The Laws of Thermodynamics PDF

THE LAWS OF THERMODYNAMICS Learning Objective Understand the basic principles in three laws of thermodynamics. Solve for heat, work, entropy and internal energy using integral calculus. Key Points The first law, also known as Law of Conservation of Energy, states that en...

THE LAWS OF THERMODYNAMICS Learning Objective Understand the basic principles in three laws of thermodynamics. Solve for heat, work, entropy and internal energy using integral calculus. Key Points The first law, also known as Law of Conservation of Energy, states that energy cannot be created or destroyed in an isolated system. The second law of thermodynamics states that the entropy of any isolated system always increases. The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches absolute zero. Terms Pressure – force per unit area Volume – the amount of space that an object occupies. Also referred to as capacity. Temperature – a measure of the average kinetic energy of the molecules of a substance. Absolute zero – the lowest temperature that is theoretically possible. Entropy – a thermodynamic property that is the measure of a system’s thermal energy per unit of temperature that is unavailable for doing useful work. Work – thermodynamic work is done when the volume changes under pressure Heat – thermal energy that is transferred between substances Heat capacity – a measure of how much heat a given substance needs to absorb in order to increase its temperature a specified amount. Internal energy – the total energy of the molecules of a substance, excluding their interactions with external fields (like earth’s gravitational field). Isothermal – a process for which temperature remains constant Isobaric – a process for which pressure remains constant Isochoric – a process for which volume remains constant Adiabatic – a process for which no heat is absorbed or released Isentropic – a process for which the entropy of a system remains constant Compilation of Notes for PHYSICS 2 1 Steady-state – a process for which the internal energy of a system remains constant Quasistatic – a process that is carried out through a series of virtually infinitesimal disturbances from equilibrium, which occurs slowly enough that the system establishes equilibrium between the infinitesimal disturbances Spontaneous – a process that occurs without being driven by an energy source Reversible – a quasistatic process that can be carried out in reverse Irreversible – any process that is not quasistatic, or any process that features frictional or dissipative forces. System or Surroundings In order to avoid confusion, scientists discuss thermodynamic values in reference to a system and its surroundings. Everything that is not a part of the system constitutes its surroundings. The system and surroundings are separated by a boundary. For example, if the system is one mole of a gas in a container, then the boundary is simply the inner wall of the container itself. Everything outside of the boundary is considered the surroundings, which would include the container itself. The boundary must be clearly defined, so one can clearly say whether a given part of the world is in the system or in the surroundings. If matter is not able to pass across the boundary, then the system is said to be closed; otherwise, it is open. A closed system may still exchange energy with the surroundings unless the system is an isolated one, in which case neither matter nor energy can pass across the boundary. A diagram of a thermodynamic system Compilation of Notes for PHYSICS 2 2 The Laws of Thermodynamics According to the zeroth law of thermodynamics, if two objects, A and B, are in thermal equilibrium with a third object, C, then the two objects, A and B, are also in thermal equilibrium with each other. The fundamental law allows us to use a thermometer to measure temperature and to establish thermal equilibrium between multiple objects. According to the first law of thermodynamics, the heat (dQ) absorbed by a system with constant mole numbers minus the work (dW) done by the system equals the change in the system’s internal energy (dU). This law expresses conservation of energy. dU = dQ – dW Note the following sign conventions:  The heat change (dQ) is positive if heat is added to (or absorbed by) the system and negative if it is released by the system.  Work (dW) is positive if work is done by the system and negative if work is done on the system. Note how those tiny words “by” and “on” make a huge difference! According to the second law of thermodynamics, the total entropy (statistical disorder) of the system plus the surroundings cannot decrease for a thermodynamic process (for a macroscopic system) that begins and ends in equilibrium states. It can remain constant or it may increase, but the overall entropy (if you include both the entropy of the system and entropy of the surroundings) cannot decrease. (The entropy of the system can decrease provided that the entropy of the surroundings increases at least as much, and vice-versa.) One consequence of the second law of thermodynamics is that it is impossible to construct a perfectly efficient heat engine. According to the third law of thermodynamics, it is impossible for any object to reach a temperature of exactly absolute zero Kelvin by performing a finite number of operations, regardless of what idealized process may be applied. In practice, we can get very close to absolute zero (a small fraction of one Kelvin), but we can never reach it exactly. Heat, Work, Entropy, and Internal Energy Internal energy (U), heat (Q), and work (W) are related through the first law of thermodynamics, which expresses conservation of energy for thermal systems. 𝑑𝑈 = 𝑑𝑄 − 𝑑𝑊 Work (W) is the integral of pressure (P) over volume (V). 𝑉 𝑑𝑊 = 𝑃𝑑𝑉 , 𝑊 = ∫ 𝑃𝑑𝑉 𝑉=𝑉𝑜 Compilation of Notes for PHYSICS 2 3 The sign of work is determined as follows:  When work is done by the system, the volume increases and work is positive.  When work is done on the system, the volume decreases and work is negative. Heat (Q) exchanges may be expressed in terms of specific heat (CV or CP), molar specific heat (CV or CP), or entropy (S) – or for a phase transition, in terms of latent heat (L). 𝑑𝑄 = 𝑚𝐶𝑣 𝑑𝑇 , 𝑑𝑄 = 𝑚𝐶𝑃 𝑑𝑇 , 𝑑𝑄 = 𝑛𝑐𝑣 𝑑𝑇 , 𝑑𝑄 = 𝑛𝑐𝑃 𝑑𝑇 , 𝑑𝑄 = 𝐿𝑑𝑚, 𝑑𝑄 = 𝑇𝑑𝑆 The entropy (S) is related to the heat and temperature by dQ = TdS. Entropy provides a measure of the statistical disorder of the sysem, and is valid only for equilibrium states. Entropy is path-independent (like internal energy in this regard, but unlike work and heat), meaning that the change in entropy can be computed for any reversible process that takes the system between the initial and final equilibrium states. 𝑄 𝑑𝑄 ∆𝑆 = 𝑆 − 𝑆𝑜 = ∫ 𝑇 𝑄−𝑄𝑜 P-V Diagrams There are two common ways to calculate the work done in thermal physics 𝑉  Express pressure as a function of volume and integrate: 𝑊 = ∫𝑉=𝑉 𝑃𝑑𝑉 𝑜  Find the area under the curve for a P-V diagram. (Recall from calculus that a definite integral equals the area under the curve.) Note that area can be positive or negative:  The area is positive when volume increases (the process goes to the right).  The area is negative when volume decreases (the process goes to the left). For a complete cycle (a closed path), the network equals the area enclosed by the path. Compilation of Notes for PHYSICS 2 4 Thermodynamic Processes Following are some common thermodynamic processes.  Volume remains constant for an isochoric process. o Since the volume doesn’t change (∆𝑉 = 0), no work is done (𝑊 = 0). An isochor is a vertical line on a P-V diagram. o Express the heat exchange in terms of the specific heat capacity at constant volume: 𝑑𝑄 = 𝑚𝐶𝑉 𝑑𝑇 or 𝑑𝑄 = 𝑛𝐶𝑉 𝑑𝑇. Since no work is done, the change in internal energy equals the heat exchanged: (∆𝑈 = 𝑄).  Pressure remains constant for an isobaric process. o Since pressure is constant, it may come out of the work integral: 𝑉 𝑉 o 𝑊 = ∫𝑉=𝑉 𝑃𝑑𝑉 = 𝑃 ∫𝑉=𝑉 𝑑𝑉 = 𝑃 (𝑉 − 𝑉𝑜 ) = 𝑃∆𝑉. An isobar is a horizontal line on a P- 𝑜 𝑜 V diagram. o Express the heat exchange in terms of the specific heat capacity at constant pressure: 𝑑𝑄 = 𝑚𝐶𝑃 𝑑𝑇 or 𝑑𝑄 = 𝑛𝐶𝑃 𝑑𝑇. The internal energy change equals ∆𝑈 = 𝑄 − 𝑊 according to the first law of thermodynamics.  A free expansion occurs under zero pressure. o No work is done (𝑊 = 0) because the pressure equals zero. Since no work is done, the change in internal energy equals the heat exchanged: ∆𝑈 = 𝑄.  Temperature remains constant for an isothermal process. o For an ideal gas or a van der Waals fluid, solve for pressure in terms of volume to 𝑉 𝑛𝑅𝑇 perform the work integral: 𝑊 = ∫𝑉=𝑉 𝑃𝑑𝑉. For example, for an ideal gas, 𝑃 =. 𝑜 𝑉 Temperature is constant for an isotherm. o Although temperature is constant for an isotherm, there is generally still heat exchanged (𝑄 ≠ 0). However, if the potential energy of the molecules is constant (it is for an ideal gas), the change in internal energy will be zero (∆𝑈 = 0) and the heat exchange will equal the work done: 𝑄 = 𝑊.  No heat is absorbed or released by the system for an adiabatic process. o If the relationship between pressure and volume is known for the adiabat, express 𝑉 pressure in terms of volume to perform the work integral: 𝑊 = ∫𝑉=𝑉 𝑃𝑑𝑉. For 𝑜 𝑃𝑜 𝑉𝑜𝑌 example, for an ideal gas, 𝑃𝑜 𝑉𝑜𝑌 = 𝑃𝑉 𝑌 or 𝑃 = 𝑉𝑌 o Since no heat is exchanged, the change in internal energy equals the negative of the work done: ∆𝑈 = −𝑊 Compilation of Notes for PHYSICS 2 5   For a complete cycle (a closed path on a P-V diagram), the net internal energy change is zero (∆𝑈𝑛𝑒𝑡 = 0) for the complete cycle, although internal energy may change for parts of the cycle. The net heat exchange equals the net work done: 𝑄𝑛𝑒𝑡 = 𝑊𝑛𝑒𝑡 Symbols and SI Units Symbol Name SI Units 𝑃 Pressure Pa 𝑉 Volume 𝑚3 𝑛 Number of moles mol 𝑅 Universal gas constant 𝐽 𝑚𝑜𝑙. 𝐾 𝑇 Absolute temperature K 𝑚 Total mass kg 𝑄 Heat J 𝐿 Latent heat of transformation 𝐽 𝑘𝑔 𝐶𝑉 Specific heat capacity at constant volume 𝐽 𝑘𝑔. 𝐾 𝐶𝑃 Specific heat capacity at constant pressure 𝐽 𝑘𝑔. 𝐾 𝑐𝑉 Molar specific heat capacity at constant volume 𝐽 𝑚𝑜𝑙. 𝐾 𝑐𝑃 Molar specific heat capacity at constant pressure 𝐽 𝑚𝑜𝑙. 𝐾 𝛾 Adiabatic index Unitless 𝑈 Internal energy J 𝑊 Work J 𝑆 Entropy 𝐽 𝐾 Compilation of Notes for PHYSICS 2 6  𝑎 A correction for intermolecular forces 𝐽. 𝑚3 𝑚2 𝑏 The volume occupied by one mole of the molecules 𝑚3 𝑚𝑜𝑙 Where Do the Ideal Gas Equations 𝑼 = 𝒏𝒄𝑽 𝑻 and 𝒄𝒑 = 𝒄𝑽 + 𝑹 come from? For an isochor, volume is constant, so no work is done: W=0. The corresponding heat change depends on the molar specific heat at constant volume: 𝑄 = 𝑛𝑐𝑉 ∆𝑇. According to the first law of thermodynamics, ∆𝑈 = 𝑄 − 𝑊. For an isochor, this reduces to , ∆𝑈 = 𝑄 = 𝑛𝑐𝑉 ∆𝑇. Since ∆𝑈 is path- independent, and since the internal energy of an ideal gas depends only on temperature, 𝑈 = 𝑛𝑐𝑉 𝑇 and , ∆𝑈 = 𝑛𝑐𝑉 ∆𝑇 for any process. For an isobar, pressure is constant. We may pull pressure out of the work integral: 𝑊 = ∫ 𝑃𝑑𝑉 = 𝑃 ∫ 𝑑𝑉 = 𝑃∆𝑉. From the ideal gas law, 𝑃∆𝑉 = 𝑛𝑅∆𝑇 for an isobar, such that 𝑊 = 𝑛𝑅∆𝑇. The corresponding heat change depends on the molar specific heat at constant pressure: 𝑄 = 𝑛𝑐𝑃 ∆𝑇. According to the first law of thermodynamics ∆𝑈 = 𝑄 − 𝑊. For an isobar, this becomes ∆𝑈 = 𝑛𝑐𝑃 ∆𝑇 − 𝑛𝑅∆𝑇. Recall that ∆𝑈 = 𝑛𝑐𝑉 ∆𝑇 for an ideal gas for any process. Therefore, 𝑛𝑐𝑉 ∆𝑇 = 𝑛𝑐𝑃 ∆𝑇 − 𝑛𝑅∆𝑇. Divide both sides of the equation by 𝑛∆𝑇. This shows that 𝑐𝑣 = 𝑐𝑝 − 𝑅, which is the same as 𝑐𝑣 = 𝑐𝑝 + 𝑐𝑝 𝑐𝑣+𝑅 3 5 5 𝑅. The adiabatic index is then 𝛾 = =. For a monatomic ideal gas, 𝑐𝑣 = 𝑅, 𝑐𝑝 = 𝑅, and 𝛾 =. 𝑐𝑣 𝑐𝑣 2 2 3 𝜸 Why does 𝑷𝒐 𝑽𝒐 = 𝑷𝑽𝜸 along an adiabat for an ideal gas? For an ideal gas, 𝑃𝑉 = 𝑛𝑅𝑇. Implicitly differentiate the ideal gas law: 𝑑(𝑃𝑉) = 𝑑(𝑛𝑅𝑇). Apply the product rule: (𝑃𝑉) = 𝑃𝑑𝑉 + 𝑉𝑑𝑃. Therefore, 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 𝑛𝑅𝑑𝑇. Apply the first law of thermodynamics: 𝑑𝑈 = 𝑑𝑄 − 𝑑𝑊. For an adiabatic process, no heat is exchanged: 𝑑𝑄 = 0. The first law simplifies to 𝑑𝑈 = −𝑑𝑊. Recall that 𝑑𝑈 = 𝑛𝑐𝑉 𝑑𝑇 and 𝑑𝑊 = 𝑃𝑑𝑉. Plug these expressions into 𝑑𝑈 = 𝑃𝑑𝑉 −𝑑𝑊 to get 𝑛𝑐𝑉 𝑑𝑇 = −𝑃𝑑𝑉. Therefore, 𝑑𝑇 = −. Plug this into the equation 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 𝑛𝑅𝑑𝑇 𝑛𝑐𝑣 𝑅 to get 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = − 𝑃𝑑𝑉. Multiply both sides of the equation by 𝑐𝑉 to get 𝑐𝑉 𝑃𝑑𝑉 + 𝑐𝑉 𝑉𝑑𝑃 = 𝑐𝑣 −𝑅𝑃𝑑𝑉. Combine like differentials: 𝑐𝑉 𝑉𝑑𝑃 = −(𝑐𝑉 + 𝑅)𝑃𝑑𝑉. Recall that 𝑐𝑝 = 𝑐𝑉 + 𝑅. Thus, 𝑐𝑉 𝑉𝑑𝑃 = 𝑏 −𝑐𝑝 𝑃𝑑𝑉. Separate variables and integrate. Recall that ln(𝑏) − ln(𝑎) = 𝑙𝑛 ( ) and 𝑎𝑙𝑛(𝑥) = ln(𝑥 𝑎 ). 𝑎 𝒄𝒗 𝑽𝒅𝑷 = −𝒄𝒑 𝑷𝒅𝑽 𝒅𝑷 𝒄𝒑 𝒅𝑽 ∫ =− ∫ 𝑷 𝒄𝒗 𝑽 𝒍𝒏(𝑷) − 𝒍𝒏(𝑷𝒐 ) = −𝜸[𝒍𝒏(𝑽) − 𝒍𝒏(𝑽𝒐 )] 𝜸 𝑷 𝑽𝒐 𝑽𝒐 𝒍𝒏 ( ) = −𝜸𝒍𝒏 ( ) = 𝒍𝒏 ( 𝜸 ) 𝑷𝒐 𝑽 𝑽 𝛾 𝑃 𝑉𝑜 𝛾 Use the rule 𝑒 ln(𝑥) = 𝑥 to get =. Cross multiply: 𝑃𝑉 𝛾 = 𝑃𝑜 𝑉𝑜 𝑃𝑜 𝑉𝛾 Compilation of Notes for PHYSICS 2 7 Strategy for problems involving Work, Heat, Entropy, or Internal Energy How you solve a thermal physics problem involving work, heat, entropy, or internal energy depends on which kind of problem it is:  There are two ways to calculate the work done for a thermodynamic process: o If you are given a P-V diagram, find the area under the curve. Area is positive if volume increases and negative if volume decreases. For a closed path, the net work equals the area enclosed by the path: in this case, area is positive for a clockwise path and negative for a counterclockwise path. o Otherwise, express pressure as a function of volume and integrate. 𝑉 𝑊 = ∫ 𝑃𝑑𝑉 𝑉=𝑉𝑜  There are a variety of ways to determine the heat exchanged. o For an isobar, pressure remains constant: 𝑑𝑄 = 𝑚𝐶𝑝 𝑑𝑇 and 𝑑𝑄 = 𝑛𝐶𝑝 𝑑𝑇 o For an isochor, volume remains constant: 𝑑𝑄 = 𝑚𝐶𝑣 𝑑𝑇 and : 𝑑𝑄 = 𝑛𝐶𝑣 𝑑𝑇 o For other thermodynamic process (such as an isotherm or adiabat), apply the first law of thermodynamics to relate the heat exchange to work and internal energy. o For a phase transition 𝑄 = 𝑚𝐿  If a problem involves an ideal gas, you may apply the ideal gas law. The internal energy and molar specific heat capacities of a monatomic or diatomic (at moderate temperature) ideal gas is given by the expressions below. 𝑷𝒐 𝑽𝒐 𝑷𝑽 𝑷𝑽 = 𝒏𝑹𝑻, = , 𝒄𝒑 = 𝒄𝒗 + 𝑹 𝑻𝒐 𝑻 𝟑 𝟑 𝑼 = 𝑵𝒌𝑩 𝑻 = 𝒏𝑹𝑻 (monatomic) 𝟐 𝟐 𝟓 𝟓 𝑼 = 𝑵𝒌𝑩 𝑻 = 𝒏𝑹𝑻 (diatomic, mod. Temp.) 𝟐 𝟐 𝟑 𝟓 𝒄𝒑 𝟓 𝒄𝒗 = 𝑹 , 𝒄𝒑 = 𝑹 , 𝜸 = = (monatomic) 𝟐 𝟐 𝒄𝒗 𝟑 𝟓 𝟕 𝒄𝒑 𝟕 𝒄𝒗 = 𝑹 , 𝒄𝒑 = 𝑹 , 𝜸 = = (diatomic, moderate temp) 𝟐 𝟐 𝒄𝒗 𝟓 To find the work done along an isotherm, solve for pressure in one of the equations below, plug it into the work integral, and treat T as a constant. For an isothermal process involving an ideal gas, the internal energy change is zero. 𝑃𝑉 = 𝑛𝑅𝑇 𝑃𝑜 𝑉𝑜 = 𝑃𝑉 (isotherm) ∆𝑈 = 0 (isotherm) Compilation of Notes for PHYSICS 2 8 To find the work done by an ideal gas along an adiabat, solve for pressure in the following equation and plug it into the work integral. 𝛾 𝑃𝑜 𝑉𝑜 = 𝑃𝑉 𝛾 (adiabat)  If a problem involves a van der Waals fluid, you may apply the van der Waals equation. To find the work done, substitute the equation for pressure into the work integral. Along an isotherm, T will be constant. 𝑅𝑇 𝑎𝑛2 𝑃= − 2 𝑉 −𝑏 𝑉 𝑛  Internal energy changes, heat, and work can be related through the first law of thermodynamics. 𝑑𝑈 = 𝑑𝑄 − 𝑑𝑊  Entropy (S) is related to heat via: 𝑑𝑄 = 𝑇𝑑𝑆 The change in entropy (∆𝑆 = 𝑆 − 𝑆𝑜 ) can be computed for any reversible process that takes the system between the initial and final equilibrium states, according to the following integral. Express the temperature in Kelvin (K) after integrating (that is, after evaluating the anti- derivative over the limits). 𝑄 𝑑𝑄 ∆𝑆 = 𝑆 − 𝑆𝑜 = ∫ 𝑇 𝑄=𝑄𝑜 Important Distinctions It is very important which quantity is constant for an isothermal, isobaric, isochoric, or adiabatic process, since the solution to a problem is much different for each process.  Temperature is constant along an isotherm  Pressure is constant along an isobar  Volume is constant along an isochor  No heat is exchanged along an adiabat Note that the math is much different for isotherms and adiabats, since temperature and heat mean two different things. Heat is generally exchanged along an isotherm even though temperature remains constant, and temperature generally changes along an adiabat even though no heat is exchanged. Compilation of Notes for PHYSICS 2 9 Example: A non-ideal gas expands isobarically from a volume of 3.0m 3 to 5.0m3 under a constant pressure of 50 kPa, as the temperature of the gas changes from 150 K to 450 K. The molar specific heat at constant pressure equals 2R. There are 30 moles of the gas. (A) Determine the work done. Perform the thermodynamic work integral. 𝑉 𝑊 = ∫ 𝑃𝑑𝑉 𝑉=𝑉𝑜 Since this example involves an isobar, the pressure is constant. This means that we may pull the pressure out of the integral. (For a process that isn’t isobaric, you can’t do this. The next example involves such a process.) 𝑉 𝑊 = 𝑃 ∫ 𝑑𝑉 = 𝑃(𝑉 − 𝑉𝑜 ) = (50,000)(5 − 3) = (50,000)(2) = 100,000𝐽 = 100 𝑘𝐽 𝑉=𝑉0 The gas does W=100,00 J = 100 kJ of work. The work done is positive (meaning that work is done by the gas) because volume increased. (B) Determine the amount of heat exchanged. Since this example involves an isobar, the pressure is constant. The heat can be found from the molar specific heat at constant pressure. 𝑑𝑄 = 𝑛𝑐𝑝 𝑑𝑇 𝑇 𝑄 = ∫ 𝑛𝑐𝑝 𝑑𝑇 𝑇=𝑇𝑜 The problem states that cp = 2R for this gas, where R is the universal gas constant. Recall that 𝐽 25 𝐽 𝑅 = 8.134 ≈ 𝑚𝑜𝑙.𝐾 3 𝑚𝑜𝑙.𝐾 𝑇 𝑇 𝑄 = ∫ 𝑛2𝑅𝑑𝑇 = 2𝑛𝑅 ∫ 𝑑𝑇 = 2𝑛𝑅(𝑇 − 𝑇𝑜 ) 𝑇=𝑇𝑜 𝑇=𝑇𝑜 25 𝑄 = 2(30)(8.314)(450 − 150) ≈ 2(30) ( ) (300) = 150,000 𝐽 = 150 𝑘𝐽 3 The gas absorbs (since Q is positive) Q = 150,000 J = 150 kJ of heat energy. (C) Determine the change in the internal energy of the gas. Apply the first law of thermodynamics, using the answers to (A) and (B). Compilation of Notes for PHYSICS 2 10  ∆𝑈 = 𝑄 − 𝑊 = 150,000 − 100,000 = 50,000 𝐽 = 50𝑘𝐽 The internal energy of the gas increases by ∆𝑈 = 50,000 𝐽 = 50𝑘𝐽. Note: You can’t use the equation U=ncvT for this problem is not an ideal gas. Example: A monoatomic ideal gas is compressed isothermally from a volume of 4.0m 3 to 1.0m3. The initial pressure is 20kPa. (A) Determine the work done. Perform the thermodynamic work integral. 𝑉 𝑊 = ∫ 𝑃𝑑𝑉 𝑉=𝑉𝑜 Unlike the previous example, this process is not isobaric, so pressure is not constant and may not come out of the integral. However, since this examples involves an ideal gas, we may use the 𝑷𝒐 𝑽𝒐 𝑷𝑽 equation =. Furthermore, since this process is isothermal, temperature is constant: T=To, so 𝑻𝒐 𝑻 𝑷𝒐 𝑽𝒐 𝑷𝑽 𝑷𝒐 𝑽𝒐 T cancels out in = , reducing the equation to 𝑷𝒐 𝑽𝒐 = 𝑷𝑽. Solve for P to get 𝑃 =. 𝑻𝒐 𝑻 𝑽 Substitute this expression for pressure into the work integral. 𝑉 𝑃𝑜 𝑉𝑜 𝑊= ∫ 𝑑𝑉 𝑉 𝑉=𝑉𝑜 The initial pressure and volume are constants, so we may pull 𝑷𝒐 𝑽𝒐 out of the integral. 𝑉 𝑑𝑉 𝑊 = 𝑃𝑜 𝑉𝑜 ∫ 𝑉 𝑉=𝑉𝑜 𝑑𝑥 𝑏 Recall from calculus that ∫ = ln(𝑥). Also recall the logarithm identity ln(𝑏) − ln(𝑎) = ln( ). 𝑥 𝑎 𝑉 𝑊 = 𝑃𝑜 𝑉𝑜 [𝑙𝑛(𝑉)]𝑉𝑉=𝑉𝑜 = 𝑃𝑜 𝑉𝑜 [ln(𝑉) − ln(𝑉𝑜 )] = 𝑃𝑜 𝑉𝑜 ln ( ) 𝑉𝑜 1 1 𝑊 = (20,000)(4) ln ( ) = (80,000) ln ( ) 𝐽 = −(80,000) ln(4) 𝐽 = −80 ln(4) 𝑘𝐽 4 4 1 Recall the logarithm identity ln ( ) = − ln(𝑥). The work done is𝑊 = −(80,000) ln(4) 𝐽 = 𝑥 −80 ln(4) 𝑘𝐽. If you use a calculator, this works out to 𝑊 = −111 𝑘𝐽. This work done is negative (meaning that work is done on the gas) because the volume decreased. (Work is done “by” the gas when it expands, and “on” the gas when it is compressed. These two-letter words “by” and “on” make a sign difference.) (B) Determine the change in the internal energy of the gas. Compilation of Notes for PHYSICS 2 11  𝟑 Since this example involves a monoatomic ideal gas, the internal energy equals 𝑼 = 𝒏𝑹𝑻. Since 𝟐 this process is isothermal, T is constant, such that ∆𝑇 = 0 and ∆𝑈 = 0. The internal energy of the gas doesn’t change: ∆𝑼 = 𝟎 (C) Determine the amount of heat exchanged. Apply the first law of thermodynamics, using the answers to (A) and (B). ∆𝑈 = 𝑄 − 𝑊 Add work to both sides of the equation. 𝑄 = ∆𝑈 + 𝑊 = 0 + [−(80,000) ln(4)] = −(80,000)ln(4)𝐽 = −80 ln(4) 𝑘𝐽 The heat exchange equals 𝑄 = −(80,000)ln(4)𝐽 = −80 ln(4) 𝑘𝐽. If you use a calculator, this works out to 𝑄 = −110 𝑘𝐽. The gas releases heat (since Q is negative). Example: A fluid expands from an initial volume of 6.0 m3 to 12.0 m3. The initial pressure is 25kPa. The fluid obeys the equation 𝑉√𝑃= const. Determine the work done. Perform the thermodynamic work integral. 𝑉 𝑊 = ∫ 𝑃𝑑𝑉 𝑉=𝑉𝑜 Note that the pressure is not constant in this problem, so we may not pull it out of the integral. Instead, we must express the pressure in terms of the volume. We can write the given equation, 𝑉√𝑃= const. (where “const. is short for “constant”), as 𝑉𝑜 √𝑃𝑜 = 𝑉√𝑃 (since both 𝑉𝑜 √𝑃𝑜 and 𝑉√𝑃 must equal the same constant). Square both sides to get 𝑉𝑜2 𝑃𝑜 = 𝑉 2 𝑃 and then divide both 𝑃𝑜 𝑉𝑜2 sides by 𝑉 2 to find that 𝑃 =. Plug this expression into the work integral. 𝑉2 𝑉 𝑃𝑜 𝑉𝑜2 𝑊= ∫ 𝑑𝑉 𝑉2 𝑉=𝑉𝑜 The initial pressure and volume are constants, so we may pull 𝑃𝑜 𝑉𝑜2 out of the integral. 𝑉 𝑉 𝑉 𝑑𝑉 1 𝑉 𝑊= 𝑃𝑜 𝑉𝑜2 ∫ 2 = 𝑃𝑜 𝑉𝑜2 ∫ ∫ 𝑉 −2 𝑑𝑉 = 𝑃𝑜 𝑉𝑜2 [−𝑉 −1 ]𝑉𝑉=𝑉𝑜 = 𝑃𝑜 𝑉𝑜2 [− ] 𝑉 𝑉 𝑉=𝑉𝑜 𝑉=𝑉𝑜 𝑉=𝑉𝑜 𝑉=𝑉𝑜 1 1 Note that = 𝑉 −2 and 𝑉 −1 = 𝑉2 𝑉 Compilation of Notes for PHYSICS 2 12  1 𝑉 1 1 1 1 𝑊 = −𝑃𝑜 𝑉𝑜2 [ ] = −𝑃𝑜 𝑉𝑜2 ( − ) = −(25,000)(6)2 ( − ) 𝑉 𝑉=𝑉𝑜 𝑉 𝑉𝑜 12 6 It’s convenient to distribute the (6)2 62 62 𝑊 = −(25,000) ( − ) = −(25,000)(3 − 6) = −(25,000)(−3) = 75,000 𝐽 = 75 𝑘𝐽 12 6 The gas does 𝑊 = 75,000 𝐽 = 75 𝑘𝐽 of work. The work done is positive (meaning that work is done by the gas) because the volume increased. (The two minus signs made a plus sign). Example: Determine the work done along the straight line path in the graph below. The work done equals the area under the P-V curve (which in this case is a straight line). On the diagram on the right, you can see that the area under the straight line path, can be divided up into a triangle and rectangle. The work done equals the area of the triangle plus the area of the rectangle. In this example, the work done is positive because the volume increases (The path goes to the right). 1 𝑊 = 𝐴𝑡𝑟𝑖 + 𝐴𝑟𝑒𝑐 = 𝑏ℎ + 𝐿𝑊 2 The triangle has a base of b = 8.0 – 2.0 = 6.0 m3 and a height of h = 30 – 10 = 20 kPa, while the rectangle has dimensions of L = 8.0 – 2.0 = 6.0 m3 and W = 10 – 0 = 10 kPa. 1 𝑊 = (6)(20,000) + (6)(10,000) = 60,000 + 60,000 = 120,000 𝐽 = 120 𝑘𝐽 2 The work done is W = 120,000 J = 120 kJ. Compilation of Notes for PHYSICS 2 13 Example: Determine the net work done for the complete cycle shown below. The net work equals the area of the triangle. The net work is negative because the path is counter clockwise. The triangle has a base of b = 8.0 – 2.0 = 6.0 m3 and a height of h = 80 – 20 = 60 kPA. 1 1 𝑊𝑛𝑒𝑡 = −𝐴𝑡𝑟𝑖 = − 𝑏ℎ = − (6)(60,000) = −180,000 𝐽 = −180 𝑘𝐽 2 2 The net work done is 𝑊𝑛𝑒𝑡 = −180,000 𝐽 = −180 𝑘𝐽 Example: A monoatomic ideal gas completes the thermodynamic cycle shoen below, where path AB is an isotherm. (A) Determine the work, heat, and internal energy change for path AB. Unlike the two previous examples, path AB is a curve. We can’t get an exact answer by finding areas of the triangles and rectangles for the curve running from A to B. However, we can get an exact answer by performing the thermodynamic work integral for this path. 𝑉𝐵 𝑊𝐴𝐵 = ∫ 𝑃𝑑𝑉 𝑉=𝑉𝐴 Compilation of Notes for PHYSICS 2 14  𝑃𝐴 𝑉𝐴 𝑃𝑉 Since this example involves ideal gas, we may use the equation =. Furthermore, since this 𝑇𝐴 𝑇 𝑃 𝐴 𝑉𝐴 𝑃𝑉 process is isothermal, temperature is constant: 𝑇 = 𝑇𝐴 , so 𝑇 cancels out in = , reducing the 𝑇𝐴 𝑇 𝑃𝐴 𝑉𝐴 equation to 𝑃𝐴 𝑉𝐴 = 𝑃𝑉. Solve for P to get 𝑃 =. Substitute this for pressure in the work 𝑉 integral. 𝑉𝐵 𝑃𝐴 𝑉𝐴 𝑊𝐴𝐵 = ∫ 𝑑𝑉 𝑉 𝑉=𝑉𝐴 The initial pressure and volume are constants, so we may pull 𝑃𝐴 𝑉𝐴 out of the integral. 𝑉𝐵 𝑑𝑉 𝑊𝐴𝐵 = 𝑃𝐴 𝑉𝐴 ∫ 𝑉 𝑉=𝑉𝐴 𝑑𝑥 𝑏 Recall from calculus that ∫ = ln(𝑥). Also recall the logarithm identity ln(𝑏) − ln(𝑎) = ln ( ). 𝑥 𝑎 𝐵 𝑉 𝑉𝐵 𝑊𝐴𝐵 = 𝑃𝐴 𝑉𝐴 [ln(𝑉)]𝑉=𝑉 = 𝑃𝐴 𝑉𝐴 [ln(𝑉𝐵 ) − ln(𝑉𝐴 )] = 𝑃𝐴 𝑉𝐴 ln ( ) 𝐴 𝑉𝐴 Read the initial and final values directly from the graph: 𝑉𝐴 = 1.0𝑚3 , 𝑃𝐴 = 80 𝑘𝑃𝑎, 𝑉𝐵 = 4.0𝑚3 , 𝑎𝑛𝑑 𝑃𝐵 = 20 𝑘𝑃𝑎 4 𝑊𝐴𝐵 = (80,000)(1) ln ( ) = (80,000) ln(4) 𝐽 = 80 ln(4) 𝑘𝐽 1 The work done is 𝑊𝐴𝐵 = ( 80,000) ln(4) 𝐽 = 80 ln(4) 𝑘𝐽. If you use a calculator, this works out to 𝑊𝐴𝐵 = 111 𝑘𝐽. The work done is positive because the volume increased (path AB goes to the right). 3 Since this example involves a monoatomic ideal gas, the internal energy equals 𝑈 = 𝑛𝑅𝑇. 2 Since this process is isothermal, T is constant, such that ∆𝑇𝐴𝐵 = 0 and ∆𝑈𝐴𝐵 = 0. The internal energy of the gas doesn’t change: ∆𝑈𝐴𝐵 = 0 Apply the first law of thermodynamics. ∆𝑈𝐴𝐵 = 𝑄𝐴𝐵 − 𝑊𝐴𝐵 Add work to both sides of the equation. 𝑄𝐴𝐵 = ∆𝑈𝐴𝐵 + 𝑊𝐴𝐵 = 0 + (80,000) ln(4) 𝐽 = 80 ln(4) 𝑘𝐽 The heat exchange equals 𝑄𝐴𝐵 = (80,000) ln(4) 𝐽 = 80 ln(4) 𝑘𝐽. If you use a calculator, this works out to 𝑄𝐴𝐵 = 111 𝑘𝐽. The gas absorbs (since 𝑄𝐴𝐵 is positive) heat. Compilation of Notes for PHYSICS 2 15  (B) Determine the work, heat, and internal energy change for path BC. Path BC is an isobar. The volume doesn’t change (𝑉𝐶 = 𝑉𝐵 ). Therefore, no work is done along path BC: 𝑊𝐵𝐶 = 0 Since the volume is constant, use the formula for molar specific heat at constant volume. 𝑑𝑄 = 𝑛𝑐𝑉 𝑑𝑇 𝑇𝐶 𝑄𝐵𝐶 = ∫ 𝑛𝑐𝑉 𝑑𝑇 𝑇=𝑇𝐵 3 Since this problem involves a monoatomic gas, 𝑐𝑉 = 𝑅, where R is the universal gas constant. 2 𝑇𝐶 𝑇𝐶 3 3 3 𝑄𝐵𝐶 = ∫ 𝑛 𝑅𝑑𝑇 = 𝑛𝑅 ∫ 𝑑𝑇 = 𝑛𝑅(𝑇𝐶 − 𝑇𝐵 ) 2 2 2 𝑇=𝑇𝐵 𝑇=𝑇𝐵 We don’t know the temperature, but that’s okay. Since this is an ideal gas, 𝑃𝐵 𝑉𝐵 = 𝑛𝑅𝑇𝐵 and 𝑃𝐶 𝑉𝐶 = 𝑛𝑅𝑇𝐶. This means that 𝑛𝑅(𝑇𝐶 − 𝑇𝐵 ) = 𝑃𝐶 𝑉𝐶 − 𝑃𝐵 𝑉𝐵. Read the initial and final values directly from the graph: 𝑉𝐵 = 4.0 𝑚3 , 𝑃𝐵 = 20 𝑘𝑃𝑎, 𝑉𝐶 = 4.0 𝑚3 , 𝑎𝑛𝑑 𝑃𝐶 = 80 𝑘𝑃𝑎. 3 3 3 3 3 𝑄𝐵𝐶 = 𝑛𝑅(𝑇𝐶 − 𝑇𝐵 ) = 𝑃𝐶 𝑉𝐶 − 𝑃𝐵 𝑉𝐵 = (80,000)(4) − (20,000)(4) 2 2 2 2 2 𝑄𝐵𝐶 = 480,000 − 120,000 = 360,000 𝐽 = 360 𝑘𝐽 The gas absorbs (since 𝑄𝐵𝐶 is positive) 𝑄𝐵𝐶 = 360,000 𝐽 = 360 𝑘𝐽 of heat energy. Apply the first law of thermodynamics. ∆𝑈𝐵𝐶 = 𝑄𝐵𝐶 − 𝑊𝐵𝐶 = 360,000 − 0 = 360,000 𝐽 = 360 𝑘𝐽 The internal energy of the gas increases by ∆𝑈𝐴𝐵 = 360,000 𝐽 = 360 𝑘𝐽. (C) Determine the work, heat, and internal energy change for path CA. Path CA is an isobar. The pressure doesn’t change (𝑃𝐴 = 𝑃𝐶 ). Perform the thermodynamic work integral. 𝑉𝐴 𝑊𝐶𝐴 = ∫ 𝑃𝑑𝑉 𝑉=𝑉𝐶 Since the pressure is constant, we may pull the pressure out of the integral. Compilation of Notes for PHYSICS 2 16  𝑉𝐴 𝑊𝐶𝐴 = 𝑃𝐶 ∫ 𝑑𝑉(𝑉𝐴 − 𝑉𝐶 ) 𝑉=𝑉𝐶 Read the initial and final values directly from the graph: 𝑉𝐶 = 4.0𝑚3 , 𝑃𝐶 = 80𝑘𝑃𝑎, 𝑉𝐴 = 1.0𝑚3 , 𝑎𝑛𝑑 𝑃𝐴 = 80 𝑘𝑃𝑎. 𝑊𝐶𝐴 = (80,000)(1 − 4) = (80,000)(−3) = −240,000 𝐽 = −240 𝑘𝐽 The work done is 𝑊𝐶𝐴 = −240,000 𝐽 = −240 𝑘𝐽. The work done is negative because the volume decreased (path CA goes to the left). Since the pressure is constant, use the formula for molar specific heat at constant pressure. 𝑑𝑄 = 𝑛𝑐𝑃 𝑑𝑇 𝑇𝐴 𝑄𝐶𝐴 = ∫ 𝑛𝑐𝑃 𝑑𝑇 𝑇=𝑇𝐶 5 Since this problem involves a monoatomic ideal gas, 𝐶𝑃 = 𝑅, where R is the universal gas constant. 2 𝑇𝐴 𝑇𝐴 5 5 5 𝑄𝐶𝐴 = ∫ 𝑛 𝑅𝑑𝑇 = 𝑛𝑅 ∫ 𝑑𝑇 = 𝑛𝑅(𝑇𝐴 − 𝑇𝐶 ) 2 2 2 𝑇=𝑇𝐶 𝑇=𝑇𝐶 We don’t know the temperature, but that’s okay. Since this is an ideal gas, 𝑃𝐴 𝑉𝐴 = 𝑛𝑅𝑇𝐴 and 𝑃𝐶 𝑉𝐶 = 𝑛𝑅𝑇𝐶. This means that 𝑛𝑅(𝑇𝐴 − 𝑇𝐶 ) = 𝑃𝐴 𝑉𝐴 − 𝑃𝐶 𝑉𝐶. Read the initial and final values directly from the graph: 𝑉𝐶 = 4.0𝑚3 , 𝑃𝐶 = 80𝑘𝑃𝑎, 𝑉𝐴 = 1.0𝑚3 , 𝑎𝑛𝑑 𝑃𝐴 = 80 𝑘𝑃𝑎. 5 5 5 5 5 𝑄𝐶𝐴 = 𝑛𝑅(𝑇𝐴 − 𝑇𝐶 ) = 𝑃𝐴 𝑉𝐴 − 𝑃𝐶 𝑉𝐶 = (80,000)(1) − (80,000)(4) 2 2 2 2 2 𝑄𝐶𝐴 = 200,000 − 800,000 = −600,000 𝐽 = −600 𝑘𝐽 The heat exchange is 𝑄𝐶𝐴 = −600,000 𝐽 = −600 𝑘𝐽. Since 𝑄𝐶𝐴 is negative, the gas releases heat energy. Apply the first law of thermodynamics, using the answers to (A) and (B). ∆𝑈𝐶𝐴 = 𝑄𝐶𝐴 − 𝑊𝐶𝐴 = −600,000 − (−240,000) = −600,000 + 240,000 ∆𝑈𝐶𝐴 = −360,000 𝐽 = −360 𝑘𝐽 The internal energy change is ∆𝑈𝐶𝐴 = −360,000 𝐽 = −360 𝑘𝐽. (D) Determine the net work, heat and internal energy change for the complete cycle. Compilation of Notes for PHYSICS 2 17 Add up the work done by each process: 𝑊𝑛𝑒𝑡 = 𝑊𝐴𝐵 + 𝑊𝐵𝐶 + 𝑊𝐶𝐴 = (80,000) ln(4) + 0 − 240,000 𝑊𝑛𝑒𝑡 = (80,000) ln(4)𝐽 − 240,000 𝐽 = 80 ln(4) 𝑘𝐽 − 240 𝑘𝐽 The net work is 𝑊𝑛𝑒𝑡 = (80,000) ln(4)𝐽 − 240,000 𝐽 = 80 ln(4) 𝑘𝐽 − 240 𝑘𝐽. If you use a calculator, this comes out to 𝑊𝑛𝑒𝑡 = −129 𝑘𝐽. As expected, the net work is negative because the path is counterclockwise. Add up the heat charges for each process: 𝑄𝑛𝑒𝑡 = 𝑄𝐴𝐵 + 𝑄𝐵𝐶 + 𝑄𝐶𝐴 = (80,000) ln(4) + 360,000 − 600,000 𝑄𝑛𝑒𝑡 = (80,000) ln(4) 𝐽 − 240,000 𝐽 = 80 ln(4) 𝑘𝐽 − 240 𝑘𝐽 The net heat charge is 𝑄𝑛𝑒𝑡 = (80,000) ln(4) 𝐽 − 240,000 𝐽 = 80 ln(4) 𝑘𝐽 − 240 𝑘𝐽. If you use a calculator, this comes out to 𝑄𝑛𝑒𝑡 = −129 𝑘𝐽. Add up the internal energy changes for each process. 𝑈𝑛𝑒𝑡 = 𝑈𝐴𝐵 + 𝑈𝐵𝐶 + 𝑈𝐶𝐴 = 0 + 360,000 − 360,000 = 0 The net internal energy change is zero: 𝑈𝑛𝑒𝑡 = 0. In fact, this will be true for the net internal energy of any complete cycle (closed path). Internal energy (unlike work and heat) is path- independent, so it only depends on the initial and final points: For a complete cycle, the initial and final positions are the same, so 𝑈𝑛𝑒𝑡 = 0. You can verify that the first law of thermodynamics, 𝑈𝑛𝑒𝑡 = 𝑄𝑛𝑒𝑡 − 𝑊𝑛𝑒𝑡 , is satisfied for the complete cycle: The left hand side is 𝑈𝑛𝑒𝑡 = 0, and the right-hand side is also zero because 𝑄𝑛𝑒𝑡 = 𝑊𝑛𝑒𝑡. Example: One mole of an ideal gas at 300 K expands to twice its initial volume during a reversible isothermal process. Determine the change in entropy. Perform the integral for entropy change for this reversible process. 𝑄 𝑑𝑄 ∆𝑆 = ∫ 𝑇 𝑄=𝑄0 From the first law of thermodynamics, 𝑑𝑈 = 𝑑𝑄 − 𝑑𝑊. For an ideal gas, 𝑑𝑈 = 𝑛𝑐𝑉 𝑑𝑇. For an isothermal process, temperature remains constant. Therefore, 𝑑𝑈 = 0 (internal energy doesn’t change) and 0 = 𝑑𝑄 − 𝑑𝑊, such that 𝑑𝑄 = 𝑑𝑊. Recall that 𝑑𝑊 = 𝑃𝑑𝑉. The upper limit of integration (which is over volume after substitution) is 2𝑉𝑜 because the problem states that the gas expands to twice its initial volume. Compilation of Notes for PHYSICS 2 18  𝑄 𝑊 2𝑉𝑜 𝑑𝑄 𝑑𝑊 𝑃 ∆𝑆 = ∫ = ∫ = ∫ 𝑑𝑉 𝑇 𝑇 𝑇 𝑄=𝑄0 𝑊=𝑊𝑜 𝑉=𝑉𝑜 𝑃 𝑛𝑅 For an ideal gas, 𝑃𝑉 = 𝑛𝑅𝑇 such that =. 𝑇 𝑉 2𝑉𝑜 2𝑉𝑜 𝑛𝑅 2𝑉𝑜 ∆𝑆 = ∫ 𝑑𝑉 = 𝑛𝑅 ∫ [ln(𝑉)]𝑉=𝑉 = 𝑛𝑅[ln(2𝑉𝑜 ) − ln(𝑉𝑜 )] 𝑉 𝑜 𝑉=𝑉𝑜 𝑉=𝑉𝑜 𝑏 Recall that ln(𝑏) − ln(𝑎) = ln ( ). 𝑎 2𝑉𝑜 ∆𝑆 = 𝑛𝑅 ln ( ) = 𝑛𝑅 ln(2) 𝑉𝑜 1 25 1 Recall that the universal gas constant is 𝑅 = 8.314 ≈. The problem states that there is just 𝑚𝑜𝑙.𝐾 3 𝑚𝑜𝑙.𝐾 one mole (𝑛 = 1.0 𝑚𝑜𝑙). 25 25 ∆𝑆 = (1)(8.314) ln(2) ≈ (1) ( ) ln(2) = ln(2) 𝐽/𝐾 3 3 25 The entropy of the system increases by ∆𝑆 ≈ ln(2) 𝐽/𝐾. If you use a calculator, this works out to 3 ∆𝑆 = 5.8 𝐽/𝐾. Compilation of Notes for PHYSICS 2 19