iii. ideal gas law (p. 334-335, 340-346)
DESCRIPTION
Ch. 10 & 11 - Gases. III. Ideal Gas Law (p. 334-335, 340-346). Quantities to Describe Gases. P: Pressure V: Volume T: Temperature (Kelvin!) n: # of moles. Avogadro’s Principle. V. n. Equal volumes of gases contain equal numbers of moles at constant temp & pressure true for any gas. - PowerPoint PPT PresentationTRANSCRIPT
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
Quantities to Describe Quantities to Describe GasesGasesQuantities to Describe Quantities to Describe GasesGases
P: Pressure
V: Volume
T: Temperature (Kelvin!)
n: # of moles
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
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:
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!
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
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
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.
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
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
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
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