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# Thermodynamics

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### Thermodynamics

3. 3. 1. Zeroth law of thermodynamics If two systems are at the same time in equilibrium with a third system, they are in equilibrium with each other. Practically, this means that all three systems are at the same temperature. A B C Since A and B are at equilibrium and B and C are at equilibrium, A and C are also at equilibrium according to the zeroth law
4. 4. 2. First law of thermodynamics The change in internal energy of a system is equal to the heat added to the system minus the work done by the system. dU = δQ – δW (U- internal energy of the system, Qheat added to the system, W- work done by the system) The first law of thermodynamics is the application of the conservation of energy principle to heat and thermodynamic processes.
5. 5. 3. Second law of thermodynamics When two initially isolated systems which are at thermal equilibriums are brought into contact they reach a common thermal equilibrium However, the second law can also be expressed in terms of the application in which it is used. For example, I. Second law in terms of heat flow: Heat flows spontaneously from hotter to colder objects but not vice versa. II. Second law in terms of heat engines: It is impossible to construct an engine which has 100% efficiency or a system in which the heat added to the system is solely used to perform work.
6. 6. 4. Third law of thermodynamics The entropy of a system approaches a constant value as the temperature approaches zero.
7. 7. Kelvin-Planck statement: It is impossible to extract an amount of heat QH from a hot reservoir and use it all to do work W. Some amount of heat QC must be exhausted to a cold reservoir.
8. 8. Refrigerator and Air ConditionerClausius StatementIt is impossible for heat to flow from a cold body to a warm body without any work ( or without the aid of an external agency) having been done to accomplish the flow. Hot Reservoir QH W QC Cold Reservoir
9. 9. Internal energy of a gasThe internal energy of a gas is the kinetic energy of thermal motion of its molecules. Important: When a cylinder of a gas is moving in a locomotive or any other moving agent, the molecules inside the cylinder move relative to that of the cylinder (molecules travel in the same speed as the cylinder). Thus the kinetic energy of each molecule is equivalent to the kinetic energy experienced by the cylinder. Therefore, the temperature and thus the internal energy of the molecules relative to the cylinder is unchanged.
10. 10. The internal energy of a gas is the sum of the kinetic energy of the molecules and the potential energy(due to intermolecular attractions) between molecules. But there are no intermolecular forces between molecules in an ideal gas. Therefore, potential energy in such an instance is zero. Therefore, Internal Energy = Kinetic Energy of molecules
11. 11. But, PV = ⅓ mNc2 and E = ½mNc2 So, PV = ⅓ × 2 × ½ mNc2 E Therefore, PV = ⅔E or, E= PV 3 2 But, PV = nRT . Therefore, E= nRT Now that E = ΔU, For a difference of temperature, Internal Energy = nRΔT (ΔU)
12. 12. Work done by gasWork done by gas is the work done to increase its volume during expansion. A Δ W = Fx Δ W = PAx Therefore, Δ W = P(ΔV) P F (PA ) X
13. 13. When heat is given to a substance it expands and does external work. In the case of solids and liquids the change in volume and hence the external work done is negligible. Therefore, there is only one specific heat for a substance. Gases experience the effect of the change of volume to a great extent. Since the volume can be controlled, gases can be used to do variable amounts of external work. Therefore, there are no. of specific heats that can be defined for a gas.
14. 14. The two most commonly used are1. Specific heat at constant volume (CV) It is the amount of heat required to raise the temperature of unit mass of a gas through 1°C, when volume is kept constant. 2. Specific heat at constant pressure (CP) It is the amount of heat required to raise the temperature of unit mass of a gas through 1°C, when the pressure of the gas is kept constant However, CP > CV and CP / CV = γ. γ is called the ratio of specific heats. At constant volume, ΔQ = n CV θ At constant pressure, ΔQ = n CP θ
15. 15. Increase / Decrease Ter m ΔQ Positive OR Increasing If heat is given from outside Negative OR Decreasin g If heat is released to outside ΔU ΔU ΔW If internal energy increases If a gas does work to outside If internal energy decreases If work is done on the gas
16. 16. Isothermal Process (ΔU = 0) 1. This usually occurs when a system is in thermal contact with a reservoir. The change occurs very slowly and the system will continually adjust to the temperature of the reservoir through heat exchange. ΔT = 0, therefore, ΔU = 0. Thus, ΔQ = ΔW The graph of such a process is as follows: P Temperature constant. Therefore the system obeys Boyle’s Law V
17. 17. 2. Adiabatic Process (ΔQ = 0) An expansion in which no heat energy enters or leaves the system. Adiabatic processes can take place if the container in which the process takes place has thermally-insulated walls or the process happens in an extremely short time. Adiabatic CoolingOccurs when the pressure of a substance is decreased as it does work on the surroundings Adiabatic HeatingOccurs when the pressure of a gas is increased from work done on it by the surroundings.
18. 18. The graph of such a process is as follows: P A (T1) B (T2) V Adiabatic process An adiabat is similar to an isotherm, except that during expansion an adiabat loses more pressure than an isotherm, so it has a steeper inclination Density of isotherms stays constant but the density of adiabats grow Isothermal Processes Each adiabat intersects each isotherm exactly once
19. 19. 3. Isochoric Process (ΔW = 0) Also called constant-volume process, iso volumetric process, and isometric process. Is a thermodynamic process during which the volume of the closed system stays constant. Since the system undergoes isochoric process, the volume is constant. Therefore, Q= mCv ΔT and Q = U (W=0 and thus PΔV = 0) The graph for such a process is as follows: and since V is constant T
20. 20. 1. Isobaric Process (Pressure Constant) Since pressure stays constant, ΔQ = nCP ΔT . 2. Isoentropic process (Entropy of system stays constant) 3. Isoenthalpic process(Enthalpy of system constant) 4. Quasistatic Process (Is a process that happens infinitely slowly)
21. 21. 1. For an isothermal process, P A(T) ΔU = 0 ΔQ = ΔW Since T is constant, ΔQ = + By applying Boyle’s law, ΔW= + P1V1 = P2 V2 B(T) 2. V ΔW = 0 ΔQ = - ΔU ΔQ = + ΔU= - P1 / T1 = P2 / T2
22. 22. 3. The work done in a process is the area under a P-V graph P PΔV- the work done by the process ( Area under the graph) P V ΔV 4. For an adiabatic process, P A (T1) ΔQ = 0 ΔW = - ΔU B (T2) V P1 V1 / T1 = P2 V2 / T2
23. 23. 5. If some system is closed the sum of internal energies should equal to zero. Since the system is closed sum of ΔU for the whole process is zero. However, the area shaded inside the process represents the net work done. 6. In a P- V diagram the internal energy acquired or released is independent of the path ΔU Internal energy between two points for a closed system is independent of the path (Same value regardless of path)
24. 24. 7. If some system is clockwise ΔQ > 0 for that system. Similarly, if a system is counterclockwise, ΔQ < 0 for that system System clockwise. Therefore, ΔQ > 0 System counterclockwise. Therefore, ΔQ < 0
25. 25. Sample QuestionUsing 3 moles of air an engine goes through the following changes in a single cycle.(ABCD) i. Isothermal expansion from a volume of 0.03m3 to 0.07m3 at 373K (A to B) ii. At constant volume cooling to 320K(B to C) iii. Isothermal compression at 320K to 0.03m3 (C to D) iv. At constant volume, compression to original volume, pressure, and temperature.(D to A) Show these changes in a P-V diagram and calculate the following: a. The heat taken during process B to C b. The heat taken during the process D to A c. The total internal energy of the engine after process completion
26. 26. P (Pa) For point A, PV= nRT P × 0.03 = 3 × 8.314 × 373 P = 3.1 × 106 Pa A 3.1 × 106 ΔU = 0 ΔQ = ΔW 1.33 × 106 1.14 ×106 B For point B, PV=nRT P × 0.07 = 3 × 8.314 × 373 P = 1.33 × 106 Pa ΔW=0 ΔQ = ΔU D C 0.03 For point D, At constant value, P1 /T1 = P2 / T2 (1.33 × 106)/ 373 = P2 / 320 P2= 1.14 × 106 Pa 0.07 V (m3 )
27. 27. By considering process B to C, ΔU = (3/2) nRΔT = (3/2) × 3 × 8.314× 373 ΔU = 13955 J However, during B to C, ΔQ = ΔU as the process is isochoric. Therefore, heat added to the system during B to C is 13995 J Similarly, same heat is added to the system during D to A as during B to C as the temperature difference(ΔT) is same.(ΔU same) Therefore, heat added to the system during D to A is also 13995 J. Since this is a closed process, the sum of internal energies should be zero.
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