Thermodynamics lecture 5

BITS Pil iBITS PilaniPilani Campus

L t 4 Ph B h i f PLecture 4 – Phase Behavior of PureSubstance, Gases,

P-v-T Surface

Projections on P‐T T‐v and P‐v planes

BITSPilani, Pilani Campus

Projections on P‐T, T‐v, and P‐v planesVarious features

Gibbs Phase Rule• Had already briefly discussed it without giving the namename • F = C – P + 2, where C is the number of components or chemical constituents, P the pnumber of phases, and F the number of degrees of freedom

F i h b f i i i bl hi h• F is the number of intensive variables which may be freely chosen, and must be so chosen before the system is in a determined statethe system is in a determined state• For C = 1, F = 3 – P, F= 2 if P = 1, 1 if P = 2, and 0 if P = 3


0 3• Does not make any reference to ‘amounts’

T-v liquid-vapor equilibrium - Quality

As one moves from left to right in the saturation region at a given T and P, the relative amount of vapor grows at the expense of the liquid., p g p qQuality x = mg/m, fraction of total mass present as vapor, varies from 0 to 1 along tie lineAt any point on the line, v = (1-x)vf + xvg = vf + xvfg where vf and vg are


g g gthe specific volumes of the liquid and vapor phases respectively, and vfg= vg - vf

State of liquid-vapor system

• T and P not independent in two phase equilibrium i f b tregions for pure substance

• Only one degree of freedom, and so only one d t b ifi d d it h t b fneeds to be specified, does it have to be one of

these?• Quality x as an ‘intensive’ property to ‘fully fix’• Quality x, as an intensive property to fully fix state of two phase system along with T or p• State postulate two independent intensive• State postulate – two independent intensive variables fix state of pure substance• Example• Example


Example: Water saturation regionTemp (C) Sat P (kPa) vf (m3/kg) vf g(m3/kg) vg (m3/kg)

95 84.6 0.001040 1.98186

100 101.3 0.001044 1.67185 1.67290

105 120.8 0.001047 1.41936

Three rigid containers of volume 1.0 m3, each contain water at 100ºC, with respective quality values 0.500, 0.001263, and 0.00010. Find the mass in each. The contents of each container are heated till they just

Container Quality v(m3/kg) Remark Final State

become homogeneous. Identify the final state in each case, also finding the final pressure.

Container Quality v(m3/kg) Remark Final State1 0.500 0.83697 > vc Sat. vapor2 0.001263 0.003155 = vc Critical


3 0.000100 0.001211 < vc Sat. liquid

Example (contd.) – final state

Temp (C) Sat P (kPa) vf (m3/kg)230 2794 9 0 001209230 2794.9 0.001209235 3060.1 0.001219231 2848 0.001211

Container 3

Temp (C) P (kPa) v (m3/kg)374 1 22089 0 003155

Container 2374.1 22089 0.003155

Temp (C) Sat P (kPa) vg(m3/kg)Temp (C) Sat P (kPa) vg(m /kg)120 198.5 0.89186125 232.1 0.77059

122 3 214 0 83697Container 1


122.3 214 0.83697

Example (contd.)• The contents of container 1, now in the saturated vaporstate at 122. C are now heated at constant volume to 150 C. Find the final pressureThe final state is in the superheated steam region

Temp (C) P (kPa) vg(m3/kg)150 200 0.93964150 300 0.63388150 233.6 0.83697


The Ideal Gas• PV = nȒT or Pṽ = ȒT

Ȓ 1 1• Ȓ is the universal gas constant 8.3145 kJ kmole-1 K-1

• n = m/M

• PV = mRT, where R is the gas constant per unit mass of a

given gas, 0.2968 kJ/kg K for N2, 0.1889 kJ/kg K for CO2 etcg g g 2 g 2

• Obeyed as a limiting law by gases at low density where

inter-molecular forces are negligibleinter-molecular forces are negligible

• What constitutes low density, and by how much does a

l d i t f id l b h i t i T d P?real gas deviate from ideal behaviour at given T and P?


Compressibility Factor Z• Z = Pv/RT• Z = 1 for ideal gas. Note below that Z → 1 as P → 0g• What do Z > 1 and Z < 1 signify?


Generalized compressibility factor chart


Thermodynamics lecture 5

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