Chapter 8 Real gases
Chapter 8 Real Gases
Physical Chemistry Real gases Homework · Section8.2:8.2.8.3 Section84:8.7.8.11 Section85:8.17.8.18 Section87:8.20.8.21 Section 8.8 :8.26 · General:835.8.38.840.8.45
Homework • Section 8.2: 8.2, 8.3 • Section 8.4: 8.7, 8.11 • Section 8.5: 8.17, 8.18 • Section 8.7: 8.20, 8.21 • Section 8.8: 8.26 • General: 8.35, 8.38, 8.40, 8.45 Physical Chemistry Real Gases
Physical Chemistry Real gases Estimate a molar volume e Estimate the molar volume of CO 2 at 500 K and 100 atm by treating it as a van der Waals gas.(the van der waals e coefficients of co are: a=3.592 atmL2mo1-2 6=4267x10-2L mo Answer. Rearrange equation( 8.2) for a molar volume RT a V-b (8.2) RT ab 73-b+ 0 P RT(8.206×10- Latmk- mo)×(500K) 0.410LmO P 100atm
Estimate a molar volume Estimate the molar volume of CO2 at 500 K and 100 atm by treating it as a van der Waals gas. (the van der Waals coefficients of CO2 are: a=3.592 atmL2mol-2 , b=4.26710-2 L mol-1 ) Answer. 1 2 1 1 0.410 100 (8.206 10 ) (500 ) − − − − = = Lmol atm LatmK mol K P RT 2 (8.2) m Vm a V b RT P − − = Rearrange equation (8.2) for a molar volume 0 3 2 − = + − + P ab V P a V P RT Vm b m m Physical Chemistry Real Gases
Physical Chemistry Real gases Estimate a molar volume e Estimate the molar volume of CO 2 at 500 K and 100 atm by treating it as a van der Waals gas.(the van der waals e coefficients of co are: a=3.592 atmL2mo1-2 6=4267x10-2L mo Answer RT r2 arab P P RT(8.206×10- Latmk- mol)×(500K) =0.410LmOl 100atm RT 6+ 0453Lmo0=3.59×10(Zmo ab =1.53×10-3(Lmol)3
Estimate a molar volume Estimate the molar volume of CO2 at 500 K and 100 atm by treating it as a van der Waals gas. (the van der Waals coefficients of CO2 are: a=3.592 atmL2mol-2 , b=4.26710-2 L mol-1 ) Answer. 1 2 1 1 0.410 100 (8.206 10 ) (500 ) − − − − = = Lmol atm LatmK mol K P RT 0 3 2 − = + − + P ab V P a V P RT Vm b m m 1 0.453 − + = Lmol P RT b 2 1 2 3.59 10 ( ) − − = Lmol P a 3 1 3 1.53 10 ( ) − − = Lmol P ab Physical Chemistry Real Gases
Physical Chemistry Real gases Estimate a molar volume Answer RT r2 arab 2+ P RT(8.206×102 Latmk mol)×(500K) 0.410Lmol 100a RT 6+ =0.453Lmol1 3.59×10-2(Lmol) 1.53×10-°(/mol) Must solve Vn3-0.453V2+(359×102) (1.53×103)=0
Estimate a molar volume Answer. 1 2 1 1 0.410 100 (8.206 10 ) (500 ) − − − − = = Lmol atm LatmK mol K P RT 0 3 2 − = + − + P ab V P a V P RT Vm b m m 1 0.453 − + = Lmol P RT b 2 1 2 3.59 10 ( ) − − = Lmol P a 3 1 3 1.53 10 ( ) − − = Lmol P ab 0.453 (3.59 10 ) (1.53 10 ) 0 3 2 2 3 − + − = − − Vm Vm Vm Must solve Physical Chemistry Real Gases
Physical Chemistry Real gases Estimate a molar volume e Estimate the molar volume of CO 2 at 500 K and 100 atm by treating it as a van der Waals gas.(the van der waals e coefficients of cO are: a=3.592 atmL2 mo1- 2. 6=4267x10-2L mo Answer Vn3-0453V2+(3.59×102)n-(1.53×10-3)=0 Solve for the molar volume =0.366Lmol1 For a perfect gas =0.410LmOl RT(8:206×102 Latmk- mol)×(500K) 0.410Lmol1 100atm
Estimate a molar volume Estimate the molar volume of CO2 at 500 K and 100 atm by treating it as a van der Waals gas. (the van der Waals coefficients of CO2 are: a=3.592 atmL2mol-2 , b=4.26710-2 L mol-1 ) Answer. Solve for the molar volume 1 0.366 − V = Lmol m For a perfect gas 1 0.410 − V = Lmol m 1 2 1 1 0.410 100 (8.206 10 ) (500 ) − − − − = = Lmol atm LatmK mol K P RT 0.453 (3.59 10 ) (1.53 10 ) 0 3 2 2 3 − + − = − − Vm Vm Vm Physical Chemistry Real Gases
Physical Chemistry Real gases Equation of State(eos) The van der Waals and R-K equations are cubic equations of state a cubic algebraic equations al ways has three roots e above the critical temperature T two of the roots are complex numbers, one will be a real number At the critical temperature t three equal real roots Below the critical temperature To three unequal real roots a cubic equation of states isotherm in the two-phase region below Te will resemble the dotted line in Fig. 8.3
Equation of State (eos) The van der Waals and R-K equations are cubic equations of state. Physical Chemistry Real Gases A cubic algebraic equations always has three roots. Above the critical temperature Tc , two of the roots are complex numbers, one will be a real number. At the critical temperature Tc , three equal real roots. Below the critical temperature Tc , three unequal real roots. A cubic equation of states isotherm in the two-phase region below Tc will resemble the dotted line in Fig. 8.3
Physical Chemistry Real gases Condensation P H HO G 4000C 374°C 3000C L+Ⅴ 200°C K Isotherms of ho
Condensation Physical Chemistry Real Gases Isotherms of H2O Vm 400 oC U R J N Y 374 oC 300 oC 200 oC H2O L + V L V L G H T S K M W P l Vm v Vm
Physical Chemistry Real gases virial equation of state PVm=RT[+B(T)P+C(T)P2+D'(T)P+1(8.5 A more convenient expansion is(in many applications B C PIm=rt1+2+g+ PV=RT first second third coefficients The virial equation is an example of a common procedure in physical chemistry, in which a simple law is treated as the first term in a series in powers of a variable(por ym)
virial equation of state [1 '( ) '( ) '( ) ] PVm = RT + B T P +C T P 2 + D T P 3 + (8.5) = 1+ + 2 + m m m V C V B PV RT A more convenient expansion is (in many applications) PVm = RT first second third coefficients The virial equation is an example of a common procedure in physical chemistry, in which a simple law is treated as the first term in a series in powers of a variable (P or Vm). Physical Chemistry Real Gases
Physical Chemistry Real gases as virial equation of state d2=0 For a perfect gas =B+2PC+…→BaP→0 For a real gas There may be a temperature at which Z>1 Z with zero slope at low pressure. ( Boyle n) The properties of the real gas Higher do coincide with those of a perfect gas as P->0 Perfect gas Pressure ower T
virial equation of state Z Pressure 1 Perfect gas Higher T Lower T For a perfect gas = B'+2PC'+ → B' as P → 0 dP dZ = 0 dP dZ For a real gas There may be a temperature at which Z→1 with zero slope at low pressure. (Boyle T) The properties of the real gas do coincide with those of a perfect gas as P→0. Physical Chemistry Real Gases
