Thermodynamics lecture 5
- 1. BITS Pil i Pilani Pilani Campus Lecture 4 – Ph L t Phase Behavior of Pure B h i fP Substance, Gases ,
- 2. P-v-T Surface Projections on P‐T, T‐v, and P‐v planes Projections on P‐T T‐v and P‐v planes Various features BITSPilani, Pilani Campus
- 3. Gibbs Phase Rule • Had already briefly discussed it without giving the name • F = C – P + 2, where C is the number of components or chemical constituents, P the p number of phases, and F the number of degrees of freedom • F is the number of intensive variables which may i h b fi i i bl hi h be freely chosen, and must be so chosen before the 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 • Does not make any reference to ‘amounts’ BITSPilani, Pilani Campus
- 4. 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 q Quality x = mg/m, fraction of total mass present as vapor, varies from 0 to 1 along tie line At any point on the line, v = (1-x)vf + xvg = vf + xvfg where vf and vg are g the specific volumes of the liquid and vapor phases respectively, and vfg = vg - vf BITSPilani, Pilani Campus
- 5. State of liquid-vapor system • T and P not independent in two phase equilibrium regions f pure substance i for b t • Only one degree of freedom, and so only one needs t be specified, d d to b ifi d does it h have t b one of to be f these? • Quality x as an ‘intensive’ property to ‘fully fix’ x, intensive fully fix state of two phase system along with T or p • State postulate – two independent intensive variables fix state of pure substance • Example BITSPilani, Pilani Campus
- 6. Example: Water saturation region Temp (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 become homogeneous. Identify the final state in each case, also finding the final pressure. Container Quality v(m3/kg) Remark Final State 1 0.500 0.83697 > vc Sat. vapor 2 0.001263 0.003155 = vc Critical 3 0.000100 0.001211 < vc Sat. liquid BITSPilani, Pilani Campus
- 7. Example (contd.) – final state Temp (C) Sat P (kPa) vf (m3/kg) 230 2794.9 2794 9 0.001209 0 001209 Container 3 235 3060.1 0.001219 231 2848 0.001211 Temp (C) P (kPa) v (m3/kg) Container 2 374.1 374 1 22089 0.003155 0 003155 Temp (C) Sat P (kPa) vg(m3/kg) 120 198.5 0.89186 Container 1 125 232.1 0.77059 122.3 122 3 214 0.83697 0 83697 BITSPilani, Pilani Campus
- 8. Example (contd.) • The contents of container 1, now in the saturated vapor state at 122. C are now heated at constant volume to 150 C. Find the final pressure The final state is in the superheated steam region Temp (C) P (kPa) vg(m3/kg) 150 200 0.93964 150 300 0.63388 150 233.6 0.83697 BITSPilani, Pilani Campus
- 9. The Ideal Gas • PV = nȒT or Pṽ = ȒT • Ȓ is the universal gas constant 8.3145 kJ kmole-1 K-1 1 1 • n = m/M • PV = mRT, where R is the gas constant per unit mass of a g given g gas, 0.2968 kJ/kg K for N2, 0.1889 kJ/kg K for CO2 etc g g • Obeyed as a limiting law by gases at low density where inter-molecular forces are negligible • What constitutes low density, and by how much does a real gas d i t f l deviate from id l b h i ideal behaviour at given T and P? t i d BITSPilani, Pilani Campus
- 10. Compressibility Factor Z • Z = Pv/RT • Z = 1 for ideal gas. Note below that Z → 1 as P → 0 g • What do Z > 1 and Z < 1 signify? BITSPilani, Pilani Campus
- 11. Generalized compressibility factor chart BITSPilani, Pilani Campus