Physics Ideal Gases and the First Law of Thermodynamics Stady Guides

Chapter N Ideal Gases and the First Law of Thermodynamics N.1 - Ideal Gases Basic Definitions The assumption behind an ideal gas is that it is a collection of non-interacting point particles. By this we mean that the size of each molecule is small compared to the fraction of the total volume occupied each molecule (the total volume divided by the number of molecules). Also we assume that the forces between the molecules can be neglected. P = Pressure SI units = Pa = N  m2 V = Volume T = Temperature N =  of molecules SI units = m 3 SI units = K Dimensionless Pressure is force per area measured in Pascals, Pa. Temperature is the absolute temperature in K. Our emphasis will be different from that of a chemistry class. In chemistry it is important to keep track of the amount of material; this is related to the number of moles n. In physics our emphasis is relating macroscopic quantities to microscopic ones; the actual number of molecules N is then more important. Ideal Gas Law There are some basic proportionalities that are satisfied. Keeping N and T constant we get an inverse proportionality between pressure and volume P ∝ 1 / V. With N and V constant we get a proportionality between pressure and temperature P ∝ T. For the same P and T we get V ∝ N. The ideal gas law follows from this. PV ∝N T The constant of proportionality is a fundamental constant called Boltzmann's constant kB P V = N kB T where Boltzmann's constant has the value kB = 1.38 × 10-23 J K . Note that Boltzmann’s constant could be viewed as a conversion between Joules and Kelvins, and temperatures could be measured in Joules. This approach is not standard at this level and will be avoided here. Example N.1 - Number of Molecules How many air molecules are in a room with dimensions, 8 m by 10 m by 3 m? Assume 1 atm of pressure and a temperature of 20 °C. Solution This is a straight-forward application of the ideal gas law. V = 8 m × 10 m × 3 m = 240 m3 , P = 1 atm = 1.013 × 105 Pa , T = 20 °C = 293 K P V = N kB T ⟹ N = PV kB T = 6.01 × 1027 2 | Chapter N - Ideal Gases and the First Law of Thermodynamics Atomic and Molecular Masses mproton ≃ mneutron ≫ melectron Atoms consist of protons, neutrons and electrons. The masses of the proton and neutron are approximately equal and both are much larger than the mass of an electron. mproton = 1.6726 × 10-27 kg mneutron = 1.6749 × 10-27 kg melectron = 9.109 × 10-31 kg For atoms the atomic number Z is the number of protons in the nucleus, which is the same as the number of electrons in a neutral atom. A, the atomic mass number, is the number of nucleons, where protons and neutrons are nucleons. Z = atomic number =  of protons =  of electrons A = atomic mass number =  of nucleons (protons and neutrons) A - Z =  of neutrons One would expect, naively, that the mass of an atom should equal the sum of its constituent parts. This is not the case. It takes energy to break up an atom into its constituent parts. Using the mass-energy equivalence of relativity, E = m c2 , it follows that since the constituent parts have more energy, they have more mass. matom = Z mproton + (A - Z ) mneutron + Z melectron -  Binding   c2 Energy In this expression, the Binding Energy is the amount of energy required to break up the atom into is constituent parts. Because this naive approach doesn't work we introduce the atomic mass unit, u, which is defined to be the approximate contribution to the mass of an atom due to each proton and neutron. u ≃ mproton ≃ mneutron Thus the approximate mass of an atom is A u. matom ≃ A u We precisely define u in terms of the carbon-12 isotope, 1  12th the mass of 12 C. u= 1 12 12 C. Carbon has Z = 6. The 12 refers to the mass number, A = 12. u is defined as mass12 C = 1.6605 × 10-27 kg = 1 6.0221 × 1023 g The mole, abbreviated mol, is an SI base unit related to this. A mole is defined as the number of atoms in 12-grams of carbon-12; this number, known as Avogadro’s number, has the approximate value NA = 6.02 × 1023  mol . The number of moles n and the number of molecules N are related by N = n NA . We can then write the ideal gas law in terms of the number of moles PV = N kB T = nNA kB T If we define the ideal gas constant R by R = NA kB = 8.314 J mol · K , then the ideal gas law has the form PV = nRT. The mass of a mole of some molecule can be related to the molecular mass in atomic mass units, u. The value for the molecular mass in u is the same as the molar mass in grams. For example a molecule of CO2 has molecular mass of 44 u, so its molar mass is 44 g / mol. mmol = NA mmolecule = NA 44 u = 44 g / mol, since NA u = g / mol. The total mass of a gas can then be written as Chapter N - Ideal Gases and the First Law of Thermodynamics | 3 mtot = N mmolecule = n mmol . Example N.2 - The Density of Air (a) Estimate the density of (dry) air at 20 °C and 1 atm, assuming air is composed entirely of nitrogen N2 . Solution P = 1 atm = 1.013 × 105 Pa , T = 20 °C = 293 K and R = 8.314 J mol · K First we need to find the molar mass. mmolecule = mN2 = 2 mN = 2 × 14 u = 28 u ⟹ mmol = 28 g / mol = 0.028 kg / mol. We can find an expression for the number of moles n using the ideal gas law and then get an expression for the mass PV = nRT ⟹ n = PV RT and mtot = n mmol = mmol PV RT Density ρ is mass per volume. The volume V will cancel. ρ= mtot V = mmol P RT = 1.16 kg  m3 (b) The accepted value for the density of dry air at 20 °C is 1.204 kg  m3 . Calculate a more accurate result for the density of dry air by taking air to be 78% N2 , 21% O2 and 1% Argon, where the atomic mass of argon is: mAr = 40 u. Solution mO2 = 2 mO = 2 × 16 u = 32 u ⟹ mmol,O2 = 32 g / mol = 0.032 kg / mol Also for Argon, which is monatomic, we have mAr = 40 u ⟹ mmol,Ar = 40 g / mol = 0.040 kg / mol. The molar mass of air can be found as a weighted average. mmol = 0.78 mmol,N2 + 0.21 mmol,O2 + 0.01 mmol,Ar ⟹ ρ = mtot V = mmol P RT = 1.20 kg  m3 N.2 - Work and the First Law Thermodynamic Work As we saw in Chapter F, the work done in one dimension is W = ∫ F ⅆ x. Consider a gas at a pressure P expanding in a piston. If the piston has a cross-sectional area of A and it expands by the infinitesimal amount ⅆ x, then the infinitesimal work done by the gas is F ⅆ x = P A ⅆ x = P ⅆ V. Here we have used F = P A and ⅆ V = A ⅆ x. One-dimensional work: An expanding gas in a piston does work on the environment. This result is more general; if any system at a pressure P expands by ⅆ V, the infinitesimal work is P ⅆ V. 4 | Chapter N - Ideal Gases and the First Law of Thermodynamics Three-dimensional case: Generally, an expanding thermodynamic system does work as it expands. The work done by a thermodynamic system is W =  P ⅆ V. When a system expands it does work on the environment. Our sign convention is that W is the work done by the system, where positive work decreases the energy of the system. The work done on the system is -W. The historical reason for this convention is that the focus was on getting work from engines, where the engine was the system. For the simple case of expansion with a constant pressure we get W = P Δ V. Generally, when the volume increases the work is positive; it is negative when volume decreases. PV-Diagrams and Thermodynamic States Suppose there is some thermodynamic system, for instance a fixed quantity of some substance. Knowledge of the pressure P and volume V uniquely specifies the thermodynamic state. For example, one could find the temperature uniquely from P and V. The equation that relates P, V and T is called the equation of state. For complex thermodynamic systems this is not a simple function that one can write down, but for an ideal gas it is just the ideal gas law. A PV-diagram is a graph of P vs. V (P is the y-axis and V is the x-axis.) Work has a simple interpretation in terms of PV-diagrams; it is the area under the curve. W =  P ⅆ V = ± Area Under Since P is always positive the sign of P ⅆ V is the same as on the sign of ⅆ V. Chapter N - Ideal Gases and the First Law of Thermodynamics | 5 P P W >0 Pf W >0 Pi Pf Pi Vi P Vf V P W <0 Pf Vi Vf V Vi V W <0 Pi Pf Pi Vf Vi V Vf Interactive Figure In thermodynamics we often deal with cycles. A cycle is a closed path in a PV-diagram. For a cycle there is a positive and a negative contribution. If the path is clockwise the positive contribution dominates and the result is positive and it just then enclosed area. When the path is counterclockwise the result is negative. W =  P ⅆ V = ± Area Enclosed cycle P P W >0 W <0 V V Example N.3 - The PV-Diagram P ( atm ) 8 2 A C D B 2 V (L) 6 (a) What is the work for different processes taking a system from point A to point B? Find WAB (the direct path), WACB and WADB . Solution Since all three cases correspond to increasing volumes we have W = +(Area under). 6 | Chapter N - Ideal Gases and the First Law of Thermodynamics P ( atm ) 8 P ( atm ) A A 8 2 C 2 B 2 P ( atm ) 2 B V (L) 6 2 A 8 D V (L) 6 W AB B 2 W ACB V (L) 6 W ADB The SI unit of work is Joule, where J = Pa m3 . Our axes have the units of atmospheres and liters. We need to convert the units of our areas from atm L to J. 1 atm = 1.013 × 105 Pa and 1 L = 10-3 m3 ⟹ 1 atm L = 101.3 J For WAB we have a trapezoid. The area of a trapezoid with a base width of b and sides with heights of h1 and h2 is 1 Area = b (h1 + h2 ). It follows that 2 WAB = + 1 2 4 L (2 atm + 8 atm) = +20 atm L = +2026 J For WACB and WADB we have rectangles. WACB = + 4 L × 8 atm = +32 atm L = +3242 J and WADB = + 4 L × 2 atm = +8 atm L = +810 J This shows the path-dependent nature of work explicitly. (b) What is the work for a process that takes the system from point B to point A, WBA ? Solution Here we have a process with decreasing volume, so W = -(Area under). P ( atm ) A 8 2 B 2 V (L) 6 W BA For WAB we have a trapezoid. The area of a trapezoid with a base width of b and sides with heights of h1 and h2 is 1 Area = b (h1 + h2 ). It follows that 2 WBA = -WAB = -2026 J (c) What is the work for the cycles ADBCA and ABDA? Solution For a cycle we have W = ± (Area enclosed), where clockwise gives positive and counterclockwise gives negative. P ( atm ) 8 2 P ( atm ) A C D B 2 8 2 V (L) 6 W ADBCA For WADBCA there is a counterclockwise path around a rectangle. A D B 2 6 W ABDA V (L) Chapter N - Ideal Gases and the First Law of Thermodynamics | 7 WADBCA = - 4 L × 6 atm = -24 atm L = -2431 J There is a clockwise path around a triangle for WABDA . WABDA = + 1 2 4 L × 6 atm = +12 atm L = +1215 J Quasi-static Processes This discussion of clear paths in a PV-diagram implies that we are undergoing a quasi-static process. A process is quasi-static if it is performed sufficiently slowly, so that at each intermediate state the system stays in thermal equilibrium. As an example of a process that is not quasi-static, consider the free expansion of a gas. Start with a container containing two regions separated by a partition. On one side of the partition is a pressurized gas and on the other side is a vacuum. When the partition is removed, the gas undergoes a free expansion. The initial and final states are in equilibrium but the intermediate states are not. When a system is not in equilibrium, pressure and temperature are not welldefined quantities. Partition Partition Removed Vacuum Initial Final The free expansion of a gas is not quasi-static. It is possible, theoretically, to do this expansion with a quasi-static process by slowly dragging a piston from initial to final positions. In this example with an expanding gas, approximating a quasi-static process need not be too slow. To be in thermal equilibrium the system must be at a uniform temperature; there can be no temperature gradients. Suppose heat is added to a solid block. To be in equilibrium every point in the block must be at the same temperature. An approximately quasi-static process here would be very slow; each time after a small amount of heat is added, one must wait for the system to equilibrate. Isothermal Expansion of an Ideal Gas The word isothermal implies the temperature is constant. For an ideal gas we can use the ideal gas law, the equation of state, to write the pressure as a function of volume. P(V ) = nRT V P P= Vi (T is constant) nRT V Vf V Interactive Figure: Work for an Isothermal Process of an Ideal Gas We can easily integrate this to get the work. 8 | Chapter N - Ideal Gases and the First Law of Thermodynamics Vf W = Vi Vf P(V ) ⅆ V = n R T  ⅆV V Vi = n R T ln(V ) Vf Vi = n R T (ln(V f ) - ln(Vi )) This gives the result W = n R T ln Vf Vi . Note that the natural log function guarantees that result of the work formula has the correct sign, since the natural log of a ratio less than one is negative. Example N.4 - Isothermal Expansion What is the work done by 50-g of oxygen O2 gas as it is compressed by tripling its pressure at a constant temperature of 20 °C? Solution m = 0.050 kg , T = 20 °C = 293 K and Pf Pi =3 The mass of O2 will tell us the number of moles n. The mass of a molecule in u gives the molar mass mO2 = 2 mO = 2 × 16 u = 32 u ⟹ mmol = 32 g / mol = 0.032 kg / mol and this gives us the number of moles. m = n mmol ⟹ n = m mmol = 1.56 mol The ideal gas law with constant n and T relates the ratio of the volumes to the ratio of the pressures. This is known as Boyle’s law. PV =nRT W = n R T ln n,T const. Vf Vi Pi Vi = P f V f ⟹ and R = 8.314 J mol · K Vf Vi = Pi Pf = 1 3 ⟹ W = -4180 J Note that increasing the pressure will decrease the volume and the work will then be negative. State Functions - Exact and Inexact Differentials A state function is a function of pressure and volume. It has a unique value at each point in a PV-diagram. Temperature is a state function, since it follows from the equation of state. f = f (V, P) is a state function. For state functions we take exact differentials. An exact differential is the infinitesimal change in some variable between two nearby states. Integrating an exact differential of a state function f gives the change in f. This is path independent, meaning that it depends only on the endpoints  ⅆ f = Δ f = f (V f , P f ) - f (Vi , Pi ) (path independent) Since a cycle ends at the same position it begins, then it follows that for a cycle the integral of an exact differential is zero If ⅆ f is an exact differential then  ⅆ f = 0 (exact differential) cycle We have seen that the work for a cycle is not zero. It follows that work is not a state function; one cannot assign a value to work at each state. Generally, for functions that are not state functions we use inexact differentials δ to represent infinitesimal values. The infinitesimal amount of work is written as δ W = P ⅆ V. For a general inexact differential δ g the integral depends on the path and thus the integral around a cycle is nonzero. Chapter N - Ideal Gases and the First Law of Thermodynamics | 9  δ g ≠ 0 (inexact differential) cycle Internal Energy and the First Law The energy of a thermodynamic system can be increased by adding thermal energy in the form of heat or by adding mechanical energy by doing work on the system. We will write the internal energy of a thermodynamic system as U. The first law of thermodynamics is Δ U = Q - W. Δ U is the change in the internal energy of the system. Q is the heat added to the system. W is the work done by the system. It follows that -W is the work done on the system. Internal energy is a state function. Since it is a state function then Δ U is independent of the path taken. Since the work is path dependent it follows from the first law that the heat is also path dependent, and not a state function. If we write the first law in its infinitesimal form then we get ⅆ U = δ Q - δ W. Since work and heat are not state functions then we use inexact differentials for them. The internal energy being a state function means that we use exact differentials for it. Example N.5 - The PV-Diagram (Continued) P ( atm ) 8 2 A C D B 2 V (L) 6 (d) What is the heat added to the system over the ABDA cycle? Solution First we need the work which we calculated in part (c). P ( atm ) 8 2 A D B 2 V (L) 6 W ABDA There is a clockwise path around a triangle for WABDA . W = WABDA = +12 atm L = +1215 J Because it is a cycle, Δ U = 0 and the first law implies that Q = W. 0 = Δ U = Q - W ⟹ Q = W = +1215 J (e) If 0.30 kcal of heat flow out of the system as it goes from point B to point A, then what is the change in internal energy for this process? Solution We are given the heat that flows out so the heat that flow in is negative and we have to convert kcal to J. Q = -0.30 kcal × 4186 J kcal = -1256 J 10 | Chapter N - Ideal Gases and the First Law of Thermodynamics P ( atm ) 8 A 2 B 2 V (L) 6 W BA We calculated the work in part (b) as the negative of the area under the trapezoid. W = WBA = -2026 J The first law of thermodynamics gives us the change in internal energy. Δ U = Q - W = +770 J