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CHAPTER 3The Importance of State Functions: Internal Energy (U) and Enthalpy (H)

How do U and H depend on experimental parameters like

P, V, and T?

Three examplesP = P(V,T)U = U(V,T)H = H(P,T)

Mathematical Properties of State Functions• U and H are state functions

• dU and dH are exact differentials

Partial derivatives

= (R/V)dT (RT/V2)dV

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Second or Higher Order Derivatives

State Functions

• The order in which the function is differentiated does not affect the outcome

• If f is a state function, f = ʃdf = ffinal – finitial. That is, the change of the state function is defined only by the initial and final states, independent of the path leading to the change

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Additional Frequently Used Properties

Cyclic rule

Reciprocal rule

• = isobaric volumetric thermal expansion coefficient

• = isothermal compressibility

Applied to any system (gas,

liquid, or solid)

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Example 3.2

• You have accidentally arrived at the end of the range of an ethanol-in-glass thermometer so that the entire volume of the glass capillary is filled. By how much will the pressure in the capillary increase if the temperature is increased by another 10 C? glass =2.00 × 105 (C)1, ethanol = 11.2 ×104 (C)1 and ethanol = 11.0 × 105 (bar)1. Will the thermometer survive your experiment?

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U = U(V,T) Dependence of U on V and T

• U is a state function, so dU is an exact differential

First law of thermodynamics(only PV work)

Dependence on T

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• Applicable to systems containing gases, liquids, or solids in a single phase (or mixed phases at a constant composition) if no chemical reactions or phase changes occur

• Interactions between molecules in the system

• From section 5.12 (second law of thermodynamics)

Internal Pressure

Dependence on V and T

• U is a state function

• Path is unimportant

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Does U depend more strongly on V or T?

• For an ideal gas, =0, thus U is independent of V but depends only on T

• This equation holds even if the volume is not constant

• Pexternal = 0, so w = 0

• Water bath and system are in thermal equilibrium, dq = 0

James Joule experiment

= 0

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van der Waals gas (Example 3.5)

P

= (R/V)dT (RT/V2)dV ideal gas

any system

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For solids and liquids

• Under most conditions encountered by chemists in the laboratory, U can be regarded as a function of T alone for all substances

H = H(P,T) Variation of H with T at const P• With only PV work

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Phase transition at const P

• Vaporization and sublimation

• Melting∆ ∆ ∆ 0

H = U + PV

H is a state function

(CP,m is independent of T)

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Example 3.7

• A 143 g sample of graphite is heated from 300 to 600 K at a constant pressure. Over this temperature range CP,m (J K-1 mol-1) has been determined to be – 12.9 + 0.1126T/K -1.947 × 10-4 T2/K2 + 1.919 10-7 T3/K3 – 7.800 × 10-11 T4/K4. Calculate H and qp.

How are CP and CV related?

at const P = Pexternal

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for ideal gas

, ,

Variation of H with P at const T

• H = U + PV

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For isothermal processes, T = 0

= 0

For Ideal Gases

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For liquids and solids

• 1 » Tso ≈ V

H dependence on T and P

• Under most conditions encountered by chemists in the laboratory, H can be regarded as a function of T alone for liquids and solids

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Summary

Any system

κ1κ

Ideal Gases

•

• dU = CVdT

• dH = CPdT1

• A gas is force through a porous plug using a piston and cylinder mechanism.

• The pistons move to maintain a constant P in each region.

• There is an appreciable drop of pressure across the plug and the temperature change of the gas is measured.

• The expansion is isenthalpic (const H)

Joule-Thomson Experiment

Adiabatic walls

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q = 0

Joule-Thomson Coefficient

• For the conditions of the experiment using N2, dT and dP< 0, so > 0

• Also it can be used to determine

0

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• Positive J-T: the attractive part of the potential is dominant

• Negative J-T: the repulsive part of the potential is dominant

Ideal and van der Waals gases

• Ideal gases, J-T = 0

• Van der Waals gases (in the limit of zero pressure)

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Liquefying Gases Using an Isenthalpic Expansion

J-T > 0

J-T < 0

J-T = 0

Contour Map

• Staring at point z, a person first moves in the positive x direction and then along the y direction. If dx and dyare sufficiently small, the change in height dz is given by

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Quiz #2

• (20 pts) Please show that for an ideal gas (PV = nRT), and . Hint: and

• (bonus 10 pts) derive b and k for a van der Waals gas