iii. ideal gas law (p. 334-335, 340-346) ch. 10 & 11 - gases

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