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P = RT

P = RT. EQUATION OF STATE FOR IDEAL GAS p = RT (11.1). [units of Kelvin or Rankine]. = unique constant for each gas. low density - if average distance between molecules is 10 diameters or more, then very weak attractive forces approximately point non-interacting particles.

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P = RT

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  1. P = RT

  2. EQUATION OF STATE FOR IDEAL GAS p = RT (11.1) [units of Kelvin or Rankine] = unique constant for each gas low density - if average distance between molecules is 10 diameters or more, then very weak attractive forces approximately point non-interacting particles

  3. p = RT Good to 1% for air at 1 atm and temperatures > 140 K (-130 oC) or for room temperature and < 30 atm At large pressures, great departure from ideal gas equation of state.

  4. cp/cv = k = 1.4 for perfect gas = k

  5. cp/cv = k = 1.4 for perfect gas = k

  6. IDEAL GAS ~ SOME HISTORY PV = constant

  7. IDEAL GAS: p = RT (eq. 1.11) R = Runiv/mmole 1662: Boyle (and Hooke)experimentally showed that: PV = const for const T;

  8. New Experiments 1662 ~ Robert Boyle

  9. Daniel Bernoulli ~ PV = const Hydrodynamics, 1738 (early theoretical approach) . . . . 2L2 L1 . . . . . . p1, n1, m1, vx1(T1) = v,L1 . . p2, n1 m1, vx1(T1) = v, L2=2L1

  10. Daniel Bernoulli ~ Hydrodynamics, 1738 PV = const ( v = constant) (system 1) If Ldoubled (system 2) but same v, then (# of collisions/sec)1 = v x (1 sec)/L (# of collisions/sec)2 = v x (1 sec)/2L (# of collisions/sec)2= ½ (# of collisions/sec)1

  11. Daniel Bernoulli p = F/A F  {# collisions / sec} p1 (# of collisions/sec)1/(L)2 p2 (# of collisions/sec)2/(2L)2 p2½ (# of collisions/sec1)/(2L)2 p2= 1/8 p1 Vol2= 8Vol1 p2Vol2 = p1Vol1 QED

  12. “ The elasticity of air is not only increased by compression but by heat supplied to it, and since it is admitted that heat may be considered as an increasing internal motion of the particles, it follows that … this indicates a more intense motion of the particles of air.” Daniel Bernoulli Here was the recipe for quantifying the idea that heat is motion – two generations before Count Rumford, but it came too early.

  13. IDEAL GAS ~ SOME HISTORY PV  <K.E.>

  14. Assume perfect elastic reflections (ideal gas) so: - 2mvx is change of x-momentum per collision. Initially assume vx is same for all particles.

  15. Force of one particle impact = Magnitude of momentum change per second due to one particle: = (mvx)/t =2mvx/(2L/vx) = mvx2/L Time between collisions, t, of particle with same wall is equal to: t = 2L/vx L

  16. Magnitude of momentum change per second due to n molecules: nmvx2/L <vx2> = <vy2> = <vz2>; <vx2> + <vy2> + <vz2> = <v2> <vx2> = 1/3 1/3nm<v2>/L

  17. Pressure = F/A = [1/3 nm<v2>/L]/L2 P = 1/3 nm<v2>/L3 PV = 1/3 nm<v2> = 2/3 n (1/2m<v2>) average kinetic energy per particle

  18. PV = 1/3 nm<v2> = 2/3 n (1/2m<v2>) Experimental T(Ko) = [2/(3kB)][avg K.E.] n = # of particles; kB = Runiv/NAvag kB=1.38x10-23 J/K PV = 2/3 n (3/2 kB T)

  19. PV = (nkB) T(Ko) PV = NmolesRunivT(Ko) n = # of particles; kB = Runiv/NAvag kB=1.38x10-23 J/K

  20. IDEAL GAS LAW ~ Different Representations

  21. pV= NmRunivT p =(1/V)Nmmmole{Runiv/mmole}T p =(m/V){Runiv/mmole}T p = {Runiv/mmole}T p = RT (11.1)

  22. Runiv pV= NmRunivT pV = nkBT p = RT pv = RT pV= MRT R =Runiv/Mol. Mass ( M = total mass )

  23. Note: It is assumed that system always in equilibrium. Assume all gases obey ideal gas law: p =  RT Not gauge pressure Kelvin (or Rankine) AIR R = Runiv/MolMass = 287.03 m2/(s2-K) R = 1716.4 ft2/(s2-R)

  24. 1ST LAW OF THERMODYNAMICS Energy Is Conserved

  25. 1st Law of Thermodynamics Q - W* = dE q - w = de e = u + V2/2 +gz (1st law came after 2nd Law and there actually is a 0th law that came after the 1st Law. All laws are based on experience.) heat is positive when added to system; work is positive when work is done by the system*

  26. 1st Law of Thermodynamics Q - W = dE q - w = de e = u + V2/2 +gz dE = is an exact differential, a state variable which depends only on initial and final states q & w depend on the process going from initial to final states; also q & w must cross boundaries

  27. dW on gas = F(-dx) = -pAdx = -pdV dQ - dW = dE (1st Law) dq – pdv = du (neglect avg. K.E. and P.E.)

  28. 2ND LAW OF THERMODYNAMICS “not knowing the Second Law or thermodynamics is analogous to never having read a work of Shakespeare” ~ C.P. Snow

  29. Imagine an ice cube on a hot griddle. First law says nothing about the flow of heat. First law allows heat to flow from ice cube to hot griddle, as long as the total energy is conserved. The Second Law tells us in which direction a process can take place. ds = qrev/T Process always in equilibrium, applied infinitely slowly, no gradients, qrev

  30. ds = dq/Treversible* (Eq. 11.8) ds = dq/T + dsirrev Change of entropy during any incremental process is equal to the actual heat divided by the temperature plus a contribution from the irreversible dissipative phenomena of viscosity, thermal conductivity, and mass diffusion [all associated with gradients] occurring within the system. These dissipative phenomena always increase the entropy.

  31. A Change in Entropy Meter Heater trickles in dq quasistatically in such a way that T remains approximately constant over small time.

  32. C L A U S I U S “I propose to name the magnitude S the entropy of the body, from The Greek word for transformation. I have intentionally formed the word entropy so as to be as similar as possible to the word energy, since both these quantities…. are so nearly related”

  33. T2> T1 For heat to flow from cold to hot object is forbidden by the second law e.g. would result in decrease in entropy, therefore is not allowed. T2 Q1=Q2 T2> T1 1/T2< 1/T1 Q/T2< Q/T1 Entropy lost = Q/T1 greater Entropy gained = Q/T2 T1 Q1

  34. “Boltzman’s tomb is the bridge between the world of appearance and its underworld of atoms.”

  35. Example of reversible process that illustrates  dq/T over cycle = 0

  36.  dq/T over cycle = 0 isothermal isothermal adiabatic isothermal adiabatic adiabatic ideal and calorically perfect gas – reversible process (no gradients)

  37. ideal and calorically perfect gas – reversible process (no gradients)

  38. Step 1 ~ Isothermal:  T=0, T=T1 pv = RT1 If perfect gas and adding q1 to va and keep at T1 what is vb? ideal and calorically perfect gas – reversible process (no gradients)

  39. Step 1 ~ Isothermal: T=0, T=T1; pv = RT1 For ideal gas u = f (T); p = RT pv = constant So for step (1-2), du = 0 du = q – w (cons. of energy) q = w q1 = ba pdv = baRTH(dv/v) = RTHln(vb/va) ideal and calorically perfect gas – reversible process (no gradients)

  40. Step 1 ~ Isothermal:  T=0, T=T1 pv = RT1 c For ideal gas u = f (T) p = RT ; pv = constant So for step (1-2), du = 0 du = q – w; q = w q1 = ba pdv = baRTH(dv/v) = RTHln(vb/va) ideal and calorically perfect gas – reversible process (no gradients)

  41. Step 2 ~ Adiabatic:  Q=0; T1 to T2 If perfect gas and expanding adiabatically from T1 to T2 and vb to vc - what is vb?

  42. Step 2 ~ Adiabatic:  Q=0; T1 to T2 du = q – w du = – w cvdT = -pdv cvdT = -pdv = -(RT/v)dv cbcvdT = -cbpdv = bcRT(dv/v) (cv/R) cb (dT/T) = bc(dv/v) ln(Tc/Tb)(Cv/R) = ln(vb/vc) ideal and calorically perfect gas – reversible process (no gradients)

  43. Step 2 ~ Adiabatic:  Q=0; T1 to T2 q = 0 ~ adiabatic du = q – w; du = – w cvdT = -pdv = -(RT/v)dv cbcvdT = -cbpdv = bcRT(dv/v) (cv/R) cb (dT/T) = bc(dv/v) ln(Tc/Tb)(Cv/R) = ln(vb/vc)

  44. Step 2 ~ Adiabatic:  Q=0; T1 to T2 ln(Tc/Tb)(Cv/R) = ln(vb/vc) (Tc/Tb)(Cv/R) = vb/vc [pv/R]c/[pv/R]b = [vb/vc](R/Cv) pc/pb = [vb/vc](R/Cv)+1 = [vb/vc](Cp-Cv)/Cv+1 = [vb/vc]k pcvck = pbvbk= constant ideal and calorically perfect gas – reversible process (no gradients)

  45. Step 3 ~ Isothermal:  T=0; T=T2 pv=RT2 = RTC For perfect gas u = f (T) P = RT So for steps (3 - 4), u = 0 u = q – w q = w q1 = dc pdv = dcRTC(dv/v) = RTCln(vd/vc)

  46. Step 4~ Adiabatic:  q=0; T2 to T1 Step 4~ Adiabatic:  q=0; T2 to T1  Q = 0 ~ adiabatic du = q – w; du = – w cvdT = -pdv = -(RT/v)dv adcvdT = -adpdv = daRT(dv/v) (cv/R) ad (dT/T) = da(dv/v) (cv/R) ln(Ta/Td)(cv/R) = ln(vd/va)  Q = 0

  47. Step#2: (Tc/Tb)(cv/R) = vb/vc (Tcold/Thot)(cv/R) = vb/vc Step#4: (Ta/Td)(cv/R) = vd/va (Thot/Tcold)(cv/R) = vd/va (Tcold/Thot)(cv/R) = va/vd isothermal isothermal a adiabatic b isothermal d adiabatic adiabatic vb/vc = va/vd c

  48. Ideal gas: p = RT q1 = ba pdv = ba RT1(dv/v) = RT1 ln(vb/va) q2 = dc pdv = dc RT2(dv/v) = RT2 ln(vd/vc) = -RT2 ln(vc/vd) vb/vc = va/vd vb/va = vc/vd q1/T1 = -RT2 ln(vb/va) = - q2/T2

  49. Step#2: (Tc/Tb)(cv/R) = vb/vc Step:#4 (Ta/Td)(cv/R) = vd/va Ideal gas: p = RT q1 = ba pdv = ba RT1(dv/v) = RT1 ln(vb/va) q2 = dc pdv = dc RT2(dV/V) = RT2 ln(vd/vc) = -RT2 ln(vc/vd) = -RT2 ln(vb/va) vb/vc = va/vd vb/va = vc/vd q1 / T1 - q2 / T2 = Rln(vb/va) - Rln(vb/va) q1 / T1 = q2 / T2 Although proved for a reversible ideal gas, true for any reversible engine

  50. QH / T1 = QC/T2 INDEPENDENT of cv and R!!!! Can be shown true for any substance for reversible process  dq/T over cycle = 0 ENTROPY DEFINED AS: dS = rev Q/T (11.8)

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