Chapter 15 Thermal Properties of Matter

Transcription

1 Chapter 15 Thermal Properties of Matter To understand the mole and Avogadro's number. To understand equations of state. To study the kinetic theory of ideal gas. To understand heat capacity. To learn and apply the first law of thermodynamics. To study thermodynamic processes. To understand the properties of an ideal gas.

2 15.1 The Mole and Avogadro s Number Because atoms and molecules are so small, any practically meaningful amount of a substance contains a huge number of atoms or molecules. Therefore, it is more convenient to use a rather huge measuring unit to count their numbers. The Avogadro s number, N A = molecules/mole, is such a measuring unit. 1 mole of a pure chemical element or compound contains N A = identical atoms or molecules. The molar mass (M) is the mass of 1 mole of a pure chemical element or compound. It is equal to the Avogadro s number multiplying the mass of an atom or molecule (m). M = N A m The total mass of a system containing n moles of a substance: m total = nm The total number of particles in n moles of a substance: N total = nn A Example: Carbon-12 ( 12 6 C) m = g (a) The molar mass M = N A m = 12.0 g (b) The total mass of 1.5 moles of 12 6 C m total = nm = 18.0 g (c) Number of atoms in 1.5 moles of 12 6 C N total = nn A =

3 15.2 Equation of State Imagine that we can work on the device on the right. We may take different actions and expect some results: If we heat it up, temperature (T) rises and the volume (V) expands; if we compress it, pressure (p) increases; if we add more gas (n) into the system, pressure (p) increases and volume expands (V), etc. Question: how are these physical quantities (T, p, V, n, etc.) related to each other for a given system? The equation of state: a mathematical equation relates these physical quantities to each other. These physical quantities are also known as state variables or state coordinates.

4 The Ideal-Gas Equation or Ideal-Gas Law For most gases, their state variables very closely obey a simple relationship: pv = nrt. R = J/(mol K) is the ideal-gas constant and is a universal constant for all gasses. T is measured in Kelvin (K). The above relationship is known as the ideal-gas equation, or, the ideal-gas law. We may re-write the ideal-gas equation, replacing n by m total /M, pv = m total RT M We may write the ideal-gas equation in terms of the density of the gas, ρ = m total /V, ρ = pm RT

5 pv Diagram T 4 > T 3 > T 2 > T 1 Ideal Gas Non-ideal gas p = nrt/v pv = nrt

6 pt Phase Diagram Triple point and critical point

7 15.3 Kinetic Theory of an Idea Gas The question: how are measurable macroscopic variables related to microscopic properties of the atoms and molecules? The idea gas: We will treat atoms or molecules as point particles undergoing rapid elastic collisions with each other and the walls of the container in the given volume. The potential energies due to all the forces are ignored. The process is to apply Newton s laws to establish the relationship between microscopic and macroscopic quantities. The goals are to understand: the pressure of an ideal gas; the ideal-gas equation; the temperature of an ideal gas; internal energy of an ideal gas; the heat capacity of an ideal gas; etc.

8 Kinetic Molecular Theory of an Idea Gas Pressure (the impulse of molecule collision with container wall) (a) momentum change in one collision event: 2m v x (b) total momentum change in time interval t: P x = 1 N A v 2 V x t 2m v x = NAmv x 2 t V (c) Force on the wall is: (d) Pressure on the wall is: F x = P x p = F x = NAmv 2 x t V 2 = Nmv x A V (15.6) (e) Molecules with a distribution of velocities? Take average. p = F x = Nm(v x 2 ) av A V = N V 1 3 m(v2 ) av = 1 V 2 3 K tr, where K tr = N[ 1 2 m(v2 ) av ] is the total kinetic energy. (f) Compare with the ideal-gas equation: pv = nrt, we have the total kinetic energy of the gas molecules, K tr = 3 nrt (15.7) 2

9 The Boltzmann Constant The total kinetic energy of all the particles in an ideal gas is K tr = 3 2 nrt. It relates the microscopic properties to measurable macroscopic quantities. The kinetic is independent of the mass of the atoms or molecules. The average kinetic energy per atom or molecule is K av = 1 2 m(v2 ) av = K 3 tr 2 = nrt = 3 ( R )T. nn A nn A 2 N A Define the Boltzmann constant then, k = R N A = J/(mol K) /mole = J/K, K av = 1 2 m(v2 ) av = 3 2 kt.... (15.8) which is independent of the details of the particles, such as the mass. We can re-write the ideal-gas equation to or pv = nrt = n kn A T, pv = NkT.(15.9)

10 Molecular Speeds in an Ideal Gas The average kinetic energy per atom or molecule: K av = 1 2 m(v2 ) av = 3 2 kt, from which we obtain the root-mean-square velocity Maxwell-Boltzmann Distribution of Molecular Speed v rms = (v 2 ) av = 3kT m, where m is the mass of an atom or molecule. Since kn A = R and mn A = M, we may re-write, v rms = 3RT M. Note: (v 2 ) av (v av ) 2 = 0

11 15.4 The Molar Heat Capacities Note: In Chapter 14, we defined the specific heat as the heat energy required to raise the temperature of 1 kg of a substance by 1 ºC (or 1 K). Now, we are going to define a quantity called the molar heat capacity. Its meaning is similar to that of the specific heat, but it is defined in different units. Question: How much heat energy Q is needed to raise n moles of a substance by a temperature T? The answer is: Q = nc T Note: (a) The heat energy is proportional to n and T. (b) The proportionality constant C is called the molar heat capacity. It is the heart energy needed to raise the temperature of 1 mol of a substance by 1 ºC (or 1 K). (c) C is in general material-dependent. Yet, it has some simple forms for an ideal gas. (d) The molar heat capacity has the units of J/(mol K). (e) Q is defined positive if it is transferred into the system, and, negative otherwise. (f) Relationship with the specific heat c: C = Mc (M is the molar mass)

12 Constant-Volume Molar Heat Capacity of an Ideal Gas Consider an ideal gas with its volume fixed, when heat energy Q is added to or remove from the ideal gas, the total kinetic energy K tr = 3 nrt is changed by the 2 same amount based on the conservation of energy (because, for an ideal gas, the potential energies due to all the forces are ignored): Q = K tr = 3 2 nr T = n(3 2 R) T. Therefore, the constant-volume molar heat capacity is C V = 3 R = 12.5 J/(mol K) 2 Note: (a) The constant-volume molar heat capacity C V given above is correct for monatomic gases and is independent of the details of the atoms, such as atomic masses. (b) For gases of diatomic molecules, C V = 5 R = 20.8 J/(mol K) 2

13 15.5 The First Law of Thermodynamics It sets the relationship between the change in the internal energy U, work done by the system W, and heat transfer. Q U W U = Q W Note: (a) Q is positive if added to the system; negative if removed from the system. (b) W is positive if it is done by the system on the surrounding; negative otherwise.

14 Work done during volume change Work: W = F x = pa x = p V Work done at constant pressure: W = p V 2 V 1 When the pressure is not a constant: W = p 1 V + p 2 V + p 3 V +. which is the area under the pv diagram V

15 Work is the area under the pv diagram Work done at constant temperature W = nrtln V 2 V 1

16 Work is the area under the pv diagram. It depends on the exact path that the system follows.

17 Example on page 483 Given: (1) from a to b, Q ab = 150 J of heat is added to the system (2) From b to d, Q bd = 600 J of heat is added to the system Find: (1) internal energy change from a to b (2) internal energy change from a to b to d (3) total heat added to system from a to c to d Solution: (1) From a to b W = 0 U ab = Q ab W = 150 J (1) From b to d, constant pressure, W bd = p V d V b = ( )( ) = 240 J U abd = Q abd W abd = ( ) ( ) = 510 J The First Law U = Q W (3) Q acd = U acd + W acd = U abd + W ac = ( )( ) = = 600 J

18 15.6 Thermodynamics Processes U = Q W Adiabatic: no heat transfer in or out of the system Q = 0 U = W Isochoric: no volume change V = constant or V = 0 or W = 0 U = Q Isobaric: no pressure change. p = constant W = p V 2 V 1 Isothermal: no temperature change. T = constant or U = 0 Q = W

19 15.7 Properties of an Ideal Gas U = Q W We learned in Section 15.4 that the constant-volume molar heat capacity for a monatomic ideal gas is C V = 3 R = 12.5 J/(mol K) 2 Question: What is the molar heat capacity for a monatomic ideal gas under a constant pressure? [ p = constant; W = p V 2 V 1 = p V ] Q = U + W = nc V T + p V Since pv = nrt, or p V = nr T We have Q = nc V T + p V = nc V T + nr T = n C V + nr T = nc p T. The constant-pressure molar heat capacity is C p = C p + R = 5 2 R