Lacey_Che131_S2011_Lect-17s

Lacey_Che131_S2011_Lect-17s - Lec-16: Ideal gas Law;...

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Unformatted text preview: Lec-16: Ideal gas Law; Dalton’s Law Roy A. Lacey, Stony Brook University; Che 131, Spring 2011 1 Parameters affecting gasses Pressure (P) Volume (V) (V) Temperature (T) Number of Moles (n) PV = const. T For fixed n Roy A. Lacey, Stony Brook University; Che 131, Spring 2011 2 Ideal equation of state PV = const. T PV = const. = nR T PV = nRT nT V∝ P ⎛ nT ⎞ V = R⎜ ⎟ ⎝P⎠ Boyle: (constant n, T) Charles: V ∝ T (constant n, P) Avogadro: V ∝ n (constant P, T). Roy A. Lacey, Stony Brook University; Che 131, Spring 2011 3 Avagadros Law PV = nR T Pf V f PVi i = const. = niTi n f Tf For a constant P and T Vf Vi = const. = ni nf Roy A. Lacey, Stony Brook University; Che 131, Spring 2011 4 Example Air occupying 12.0dm3at a pressure of 98.9 kPa is compressed to occupying 12 pressure of 98 kPa compressed to a pressure of 119.0kPa what is the new volume if the temperature remains constant remains constant Vi = 12.0 dm3 Pi = 98.9 kPa Vf = ?? Pf = 119.0 kPa kP PV=nRT (n and are constants (n, T and R are constants ) ⇒ PV = Constant PV Constant PiVi = PfVf Vf = Vi × Pi 98.9 kPa = 12.0 dm3 × Pf 119.0 kPa = 9.97 dm3 Roy A. Lacey, Stony Brook University; Che 131, Spring 2011 5 Example 785 L of O2 is obtained from a tank at 1 atm and 210C. of is obtained from tank at atm and 21 What would be the volume of gas at 28 0C. Vi = 785 L Pi = 1 atm Ti = 294 K 785 atm 294 Vf = ?? Pf = 1 atm Tf = 301 K PV PV=nRT V = Constant T T ⇒ Vf = Vi × f Ti (n, P and R are constants ) ⇒ Vi Vf = Ti Tf Vf = Vi × Tf 301 K = 785 L × Ti 294 K = 8 04 L Roy A. Lacey, Stony Brook University; Che 131, Spring 2011 6 Example 39.8 mg of caffeine gives 10.1 cm3 of N 2 gas at 23 oC and 10 23 746 mm Hg. What is the volume of N2 at 0 oC and 760 mmHg. Vi = 10.1 cm Vf = ?? 3 Pi = 746 mm Hg Ti = 296 K Pf = 760 mm Hg Tf = 273 K PV=nRT (n, and R are constants ) ⇒ PV = Constant T Pi Vi Pf Vf = Ti Vf = Vi × Tf ⇒ Vf = Vi × Pi T × Pf T f i Pi T 746 mm Hg 273 K × = 10.1 cm3 × × Pf T 760 mm Hg 296 K f i = 9.14 cm3 Roy A. Lacey, Stony Brook University; Che 131, Spring 2011 7 Daltons Law For a mixture of gases in a container PTotal = P1 + P2 + P3 + . . . Roy A. Lacey, Stony Brook University; Che 131, Spring 2011 8 Daltons Law Mole Mole Fraction: the ratio of the number of moles of a given component in a mixture to the total number of moles in a mixture. Χ1 = n1 nTOTAL n1 n1 + n2 + n3 + ••• = • Mole Fraction in terms of pressure (n = PV/RT) Fraction in terms of pressure PV/RT Χ1 = P1(V/RT) P1(V/RT) + P2(V/RT) + P3(V/RT) + ••• Roy A. Lacey, Stony Brook University; Che 131, Spring 2011 9 Daltons Law Mole Mole Fraction: the ratio of the number of moles of a given component in a mixture to the total number of moles in a mixture. Χ1 = Χ1 = Χ1 = P1(V/RT) P1(V/RT) + P2(V/RT) + P3(V/RT) + ••• ••• P1 P1 + P2 + P3 + ••• n1 nTOTAL = = P1 PTOTAL P1 PTOTAL Roy A. Lacey, Stony Brook University; Che 131, Spring 2011 10 Partial pressures Gasses are Typically collected over water PT = PO2 + PH2O Roy A. Lacey, Stony Brook University; Che 131, Spring 2011 11 ...
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This note was uploaded on 02/21/2012 for the course CHE 131 taught by Professor Kerber during the Spring '08 term at SUNY Stony Brook.

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