Equations of state Active study of gases is done by changing pressure, volume, temperature, or quantity of material and observing the result. pV/nT (J/mol K)
8.60 H2 8.40 N2 8.20 CO 8.00 O 7.80 2
10 20 30 P (atm)
Ideal-gas
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In reality, there are no gases that fit this definition perfectly. We assume that gases are ideal to simplify our calculations
Ideal gas regime: low pressure/high temperatures/low densities
Ideal-gas equation
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pV nRT ¡
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mtotal nM
%
© ¤ m pV = total RT
M
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mtotal /V
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© ¤ ρ = pM
RT
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Ideal-gas equation
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pV nRT ¡
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¥ = , N nN A +
% R
© ¤ pV = N T = NkT N A
The constant term R/NA is referred to as Boltzmann's constant, in honor of the Austrian physicist Ludwig Boltzmann (1844–1906), and is represented by the symbol k:
Ideal-gas equation (contd.)
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T1 T2
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© pV = const . Boyle’s law
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CPS question A quantity of an ideal gas is contained in a balloon. Initially the gas temperature is 27°C. You double the pressure on the balloon and change the temperature so that the balloon shrinks to one-quarter of its original volume. What is the new temperature of the gas?
A. 54°C B. 27°C C. 13.5°C D. –123°C E. –198°C CPS question p This pV –diagram shows three possible states of a certain amount of an ideal gas. 3
Which state is at the highest 2 temperature? 1 A. state #1 B. state #2 V O C. state #3 D. Two of these are tied for highest temperature. E. All three of these are at the same temperature.
Summary:Equations of state
pV = nRT
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¡ ¤ ¥ © ¥ © pV = NkT
-23 k: Boltzmann’s constant ; k =R/N A=1.38x10 J/K Kinetic theory of an ideal gas
Pressure of a gas
The pressure that a gas exerts is caused by the impact of its molecules on the walls of the container. Consider a cubic container of volume V containing N molecules with a speed v Consider a gas molecule colliding elastically with the right wall of the container and rebounding from it.
The force on the wall is obtained using Newton’s second law as follows , ∆P F = , ∆P: change in momentum ∆t And the pressure: 1 1 ∆P P = F = , A A ∆t
Total change in momentum ∆P of all the molecules during a time interval ∆t = Change in momentum of one molecule, times the number of molecules that hit the wall during the interval ∆t: Change in momentum of one molecule: ∆ = − − = p mv x ( mv x ) 2mv x
Change in momentum of all the molecules: ∆ = × P 2( mv x ) Nhit
N 1 N N = v ∆tA ∆P = mv 2∆tA hit V x 2 V x
Finally, the pressure is given by: 1 ∆P N P = = mv 2 A ∆t V x 2 2 Particles move in random directions, so we replace vx by ( vx )av N P = m(v2 ) V x av
or, equivalently 1 PV = 2N( mv 2 ) 2 x av
Molecular interpretation of temperature 1 PV = NkT = 2N( mv 2 ) 2 x av or 1 1 ( mv 2 ) = kT 2 x av 2 There is nothing special about the x direction, in general: 2 = 2 = 2 (vx )av (vy )av (vz )av and 2 = 2 + 2 + 2 = 2 (v )av (vx )av (vy )av (vz )av (3 vx )av
Therefore: The average kinetic energy of a molecule is: 1 3 E = ( mv 2 ) = kT Kin 2 av 2 The absolute temperature is a measure of the average translational kinetic energy of the molecules CPS question
An ideal gas is trapped in a chamber, inside a thermally insulated container (see figure). When the partition is broken or removed, the gas expands and fills the entire volume of the container. As a result, the final temperature of the gas after the expansion is: A. Lower than the initial temperature B. Higher than the initial temperature C. The temperature remains constant
Equipartition theorem 1 3 E = ( mv 2 ) = kT Kin 2 av 2
When a substance is in equilibrium, there is an average energy of 1/2 kT per molecule, or 1/2RT per mole, associated with each degree of freedom
z y
x Molar heat capacity
Kinetic energy per mole: 3 E = RT Kin 2 = dQ dE Kin
3 C dT = RdT V 2
¡ 3 C = R V 2
¢ 3 C = .8( 314 J/mol.K ) = 12 47. J/mol.K V 2
Molar heat capacity
Kinetic energy per mole: 3 E = RT Kin 2 = dQ dE Kin
3 C dT = RdT V 2
¡ 3 C = R V 2
¢ 3 C = .8( 314 J/mol.K ) = 12 47. J/mol.K V 2 Works well for monoatomic gases… what’s wrong with the others…? Heat absorption into degrees of freedom
A molecule can absorb energy in translation, and also in rotation and vibrations in its structure
CPS question The molar heat capacity at constant volume of diatomic
hydrogen gas (H 2) is 5 R/2 at 500 K but only 3 R/2 at 50 K. Why is this?
A. At 500 K the molecules can vibrate, while at 50 K they cannot. B. At 500 K the molecules cannot vibrate, while at 50 K they can. C. At 500 K the molecules can rotate, while at 50 K they cannot. D. At 500 K the molecules can rotate, while at 50 K they cannot. Heat capacity of solids Solids can absorb energy in the vibrational modes. A 3-dimensional solid has 3 vibrational degrees of freedom+3 translational… = CV 3R
¡ = = CV .8(3 314 J/mol.K ) 24 9. J/mol.K
Dulong and Petit’s law
(Valid at high temperature due to
quantum nature of the vibrations)
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2kT 2RT v = = mp m M
The fraction of molecules dN with speeds between v and v+dv is given by dn = f (v)dv The average speeds are given by: ∞ = = 8kT vave vf (v)dv ∫0 πm ∞ 2 2 3kT v ave = v f (v)dv = ∫0 m