iii. ideal gas law (p. 334-335, 340-346) ch. 10 & 11 - gases
TRANSCRIPT
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III. Ideal Gas Law(p. 334-335, 340-
346)
III. Ideal Gas Law(p. 334-335, 340-
346)
Ch. 10 & 11 - Ch. 10 & 11 - GasesGases
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Quantities to Describe Quantities to Describe GasesGasesQuantities to Describe Quantities to Describe GasesGases
P: Pressure
V: Volume
T: Temperature (Kelvin!)
n: # of moles
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kn
VV
n
Avogadro’s PrincipleAvogadro’s PrincipleAvogadro’s PrincipleAvogadro’s Principle
Equal volumes of gases contain equal numbers of moles• at constant temp & pressure• true for any gas
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PV
TVn
PVnT
Ideal Gas LawIdeal Gas LawIdeal Gas LawIdeal Gas Law
= kUNIVERSAL GAS
CONSTANTR=0.0821 Latm/molK
R=8.315 dm3kPa/molK
= R
You don’t need to memorize these values!
Merge the Combined Gas Law with Avogadro’s Principle:
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Ideal Gas LawIdeal Gas LawIdeal Gas LawIdeal Gas Law
UNIVERSAL GAS CONSTANT
R=0.0821 Latm/molKR=8.315
dm3kPa/molK
PV=nRT(listen to song!!!)
You don’t need to memorize these values!
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Ideal Gas Constant, RIdeal Gas Constant, RIdeal Gas Constant, RIdeal Gas Constant, R
We know that:• 1 mol of a gas occupies 22.4 L at
STP (273.15 K and 101.325 kPa)
R = PV = (101.325kPa)(22.4L) Tn (273.15K)(1mol)
R = 8.31 L·kPa/mol·K
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Ideal Gas Constant, RIdeal Gas Constant, RIdeal Gas Constant, RIdeal Gas Constant, R
Units of numerator depend on:
• Unit of volume and pressure
• Common units of R
Numerical Value
Units
62.4 L·mmHg
mol·K
0.0821 L·atm
mol·K
8.314 J
mol·K
8.314 L·kPa
mol·K
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GIVEN:
P = ? atm
n = 0.412 mol
T = 16°C = 289 K
V = 3.25 LR = 0.0821Latm/molK
WORK:
PV = nRT
P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K
P = 3.01 atm
Ideal Gas Law ProblemsIdeal Gas Law ProblemsIdeal Gas Law ProblemsIdeal Gas Law Problems Calculate the pressure in atmospheres of
0.412 mol of He at 16°C & occupying 3.25 L.
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GIVEN:
V = ?
n = 85 g
T = 25°C = 298 K
P = 104.5 kPaR = 8.315 dm3kPa/molK
Ideal Gas Law ProblemsIdeal Gas Law ProblemsIdeal Gas Law ProblemsIdeal Gas Law Problems
Find the volume of 85 g of O2 at 25°C and 104.5 kPa.
= 2.7 mol
WORK:
85 g 1 mol = 2.7 mol
32.00 g
PV = nRT(104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K
V = 64 dm3
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MM from IGL from IGLMM from IGL from IGL
a) If the P, V, T, and mass are known for a gas sample, then n can be calculated using IGL
Then, the molar mass is found by dividing the mass by the number of moles
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MM from IGL from IGLMM from IGL from IGL
b) The number of moles (n) is equal to mass (m) divided by molar mass (M)
g ÷ g = g x mol = mol mol g
c) Substitute m/M for n in IGL: PV = mRT OR M = mRT
M PV
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D from IGLD from IGLD from IGLD from IGL
Density, D, is m/V which results in:
M = DRT
P
Rearranging for D:
D = MP
RT