -C v for an ideal gas. From the definition of C p, the amount of heat transfer into the gas at constant pressure is:  From the first law of thermodynamics,  Then  For infinitesimal changes, this becomes  Using Equation 7a for dE int, we obtain  The pressure, volume, and temperature for an ideal gas are related by  Taking the differentials of both sides of the ideal-gas law, we obtain  For a constant-pressure process dP=0, so  Substituting nRdT for PdV in Eq.8 gives  Therefore, and Figure 5: Heat is absorbed and the pressure remains constant. The gas expands, thus doing work on the piston. (7a) (7b) (8) (9)

9. ©2008 by W.H. Freeman and Company  if the internal energy of a gas consists of translational kinetic energy only, we have.  The heat capacities are then and Figure 6: The piston is held in place by pins. Heat is absorbed at constant volume, so no work is done and all the heat is transferred into the internal energy of the gas. (10) (11) (12) Cp for an ideal monatomic gas Cv for an ideal monatomic gas

10. ©2008 by W.H. Freeman and Company

11.  The kinetic energy of a diatomic molecule is therefore  The total internal energy of n moles of such a gas is then  And the heat capacity at constant volume is (13) (14) Figure 7: Rigid-dumbbell model of a diatomic molecule.

12. ©2008 by W.H. Freeman and Company  Most solids have molar heat capacities approximately equal to 3R:  The total energy of an atom in a solid is:  the internal energy of a mole of a solid is: Which means that c’ is equal to 3R. Figure 8: Model of a solid in which the atoms are connected to each other by springs. The internal energy of the molecule consists of the kinetic and potential energies of vibration. (15) (16)

13.  The first law of thermodynamics gives Where we have used dE int =C v dT (Equation 7a), dQ m =0 (the process is adiabatic), and dW on =-PdV (Equation 4). Then, substituting for P using P=nRT/V, we obtain: Separating variables by dividing both sides by TC v, we obtain Integration gives Simplifying, Thus, Where the constants in the two preceding equations are not the same constant. Equation 18 can be rewritten by noting that C p -C v =nR, so Where  is the ratio of the heat capacities: Therefore, We can eliminate T from Equation 21 using PV=nRT. We then have or Figure 9: Quasi-static adiabatic compression of an ideal gas. The dashed lines are the isotherms for the initial and final temperatures. The curve connecting the initial and final states of the adiabatic compression is steeper than the isotherms because the temperature increases during the compression. Quasi-Static Adiabatic Process (17) (18) (19) (20) (21) (22)

14. Clouds from its rising moist air cools due to adiabatic expansion of the air. Cooling causes water vapor to condense into liquid droplets. (Will and Deni McIntyre/Photo Researchers.)  Equation 22 relates P and V for adiabatic expansions and compressions. Solving PV=nRT (the ideal gas equation) for V, then substituting the resulting expression for V into Equation 22, and then simplifying, gives  The work done on the gas in an adiabatic compression can be calculated from the first law of thermodynamics: Because, we have Then Where we have assumed that C v is constant. We note that the work done on the gas depends only on the change in the temperature of the gas. During an adiabatic compression, work is done on the gas, and its thermal energy and temperature increase. During a quasi-static adiabatic expansion, work is done by the gas, and the internal energy and temperature decrease. We can use the ideal-gas law to write Equation 24 in terms of the initial and final values of the pressure and volume. If Ti is the initial and Tf is the final temperature, we have for the work done Using PV=nRT, we obtain Using Eq.19 to simplify this expression, we have Adiabatic Work on Ideal Gas (23) (24) (25)