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Thermodynamics and Chemical Potential: A Comprehensive Set of Exercises with Solutions - P, Assignments of Physical Chemistry

A series of exercises covering fundamental concepts in thermodynamics and chemical potential, including maxwell relations, helmholtz free energy, clapeyron equation, and gibbs-helmholtz equation. Each exercise is accompanied by a detailed solution, providing a comprehensive guide for students to understand and apply these concepts. The exercises are designed to enhance problem-solving skills and deepen understanding of thermodynamic principles.

Typology: Assignments

2023/2024

Uploaded on 11/07/2024

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Chem 113A Homework #4
Solutions
1. Use the Maxwell relations to prove the following relations.
(๐œ•๐‘†
๐œ•๐‘‰)๐‘‡=๐›ผ
๐œ…๐‘‡
โˆ’(๐œ•๐‘†
๐œ•๐‘)๐‘‡=๐‘‰๐›ผ
Starting with ๐‘‘๐ป=๐‘‡๐‘‘๐‘†+๐‘‰๐‘‘๐‘
๐œ•2๐ป
๐œ•๐‘†๐œ•๐‘=๐œ•2๐ป
๐œ•๐‘๐œ•๐‘† โ†’ ๐œ•
๐œ•๐‘†(๐œ•๐ป
๐œ•๐‘)๐‘†=๐œ•
๐œ•๐‘(๐œ•๐ป
๐œ•๐‘†)๐‘ โ†’ (๐œ•๐‘‰
๐œ•๐‘†)๐‘=(๐œ•๐‘‡
๐œ•๐‘)๐‘†
By definition ๐›ผ and ๐œ…๐‘‡ are given by
๐›ผ=1
๐‘‰(๐œ•๐‘‰
๐œ•๐‘‡)๐‘ ๐‘Ž๐‘›๐‘‘ ๐œ…๐‘‡=โˆ’1
๐‘‰(๐œ•๐‘‰
๐œ•๐‘)๐‘‡
Thus,
(๐œ•๐‘‰
๐œ•๐‘‡)๐‘=๐›ฝ๐‘‰
Starting with ๐‘‘๐บ=โˆ’๐‘†๐‘‘๐‘‡+๐‘‰๐‘‘๐‘
๐œ•2๐บ
๐œ•๐‘‡๐œ•๐‘=๐œ•2๐บ
๐œ•๐‘๐œ•๐‘‡ โ†’ ๐œ•
๐œ•๐‘‡(๐œ•๐บ
๐œ•๐‘)๐‘‡=๐œ•
๐œ•๐‘(๐œ•๐บ
๐œ•๐‘‡)๐‘ โ†’ (๐œ•๐‘‰
๐œ•๐‘‡)๐‘=โˆ’(๐œ•๐‘†
๐œ•๐‘)๐‘‡
โˆดโˆ’(๐œ•๐‘†
๐œ•๐‘)๐‘‡=(๐œ•๐‘‰
๐œ•๐‘‡)๐‘=๐›ฝ๐‘‰
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Chem 113A Homework # 4

Solutions

  1. Use the Maxwell relations to prove the following relations.

๐‘‡

๐‘‡

๐‘‡

Starting with ๐‘‘๐ป = ๐‘‡๐‘‘๐‘† + ๐‘‰๐‘‘๐‘

2

2

๐‘†

๐‘

๐‘

๐‘†

By definition ๐›ผ and ๐œ…

๐‘‡

are given by

๐‘

๐‘‡

๐‘‡

Thus,

๐‘

Starting with ๐‘‘๐บ = โˆ’๐‘†๐‘‘๐‘‡ + ๐‘‰๐‘‘๐‘

2

2

๐‘‡

๐‘

๐‘

๐‘‡

๐‘‡

๐‘

  1. Show that

[

]

๐‘‰

  1. Solve the differential equation derived in Problem 2 by integration from T 1

to T 2

to obtain an

expression that would allow you to relate โˆ†๐ด at the two temperatures. This is similar to what

was done in lecture for โˆ†๐บ.

  1. It is possible to derive an expression for the chemical potential in terms of the Helmholtz free

energy ๐ด = ๐‘ˆ โˆ’ ๐‘‡๐‘†. To accomplish this begin with the expression for ๐‘‘๐บ for two species 1

and 2.

1

2

๐‘–

๐‘–

2

๐‘–= 1

The Helmholtz energy is the Legendre transform ๐ด = ๐บ โˆ’ ๐‘๐‘‰. First find ๐‘‘๐ด, then plug the

above expression for ๐‘‘๐บ into your result for ๐‘‘๐ด. The new expression for ๐‘‘๐ด is a function

of ๐‘‡, ๐‘‰, ๐‘›

1

, and ๐‘›

2

. Using this expression for ๐‘‘๐ด define the quantity ๐œ‡

๐‘–

(๐‘– = 1 , 2 ) in terms of

partial molar quantities of A.

๐‘‘๐ด = ๐‘‘(๐บ โˆ’ ๐‘ƒ๐‘‰) = ๐‘‘๐บ โˆ’ ๐‘‘(๐‘ƒ๐‘‰) = ๐‘‘๐บ โˆ’ ๐‘ƒ๐‘‘๐‘‰ โˆ’ ๐‘‰๐‘‘๐‘ƒ = โˆ’๐‘†๐‘‘๐‘‡ + ๐‘‰๐‘‘๐‘ƒ + โˆ‘ ๐œ‡

๐‘–

๐‘‘๐‘›

๐‘–

2

๐‘–= 1

โˆ’ ๐‘ƒ๐‘‘๐‘‰ โˆ’ ๐‘‰๐‘‘๐‘ƒ

skater of mass 85 kg is balanced on one skate, what pressure is exerted at the

interface of the skate and the ice?

(c) What is the melting point of ice under this pressure?

(d) If the temperature of the ice is - 5.0 ยฐC, do you expect melting of the ice to occur

at the ice-skate interface?

No, the lowering of the melting temperature is less than the temperature of the

ice.

  1. The change in enthalpy is given by ๐‘‘๐ป = ๐ถ ๐‘

๐‘‘๐‘‡ + ๐‘‰๐‘‘๐‘, which implies that for enthalpy

changes โˆ†๐ป we have the differential ๐‘‘(โˆ†๐ป) = โˆ†๐ถ

๐‘

๐‘‘๐‘‡ + โˆ†๐‘‰๐‘‘๐‘. The Clapeyron equation

relates ๐‘‘๐‘ and ๐‘‘๐‘‡ at equilibrium, and so in combination the two equations can be used to

find how โˆ†๐ป changes along a phase boundary as the temperature changes and the two

phases remain in equilibrium. Show that ๐‘‘

๐‘

ln ๐‘‡

Begin with the Clapeyron Equation

Plug this into the equation for ๐‘‘(โˆ†๐ป)

๐‘

๐‘

๐‘

Thus,

๐‘

Now, we take the equation we are trying to prove and manipulate it into something

more intuitive

๐‘

๐‘‘(ln ๐‘‡) =

๐‘

๐‘

This is a better representation of what we want to prove. Begin by evaluating the

derivative

2

2

Next we plug our expression for ๐‘‘

/๐‘‘๐‘‡ obtained above

๐‘

2

๐‘

2

2

๐‘

Q.E.D.

  1. Calculate ๏„ r

Gยฐ(375 K) for the reaction 2 ๐ถ๐‘‚

2

2

(๐‘”) from the value of

r

Gยฐ( 298 K), ๏„

r

Hยฐ( 298 K) and the Gibbs-Helmholtz equation.

๐‘Ÿ

ยฐ

๐‘Ÿ

ยฐ

2

๐‘Ÿ

ยฐ

๐‘Ÿ

ยฐ

๐‘Ÿ

ยฐ

2

๐‘Ÿ

ยฐ

According to the Gibbs-Helmholtz Equation

๐‘Ÿ

2

2

๐‘Ÿ

1

1

๐‘Ÿ

2

1

๐‘Ÿ

ยฐ

2

2

1

๐‘Ÿ

ยฐ

1

๐‘Ÿ

ยฐ

2

1

๐‘Ÿ

ยฐ

2