1 / 31

CHAPTER 16 : THERMODYNAMICS

CHAPTER 16 : THERMODYNAMICS. 16.1 Work Heat Form of energy which is transferred from one body to another body at lower temperature, by virtue of the temperature difference between the bodies. Internal energy

fred
Download Presentation

CHAPTER 16 : THERMODYNAMICS

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CHAPTER 16 : THERMODYNAMICS 16.1 Work Heat • Form of energy which is transferred from one body to another body at lower temperature, by virtue of the temperature difference between the bodies. Internal energy • Total energy content of a system. It is the sum of all forms of energy possessed by the atoms and molecules of the system.

  2. 16.1 Work System • The collection of matter within prescribed and identifiable boundaries. Surroundings • Everything else in the environment. Thermodynamics system • System that can interact with its surroundings or environment in at least two ways, one of which is heat transfer.

  3. pA dx A 16.1 Work Work • Assume a system in a piston that contains fluid, the work done by the system as its volume changes.

  4. 16.1 Work • When the piston moves out an infinitesimal distance dx, the work dW done by this force is dw = Fdx = pAdx A : cross-sectional area of the cylinder p : pressure exerted by the system at the surface of the piston. F : total force exerted by the system on the piston

  5. 16.1 Work • But dV=Adx, therefore dW=pdV. • The work done in a volume change from V1 to V2,

  6. p p V V (a) p vs V at changes pressure (a) p vs V at constant pressure Graph p versus V

  7. 16.2 First Law of Thermodynamics • The internal energy of a system changes from an initial value U1 to a final value U2 due to heat, Q and work, W:

  8. Cyclic process • The change in internal energy depends on the gas temperature. • It also depends on the initial and final state. • Cyclic process : Initial state = Final state U1 = U2 Q = W

  9. Example: Two moles of the monatomic argon gas expand isothermally at 298 K, from an initial volume of 0.025 m3 to a final volume of 0.050 m3 (Argon is an ideal gas), find (a) work done by the gas (b) the heat supplied to the gas. (a) W=+3400 J, (b) Q =+3400 J

  10. 16.3 Thermodynamic Processes Isothermal process Isochoric process Isobaric process Adiabatic process

  11. p pi pf Work,W V Vi Vf Graph Isothermal process • The work done by the system is the area under the curve of a pV-diagram as a graph shown below..

  12. Isothermal process

  13. Isochoric process • a constant-volume process, where there is no work done in the system, W = 0. • The first law of thermodynamic will be, U2-U1 = ΔU = Q.

  14. Graph Isochoric process • The graph pV of an isochoric process is shown below. • p v

  15. Isothermal process • a constant-temperature process. • Heat transfer and work in the system must not change and the internal energy will be zero, ΔU=0. • The changes in pressure,p and volume,V will take place.

  16. Isobaric process • Isobaric process is a constant-pressure process. W = p(V2 – V1) The first law of thermodynamic will be, Q= ΔU +W = ΔU + p( V2-V1)

  17. p V Isobaric process carried out at a constant pressure Graph Isobaric process

  18. Adiabatic process • a process with no heat transfer to or from a system, Q=0. •  ΔU = -W

  19. Example The temperature of three moles of a monatomic ideal gas is reduced from Ti = 540 K to Tf = 350 K by two different methods. In the first method 5500 J of heat flows into the gas, while in the second method, 1500 J of heat flows into it. In each case, find (a) the change in the internal energy of the gas and (b) the work done by the gas.

  20. 16.4 Molar Specific Heat Capacities At Constant Pressure And Volume Molar specific heat capacities of constant pressure, Cp When an n mole of gas undergoes a heating process, the equation that relates Q and the temperature changes, dT is written as ; Q = nCdT ……………..(1) Whereas Q = heat n = number of mole C = molar specific heat capacities dT = temperature changes

  21. When the gas is heated at constant pressure, the (1) equation will be; Q = nCpdT Or Cp = Where Cp = molar specific heat capacities at constant pressure.

  22. Molar specific heat capacities of constant volume, Cv • Hence, when an n mole of gas is heated at constant volume, the (1) equation will be; Q= nCvdT Or Where Cv = molar specific heat capacities at constant volume

  23. The universal gas constant, R the substraction between Cp and Cv that is; • Cp-Cv = R • Where R = 8314 J mol-1 K

  24. Example • A certain perfect gas has specific heat capacities as follows: • Cp = 0.84 kJ / mol K and Cv = 0.657 kJ / mol K • Calculate the gas constant.

  25. 16.5 Specific Heat At Constant Pressure And Volume

  26. Example • A 2.00 mol sample of an ideal gas with γ = 1.40 expands slowly and adiabatically from a pressure of 5.00 atm and a volume of 12.0 L to a final volume of 30.0 L. • (a) What is the final pressure of the gas? • (b) What are the initial and final temperature?

  27. THERMODYNAMICS THE END

More Related