Chapter 2

The First Law

Unit 5 state function and exact differentials

Spring 2009

State function and Path function

State function

a property that is independent of how a sample is prepared.

example : T, P, U, H

Path function

a property that is dependent on the preparation of the state.

depends on the path between the initial and final states

example : W, q

Example 2.7
Calculating work, heat, and internal energy

Consider a perfect gas inside a cylinder fitted with a piston. Let the initial state be T, Vi and the final state be T,Vf. The change of state can be brought about in many ways, of which the two simplest are the following:

Path 1, in which there is free expansion against zero external pressure;

Path 2, in which there is reversible, isothermal expansion.

Calculate w, q, and U and DHfor each process.

Example 2.7

Path 1isothermal free expansion

Isothermal DU=0, DH=0

DU=q+w = 0q=-w

free expansion w = 0, q=0

Path 2isothermal reversible expansion

Isothermal DU=0, DH=0

DU=q+w = 0q=-w

reversible expansion

Self Test 2.8

Calculate the values of q, w, and U, DH for an irreversible isothermal expansion of a perfect gas against a constant nonzero external

Irreversible isothermal expansion

Isothermal DU=0, DH=0

DU=q+w = 0q=-w

Irreversible expansion w = - Pex DV , q= Pex DV

Change in internal energy, DU

Change in internal energy, DU

Internal pressure

Constant-pressure heat capacity

Internal pressure

The variation of the internal energy of a substance as its volume is changed at constant temeperature.For a perfect gas pT = 0 For real gases

attractive forcepT > 0

repulsive forcepT < 0

Internal pressure

Joule experiment

Expands isothermally against vacuum (pex=0)

w=0, q=0 so DU=0

and pT=0

DU at constant pressure

Expansion coefficient (a):

the fraction change in volume with a rise in temperature

Isothermal compressibility (kT):

the fractional change in volumewhen the pressure increases in small amount

E 2.32 b

The isothermal compressibility of lead at 293 K is 2.21 106 atm1. Calculate the pressure that must be applied in order to increase its density by 0.08 per cent.

Example 2.8
Calculating the expansion coefficient of a gas

Derive an expression for the expansion coefficient of a perfect gas.

DU at constant pressure

For perfect gas pT = 0,

Change in enthalpy, DH

(chain relation)

Joule-Thomson coefficient

m =

Joule-Thomson coefficient, m

A vapour at 22 atm and 5C was allowed to expand adiabatically to a final pressure of 1.00 atm; the temperature fell by 10 K. Calculate the JouleThomson coefficient, , at 5C, assuming it remains constant over this temperature range.

Joule-Thomson coefficient, m

For perfect gases m = 0For real gases

m > 0gas cools on expansion

m < 0gas heats on expansion

Inversion temperature

Exercise 2.29a

When a certain freon used in refrigeration was expanded adiabatically from an initial pressure of 32 atm and 0C to a final pressure of 1.00 atm, the temperature fell by 22 K. Calculate the JouleThomson coefficient, , at 0C, assuming it remains constant over this temperature range.

Joule-Thomson effect
Cooling by isenthalpic expansion

Adiabatic process q=0, DU=wPi > PfOn the left

isothermal irreversible compression

Pi,Vi,Ti Pi,0,Ti

w1= -pi ( 0 - Vi )= pi Vi

On the right

isothermal irreversible expansion

Pf,0,Tf Pf,Vf,Tf

w2= -pf ( Vf - 0 )= -pf Vf

Joule-Thomson effect
Cooling by isenthalpic expansion

w = w1 + w2 = pi Vi - pf Vf w = DU=Uf -Ui = pi Vi - pf Vf Uf + pf Vf = Ui + pi Vi Hf = Hi

Joule-Thomson effect is an isenthalpic process

Isothermal Joule-Thomson coefficient

Liquefaction of gases

Liquefaction of gases

Review 1

Define internal pressure pTProve that, for ideal gas, pT = 0

Review 2

Define Expansion coefficient aDefine Isothermal compressibility kTProve that for ideal gas

a= 1/T

kT= 1/p

Review 3

Define Joule-Thomsom coefficientProve that Joule-Thomson experiment is an isentahlpic process. Explain the principle of using Joule-Thomson effect to liquefy gases.

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