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Unformatted text preview: Lec16: 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.
 Spring '08
 Kerber
 Chemistry, Mole

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