1. 1 The Ideal Gas

2. 2 Ideal gas equation of state Property tables provide very accurate information about the properties. It is desirable to have simple relations among the properties that are sufficiently general and accurate. Any equation that relates P-v-T are called Equation of state. Property relations that involve other properties of a substance at equilibrium states are also referred to as equations of state. The simplest and best-known equation of state for substances in gas phase is the ideal-gas equation of state.

3. 3 Ideal gas equation of state R is the Gas constant R= R u / M R u is universal gas constant R u = 8.314 KJ/Kmol. K M is Molar Mass of the gas (molecular weight) The values of R and M are given in Table A-1 for several substance. The vapor phase of a substance is called gas when it is above the critical temperature. Vapor implies a gas that is not far from a state of condensation. In 1802, Charles and Lussac experimentally determined the following (Ideal gas equation of state): M is the mass of one mole (kmole) of substance in grams (kilograms)

4. 4 P-v-T relation for ideal gases The properties of an ideal gas for a fixed mass at two different states are related to each other by

5. 5 Ideal gas An ideal gas is an imaginary substance that obeys the P-v-T relation. The aforementioned relation approximates the behavior of real gas at low densities. At low pressure and high temperature, the density of a gas decreases, and the gas behaves like an ideal gas. In the range of practical interest, many familiar gases such as air, nitrogen, oxygen, hydrogen, helium, Aragon, neon, krypton, and even heavier gases such as carbon dioxide can be treated as ideal gases with negligible error (often less than 1 %). Dense gases such as water vapor in steam power plants and refrigerant vapor in refrigerators, however, should not be treated as ideal gases. Instead, the property tables should be used for these substances.

6. 6 Figure on the right shows the percentage of error involved in assuming steam to be an ideal gas, and the region where steam can be treated as an ideal gas. Is steam an Ideal Gas? It depends !!!!!

7. 7 Is Water vapor an ideal gas? At pressure below 10 KPa, water vapor can be treated as an ideal gas, regardless of its temperature, with negligible error (less than 0.1%). At higher pressures, however, the ideal-gas assumption yields unacceptable errors. In air-conditioning applications, where the pressure of the water vapor is very low (ideal gas relations can be used) In steam power plant applications, they should not be used

8. 8 Compressibility Factor Z Real gases deviate from ideal-gas behavior significantly at state near the saturation region and the critical point. This deviation at a given temperature and pressure can accurately accounted for by introducing a correction factor called the compressibility factor For real gases Z can be grater than or less than unity. Real Gas Z > 1 Z < 1 Ideal Gas Z = 1

9. 9 Compressibility Factor Z for Nitrogen

10. 10 Compressibility Factor Z for H 2 Similar charts have been prepared for different gases and they are found to be qualitatively similar. Can quantitative similarity be achieved?

11. 11 Generalized Compressibility chart This can be done if the coordinates are modified such that P R = P/P cr and T R = T/T cr where P R and T R are the reduced pressure and reduced temperature. The Z factor is the same for all gases at the same reduced pressure and temperature. This is called the principle of corresponding states.

12. 12 Observations B- It can be noticed that at many states the compressibility factor Z is approximately unity : A- The deviation of a gas from unity is greatest in the vicinity of the critical point. 1.At very low pressure (P R <<1), 2.At high temperatures (T R >2) except when P R >>1.

13. 13 When P and v or T and v are given, generalized chart can still be used Pseudo reduced specific volume Lines of constant v R are available on the compressibility chart (Figure A-30 pp. 867-869) 2- T R and v R 3-P R and v R v R is pseudo reduced specific volume defined as:

14. 14 Example 2-11 Determine the specific volume of refiregerant-134 at 1 MPa and 50  C, using: A. Thermodynamic tables B. The ideal gas low C. The generalized compressibility chart. Find the percentage error in the values obtained in B and C compared to the value obtained in part A.

15. 15 Equation of state Equation of State predicts the P-v-T behavior of a gas quite accurately

16. 16 Specific Heats

17. 17 Importance We know it takes more energy to warm up some materials than others For example, it takes about 10 times as much energy to warm up 1 kg of water, as it does to warm up the same mass of iron. it is desirable to have a property that will enable us to compare the energy storage capabilities for different substances.

18. 18 Specific Heat (heat capacity) Definition: It is the energy required to raise the temperature of a unit mass by one degree Units  kJ/(kg 0 C) or kJ/(kg K)  cal/(g 0 C) or cal/(g K)  Btu/(lb m 0 F) or Btu/(lb m R) Simple mathematical definition:

19. 19 Constant volume and Constant pressure specific heats, C v and C p C v can be viewed as energy required to raise the temperature of a unit mass by one degree as the volume is maintained constant. C p can be viewed as energy required to raise the temperature of a unit mass by one degree as the pressure is maintained constant. Cp > Cv

20. 20 Mathematical form of C v E=U+KE +PE d  dU  U But dE  mC v dT du  C v dT Consider constant volume system. Heat it from T 1 to T 2.

21. 21 We’ll worry about the math later, but… h includes the internal energy and the work required to expand the system boundaries Mathematical form of C p

22. 22  C p is always bigger than C v. This is because i  C p is always bigger than C v. This is because it takes more energy to warm up a constant pressure system due to the system boundaries expansion.  That is you need to provide the energy to Observations  increase the internal energy  do the work required to move the system boundary

23. 23 Observations (continued)  Both are expressed in terms of u or h, and T, which are properties and thus C v and C p are also properties.  Because they are properties, they are independent of the type of process!!

24. 24 Recall that : Hence, C v is at most function of T for an ideal gas. C v dependence on T for an ideal gas. Joule found experimentally that the internal energy of an ideal gas is a function of temperature only The partial derivative becomes ordinary derivative for an ideal gas..

25. 25 Recall that C p dependence on T for an ideal gas. Hence, C p is at most function of T for an ideal gas. The partial derivative becomes ordinary derivative for an ideal gas.

26. 26 Internal energy and Enthalpy as functions of Cv and Cp for an ideal gas.

27. 27 To carry out these integrations, we need Cv and Cp as functions of T. Analytical expressions are available in Table A-2c. In this table, Cp is given as Cp = a + bT + cT 2 + dT 3 The constants a, b, c, and d are tabulated for various gases.

28. 28 This is inconvenient!! Only do it if you really need to be very accurate!! Isn’t there an easier way? Method 1

29. 29 T 0 =0 K was chosen to be an arbitrary reference. This choice has no effect on The u and h data are given in KJ/kg for air. Other gases in KJ/Kmol. Method 2 These integrations of Eqs. 2-34 and 2-35 were tabulated in Table A-17 page 849.

30. 30 Method 3 The variation of specific heat for gases with complex molecules are higher and increase with temperature. The variation of specific heats is smooth and can be approximated as linear over small temperature interval (a few hundred degrees or less)

31. 31 Method 3 Assume Cp and Cv is constant over a short temperature range (a few hundred degrees or less). The constant specific heats are evaluated at the average temperature (T 1 +T 2 )/2. u 2 -u 1 =C v,av (T 2 -T 1 ) h 2 -h 1 =C p,av (T 2 -T 1 )

32. 32 Three Ways to Calculate  u

33. 33 Example Air at 300 K and 200 kpa is heated at constant pressure to 600 K. Determine the change in the internal energy of air per unit mass using:  Data from air tables (Table A-17)  The functional form of the specific heat (Table A-2c)  The average specific heat value (Table A- 2b).

34. 34 The previous relations are not restricted to any kind of process. The presence of constant volume specific heat in an equation should not lead to the concept that this equation is valid only for constant volume process. The constant volume or constant pressure part of the name defines only how they are measured for each substance (see figure). Once we have Cv or Cp as function of T, we can perform the integration for any process. Important Observation

35. 35 Recall that h=u+RT  dh = du + RdT  C p dT=C v dT+ RdT  C p =C v + R Specific heat relations of an Ideal Gas C p is modeled in the Appendix A-2(c) as a function of temperature – so you could calculate dh, but what if you want to calculate du? You’d need C v. You’d need C v. There is no corresponding Cv table !!

36. 36 Specific Heat Ratio k also varies with temperature, but this variation is very mild. k = 1.4 for diatomic gases (like air) k = 1.667 for noble gases

37. 37 The specific volume of incompressible substances remain constant during a process. The Cv and Cp values of incompressible substances are identical and are denoted by C. C p = C v = C Specific heats of Solids and Liquids

38. 38 Internal energy of Solids and Liquids

39. 39 But dv is 0 if the system is incompressible 0 Small for solids Enthalpy of Solids

40. 40 Enthalpy of Liquids  We have two cases: Constant pressure process, Constant temperature process,