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CHAPTER 2

CHAPTER 2. PERFECT GAS (GAS UNGGUL). Definition Of Perfect Gases. Did you know, one important type of fluid that has many applications in thermodynamics is the type in which the working temperature of the fluid remains well above the critical temperature of the fluid?

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CHAPTER 2

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  1. CHAPTER 2 PERFECT GAS (GAS UNGGUL)

  2. Definition Of Perfect Gases Did you know, one important type of fluid that has many applications in thermodynamics is the type in which the working temperature of the fluid remains well above the critical temperature of the fluid? In this case, the fluid cannot be liquefied by an isothermal compression, i.e. if it is required to condense the fluid, then cooling of the fluid must first be carried out. In the simple treatment of such fluids, their behavior is likened to that a perfect gas. Although, strictly speaking, a perfect gas is an ideal which can never be realized in practice. The behavior of many ‘permanent’ gases, e.g. hydrogen, oxygen, air etc is very similar to the behavior of a perfect gas to a first approximation.

  3. Perfect Gases A perfect gas is a collection of particles that: Are in constant, random motion, Have no intermolecular attractions (which leads to elastic collisions in which no energy is exchanged or lost), Are considered to be volume-less points.

  4. Perfect Gases The principle properties used to define the state of a gaseous system and its SI units (Systems International are as follow respectively: Pressure (P), Pascal (Pa). Volume (V) , m3 for volume (although liters and cm3 are often substituted), Temperature (T). and the absolute scale of temperature or Kelvin (K).

  5. Boyle’s Law and Charles’ Law Two of the laws describing the behaviour of a perfect gas are Boyle’s Law and Charles’ Law.

  6. P P 1/V 1/V Boyle’s Law If the process is represented on a graph having axes of pressure P and volume V, the results will be as shown in Fig. 3.1-2. The curve is known as a rectangular hyperbola, having the mathematical equation xy = constant. The Boyle’s Law may be stated as follows: Provided the temperature T of a perfect gas remains constant, then volume, V of a given mass of gas is inversely proportional to the pressure P of the gas, i.e. P 1/V (as shown in Fig. 3.1-1), or P x V = constant if temperature remains constant. Figure 3.1-1 Graph P 1/V If a gas changes from state 1 to state 2 during an isothermal process, then P1V1 = P2V2 = constant Figure 3.1-2P-V graph for constant temperature

  7. P 1 2 V 0 V2 V1 Figure 3.2 P-V graph for constant pressure process Charles’ Law If gas changes from state 1 to state 2 during a constant pressure process, then The Charles’s Law may be stated as follows: Provided the pressure P of a given mass of gas remains constant, then the volume V of the gas will be directly proportional to the absolute temperature T of the gas, i.e. VT, or V = constant x T. Therefore V/T = constant, for constant pressure P. If the process is represented on a P – V diagram as before, the result will be as shown in Fig. 3.2.

  8. Universal Gases Law (P1V1)/T = ( P2V2)/T = constant • No gases in practice obey this law rigidly, but many gases tend towards it. An imaginary ideal that obeys the law is called a perfect gas Charles’ Law gives us the change in volume of a gas with temperature when the pressure remains constant. Boyle’s Law gives us the change in volume of a gas with pressure if the temperature remains constant characteristic equation of state of a perfect gas • The relation which gives the volume of a gas when both temperature and the pressure are changed is stated in the following equation When the state is changing from 1 to state 2

  9. The constant, R (gas constant) The constant, R, is called the gas constant. The unit of R is Nm/kg K or J/kg K. Each perfect gas has a different gas constant. The characteristic equation is usually written as PV = RT Or for m kg, occupying V m3, PV = mRT

  10. Kilogram-mole Another form of the characteristic equation can be derived using the kilogram-mole as a unit. The kilogram-mole is defined as a quantity of a gas equivalent to m kg of the gas, where M is the molecular weight of the gas (e.g. since the molecular weight of oxygen is 32, then 1 kg mole of oxygen is equivalent to 32 kg of oxygen). From the definition of the kilogram-mole, for m kg of a gas we have,m = nM, (where n is the number of moles). Note: Since the standard of mass is the kg, kilogram-mole will be written simply as mole. Substituting for m from equation m = nM in PV = mRT, we have PV = nMRT or

  11. Avogadro’s hypothesis Now Avogadro’s hypothesis states that the volume of 1 mole of any gas is the same as the volume of 1 mole of any other gas, when the gases are at the same temperature and pressure. Therefore V/n is the same for all gases at the same value of P and T. That is the quantity PV/nT is constant for all gases. This constant is called the universal gas constant, and is given the symbol Ro. universal gas constant, Ro or since MR = Rothen,

  12. Universal Gas Constant, Ro ? Untuksemuajenis gas Experiment has shown that the volume of 1 mole of any perfect gas at 1 bar and 1 oC is approximately 22.71 m3. Therefore from equation

  13. Universal Gas Constant, Ro Untuksemuajenis gas Experiment has shown that the volume of 1 mole of any perfect gas at 1 bar and 1 oC is approximately 22.71 m3. Therefore from equation

  14. Universal Gas Constant, Ro Untuksemuajenis gas Experiment has shown that the volume of 1 mole of any perfect gas at 1 bar and 1 oC is approximately 22.71 m3. Therefore from equation Note: Please refer to last page of Thermodynamics and transport Properties of Fluid (STEAM TABLE)

  15. Universal Gas Constant, Ro Untukjenis gas tertentu, Oksigen Untuksemuajenis gas Experiment has shown that the volume of 1 mole of any perfect gas at 1 bar and 1 oC is approximately 22.71 m3. Therefore from equation Note: Please refer to last page of Thermodynamics and transport Properties of Fluid (STEAM TABLE) From equation below, the gas constant for any gas can be found when the molecular weight is known, e.g. for oxygen of molecular weight 32, the gas constant is

  16. Example 3.3 0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300 kN/m2 and a temperature of 45 oC. The gas is compressed until the pressure reaches 1.27 MN/m2 and the temperature is 83oC. If the gas is assumed to be a perfect gas, determine: the mass of gas (kg) the final volume of gas (m3) Given: R = 0.29 kJ/kg K

  17. Example 3.3 0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300 kN/m2 and a temperature of 45 oC. The gas is compressed until the pressure reaches 1.27 MN/m2 and the temperature is 83oC. If the gas is assumed to be a perfect gas, determine: the mass of gas (kg) the final volume of gas (m3) Given: R = 0.29 kJ/kg K Solution to Example 3.3 From the question V1= 0.046 m3 P1 = 300 kN/m2 T1 = 45 + 273 K = 318 K P2= 1.27 MN/m2 = 1.27 x 103kN/m2 T2 = 83 + 273 K = 356 K R = 0.29 kJ/kg K

  18. Example 3.3 0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300 kN/m2 and a temperature of 45 oC. The gas is compressed until the pressure reaches 1.27 MN/m2 and the temperature is 83oC. If the gas is assumed to be a perfect gas, determine: the mass of gas (kg) the final volume of gas (m3) Given: R = 0.29 kJ/kg K Solution to Example 3.3 From the question V1= 0.046 m3 P1 = 300 kN/m2 T1 = 45 + 273 K = 318 K P2= 1.27 MN/m2 = 1.27 x 103kN/m2 T2 = 83 + 273 K = 356 K R = 0.29 kJ/kg K • From equation PV = mRT

  19. Example 3.3 0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300 kN/m2 and a temperature of 45 oC. The gas is compressed until the pressure reaches 1.27 MN/m2 and the temperature is 83oC. If the gas is assumed to be a perfect gas, determine: the mass of gas (kg) the final volume of gas (m3) Given: R = 0.29 kJ/kg K Solution to Example 3.3 From the question V1= 0.046 m3 P1 = 300 kN/m2 T1 = 45 + 273 K = 318 K P2= 1.27 MN/m2 = 1.27 x 103kN/m2 T2 = 83 + 273 K = 356 K R = 0.29 kJ/kg K • a) From equation PV = mRT

  20. Example 3.3 0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300 kN/m2 and a temperature of 45 oC. The gas is compressed until the pressure reaches 1.27 MN/m2 and the temperature is 83oC. If the gas is assumed to be a perfect gas, determine: the mass of gas (kg) the final volume of gas (m3) Given: R = 0.29 kJ/kg K Solution to Example 3.3 From the question V1= 0.046 m3 P1 = 300 kN/m2 T1 = 45 + 273 K = 318 K P2= 1.27 MN/m2 = 1.27 x 103kN/m2 T2 = 83 + 273 K = 356 K R = 0.29 kJ/kg K • a) From equation PV = mRT b)From equation, the constant volume process i.e. V1 = V2

  21. Example 3.3 0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300 kN/m2 and a temperature of 45 oC. The gas is compressed until the pressure reaches 1.27 MN/m2 and the temperature is 83oC. If the gas is assumed to be a perfect gas, determine: the mass of gas (kg) the final volume of gas (m3) Given: R = 0.29 kJ/kg K Solution to Example 3.3 From the question V1= 0.046 m3 P1 = 300 kN/m2 T1 = 45 + 273 K = 318 K P2= 1.27 MN/m2 = 1.27 x 103kN/m2 T2 = 83 + 273 K = 356 K R = 0.29 kJ/kg K • a) From equation PV = mRT b)From equation, the constant volume process i.e. V1 = V2

  22. Specific Heat Capacity (Muatan/KapisititentuHaba) The specific heat capacities of any substance is defined as the amount of heat energy required to raise the unit mass through one degree temperature raise. Mass of substance 1|||||2|||||3||||4|||| Q=?

  23. Specific Heat Capacity at Constant Volume (Cv) If 1 kg of a gas is supplied with an amount of heat energy sufficient to raise the temperature of the gas by 1 degree whilst the volume of the gas remains constant, then the amount of heat energy supplied is known as the specific heat capacity at constant volume, and is denoted by Cv. The unit of Cvis J/kg K or kJ/kg K.

  24. P P2 2 P1 1 V V1 = V2 Specific Heat Capacity at Constant Volume (Cv) For a reversible non-flow process at constant volume, we have: dQ = mCvdT-------equation (1) For a perfect gas the values of Cvare constant for any one gas at all pressures and temperatures. Equations (1) can then be expanded as follows : Heat flow in a constant volume process, Q12 = mCv(T2 – T1) Also, from the non-flow energy equation: Q – W = (U2 – U1) mcv(T2 – T1) – 0 = (U2 – U1) • (U2 – U1) = mCv(T2 – T1) • i.e.dU = Q • Note: • In a reversible constant volume process, no work energy transfer can take place since the piston will be unable to move i.e. W = 0.

  25. Example 3.4 3.4 kg of gas is heated at a constant volume of 0.92 m3 and temperature 17 oC until the temperature rise to 147 oC. If the gas is assumed to be a perfect gas, determine: The heat flow during the process The beginning pressure of gas The final pressure of gas Given Cv = 0.72 kJ/kg K R = 0.287 kJ/kg K

  26. Example 3.4 3.4 kg of gas is heated at a constant volume of 0.92 m3 and temperature 17 oC until the temperature rise to 147 oC. If the gas is assumed to be a perfect gas, determine: The heat flow during the process The beginning pressure of gas The final pressure of gas Given Cv = 0.72 kJ/kg K R = 0.287 kJ/kg K Solution to Example 3.4 From the question, m = 3.4 kg V1= V2 = 0.92 m3 T1 = 17 + 273 K = 290 K T2 = 147 + 273 K = 420 K Cv = 0.72 kJ/kg K R = 0.287 kJ/kg K

  27. Example 3.4 3.4 kg of gas is heated at a constant volume of 0.92 m3 and temperature 17 oC until the temperature rise to 147 oC. If the gas is assumed to be a perfect gas, determine: The heat flow during the process The beginning pressure of gas The final pressure of gas Given Cv = 0.72 kJ/kg K R = 0.287 kJ/kg K Solution to Example 3.4 From the question, m = 3.4 kg V1= V2 = 0.92 m3 T1 = 17 + 273 K = 290 K T2 = 147 + 273 K = 420 K Cv = 0.72 kJ/kg K R = 0.287 kJ/kg K a) Fromequation: Q12 = mCv(T2 – T1) = 3.4 x 0.72(420 – 290) = 318.24 kJ

  28. Example 3.4 b) From equation : PV = mRT Hence for state 1, P1V1 = mRT1 3.4 kg of gas is heated at a constant volume of 0.92 m3 and temperature 17 oC until the temperature rise to 147 oC. If the gas is assumed to be a perfect gas, determine: The heat flow during the process The beginning pressure of gas The final pressure of gas Given Cv = 0.72 kJ/kg K R = 0.287 kJ/kg K Solution to Example 3.4 From the question, m = 3.4 kg V1= V2 = 0.92 m3 T1 = 17 + 273 K = 290 K T2 = 147 + 273 K = 420 K Cv = 0.72 kJ/kg K R = 0.287 kJ/kg K a) Fromequation: Q12 = mCv(T2 – T1) = 3.4 x 0.72(420 – 290) = 318.24 kJ

  29. Example 3.4 b) From equation : PV = mRT Hence for state 1, P1V1 = mRT1 3.4 kg of gas is heated at a constant volume of 0.92 m3 and temperature 17 oC until the temperature rise to 147 oC. If the gas is assumed to be a perfect gas, determine: The heat flow during the process The beginning pressure of gas The final pressure of gas Given Cv = 0.72 kJ/kg K R = 0.287 kJ/kg K Solution to Example 3.4 From the question, m = 3.4 kg V1= V2 = 0.92 m3 T1 = 17 + 273 K = 290 K T2 = 147 + 273 K = 420 K Cv = 0.72 kJ/kg K R = 0.287 kJ/kg K • c) For state 2,P2V2 = mRT2 a) Fromequation: Q12 = mCv(T2 – T1) = 3.4 x 0.72(420 – 290) = 318.24 kJ

  30. Specific Heat Capacity at Constant Pressure (Cp) If 1 kg of a gas is supplied with an amount of heat energy sufficient to raise the temperature of the gas by 1 degree whilst the pressure of the gas remains constant, then the amount of heat energy supplied is known as the specific heat capacity at constant pressure, and is denoted by Cp. The unit of Cp is J/kg K or kJ/kg K.

  31. For a reversible non-flow process at constant pressure, we have:dQ = mCpdT--------------(2) Specific Heat Capacity at Constant Pressure (Cp) For a perfect gas the values of Cpare constant for any one gas at all pressures and temperatures. Equation (2) can then be expanded as follows: Heat flow in a reversible constant pressure process Q = mCp(T2 – T1)

  32. Relationship Between The Specific Heats Let a perfect gas be heated at constant pressure from T1 to T2. With reference to the non-flow equation: Q = U2 – U1 + W, The equation for a perfect gas: U2 – U1= mCv(T2– T1), (refer to heat flow@ constant volume) Hence,Q = mCv(T2– T1) + W In a constant pressure process, the work done by the fluid is given by the pressure times the change in volume, i.e. W = P(V2 – V1). (refer to heat flow@ constant pressure) Then using equation PV = mRT, we have: W = mR(T2 – T1) Thereforesubstituting, Q = mCv(T2 – T1) + mR(T2 – T1) = m(Cv + R)(T2 – T1) But for a constant pressure processfromequation: Q = mCp(T2 – T1) Hence, by equating the two expressions for the heat flow Q, we have: mCp(T2 – T1) = m(Cv + R)(T2 – T1) Cp = Cv + R Alternatively, it is usually written as: R = Cp - Cv

  33. i.e.  = Specific Heat Ratio ()(symbol of gamma) • The ratio of the specific heat at constant pressure to the specific heat at constant volume is given the symbol  (gamma), • i.e. Note that since Cp - Cv=R, it is clear that Cp must be greater than Cv for any perfect gas. It follows therefore that the ratio Cp/Cv =  , is always greater than unity

  34. Also from equation Cp = Cvhence substituting in equation at , become as the following equation: 7 Relationships between Cp , Cv, R, and  1 Cp - Cv=R 2 Dividing through by Cv 6 The Derivations 3 4 Therefore using equation then, OR 5 6 8

  35. Example 3.5 Solution to Example 3.5 • From equation R = Cp – Cv i.e. R = 0.846 – 0.657 = 0.189 kJ/kg K or R = 189 Nm/kg K From equation M = Be careful with the unit conversion…….. What is the unit of Rostated in Steam Table A certain perfect gas has specific heat as follows: Cp = 0.846 kJ/kg K and Cv= 0.657 kJ/kg K Find the gas constant and the molecular weight of the gas. J/kg K i.e.M = J/kg K

  36. Summary- Ideal Gas Cp - Cv=R Relationship between Cp and Cv 3 1 PV = mRT Where: P-Pressure(N/m2) V –volume (m3) m-Mass (kg) R-Gas constant (Nm/kg K or J/kg K) T- Temperature (Kelvin) 2 4 Ro-Universal Gas Constant, which refer to Last page of Steam table M-molecular weight which refer to periodic table Ratio between Cp and Cv

  37. Thank you

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