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, V i and the final state be T,V f. 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  Hfor each process.

Example 2.7 Path 1isothermal free expansion Isothermal  U=0,  H=0  U=q+ w = 0q=- w free expansion w  q  Path 2isothermal reversible expansion Isothermal  U=0,  H=0  U=q+ w = 0q=- w reversible expansion

Self Test 2.8 Calculate the values of q, w, and ∆U,  H for an irreversible isothermal expansion of a perfect gas against a constant nonzero external Irreversible isothermal expansion Isothermal  U=0,  H=0  U=q+ w = 0q=- w Irreversible expansion w  P ex  V  q  P ex  V

Change in internal energy,  U

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  T  For real gases attractive force  T  repulsive force  T 

Internal pressure

Joule experiment Expands isothermally against vacuum (p ex =0) w=0, q=0 so  U=0 and  T =0

 U at constant pressure Expansion coefficient  the fraction change in volume with a rise in temperature Isothermal compressibility    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 × 10 −6 atm −1. 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.

 U at constant pressure For perfect gas   

Change in enthalpy,  H (chain relation) Joule-Thomson coefficient  =

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

Joule-Thomson coefficient,  For perfect gases  = 0 For real gases  0gas cools on expansion  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 0°C to a final pressure of 1.00 atm, the temperature fell by 22 K. Calculate the Joule–Thomson coefficient, µ, at 0°C, assuming it remains constant over this temperature range.

Joule-Thomson effect Cooling by isenthalpic expansion Adiabatic process q=0,  U= w P i > P f On the left isothermal irreversible compression P i,V i,T i → P i,0,T i w 1 = -p i ( 0 - V i )= p i V i On the right isothermal irreversible expansion P f,0,T f → P f,V f,T f w 2 = -p f ( V f - 0 )= -p f V f

Joule-Thomson effect Cooling by isenthalpic expansion w = w 1 + w 2 = p i V i - p f V f w =  U=U f -U i = p i V i - p f V f U f + p f V f = U i + p i V i H f = H i Joule-Thomson effect is an isenthalpic process

Isothermal Joule-Thomson coefficient

Liquefaction of gases

Review 1 Define internal pressure  T Prove that, for ideal gas,  T = 0

Review 2 Define Expansion coefficient  Define Isothermal compressibility  T Prove that for ideal gas  = 1/T  T = 1/p

Review 3 Define Joule-Thomsom coefficient Prove that Joule-Thomson experiment is an isentahlpic process. Explain the principle of using Joule-Thomson effect to liquefy gases.