derivationthe way it really is. the way we choose to view it a.p. chem. ch. 5 boyle’s law pv=k...

Derivation

The W

ay It Really Is.

The way we choose to view it

A.P. Chem. Ch. 5

Boyle’s Law PV=k

Charles’ Law V/T=k

Avog. Law V/n=k

Pressure: Force Area

Devices

BarometerManometer

Units mm, torr, atm, Pa, kPa, N/cm2

PV= nRT0.0821 L atm/mol K 8.31 J/ mol K

KMT

Dimensionless Pts.

In constant motion

Colliding 100% elast.Creating pressure

w/o influenceIn such a way that Temp is dir. Prop. To average KE.

P = 2 nNA1/2mu2

P = 2 nKEper mol

3V

3VKEper mol = (3/2)RT

P = dRT/MM

Dalton’s Law Ptot= Pa+ Pb…

Graham’s Law Ra/Rb = (MMb/MMa)1/2

Gases

aPtot= Pa

urms= (3RT/MM)1/2

Stoich.

22.4 L = 1 mole

At STP

Non- STP

P= nRT – a(n/V)2

V-nb Volumes of particles

Particle interactions

Due to Avogadro’s Hypothesis.`

Format of the Test•Option 1 – The marathon problem (1 system; many parts) 70 points

•Option 2 – The regular test (single parts; many systems) 70 points

13 multiple choice [~20 minutes]2 free response [~25 minutes]

Problems to emphasize in your HW:43, 61, 67, 83, 85

The average kinetic energy of the gas molecules is

(A) greatest in container A

(B) greatest in container B

(C) greatest in container C

(D) the same for all three containers

Container (3.0 L)

A B C

Gas Methane Ethane Butane

Formula CH4 C2H6 C4H10

Molar Mass (g/mol)

16 30. 58

Temp. (ºC) 27 27 27

Pressure (atm)

2.0 4.0 2.0

How would you calculate the average KE per mole of gas?

Avg. KEmol = 1.5 (8.31 J/mol K) 300.K

Avg. KEmol = 3740 J/mol

The average velocity of the gas molecules is





Container (3.0 L)

A B C



Molar Mass (g/mol)

16 30. 58

Temp. (ºC) 27 27 27

Pressure (atm)

2.0 4.0 2.0

How would you calculate the velocity of the methane gas?

urms = [3(8.31J/mol K)(300. K)/0.016 kg]½

urms = 680 m/s

The number of gas molecules is





Container (3.0 L)

A B C



Molar Mass (g/mol)

16 30. 58

Temp. (ºC) 27 27 27

Pressure (atm)

2.0 4.0 2.0

How would you calculate the number of particles of ethane?

n = (4.0 atm)(3.0L)(0.0821 L atm/mol K)(300. K)

n = 0.49 mol

The density of the gas is





Container (3.0 L)

A B C



Molar Mass (g/mol)

16 30. 58

Temp. (ºC) 27 27 27

Pressure (atm)

2.0 4.0 2.0

How would you calculate the density of the ethane gas?d = (4.0 atm)(30. g/mol)

(0.0821 L atm/mol K)(300. K)

d = 4.9 g/L

If the containers were opened simultaneously the diffusion rates of the gas molecules out of the containers through air would be





Container (3.0 L)

A B C



Molar Mass (g/mol)

16 30. 58

Temp. (ºC) 27 27 27

Pressure (atm)

2.0 4.0 2.0

Calculate how much faster the methane diffuses compared to the ethane.

(ratemeth/rateeth) = (30.g/mol/16g/mol)½

(ratemeth/rateeth) = 1.4 (times faster)

Consider the production of water at 450. K and 1.00 atm: 2 H2(g) + O2(g) --- > 2 H2O(g)

If 300.0 mL of hydrogen is mixed with 550.0 mL of oxygen, determine the volume of the reaction mixture when the reaction is complete.

You Have:nH2= 0.00812 molnO2= 0.0149 mol

You Need:Twice as many moles of H2 than O2 (from balanced reaction equation)

Consequence:H2 is the L.R.

Since H2 is the L.R. then 300.0 mL of it is used and only 150 mL of the O2 is used. (2:1 mole ratio can be

viewed as 2:1 volume ration as well due to

Avogadro’s Hypothesis) Also, 300.0 mL of H2O will be produced.

The reaction mixture remaining will be 600 mL in volume from the 300.0mL of water made and the 300.0 mL of O2 unused.

UF6= 352 g

Urms= [3(8.31Jmol-1K-1)(330.0K)/0.352kg]½

Urms= 153 m/s

1.03 mg O2 = 3.22 x 10-5 mol O2 P O2 = 0.00380 atm

0.41 mg He = 1.0 x 10-4 mol He P He = 0.012 atm

P tot = 0.016 atm

XO2 (Ptot) = PO2

XO2 = PO2

/(Ptot) = 0.24

XHe = 0.76

P = dRT/MM

MMair = 29 g/mol

dair@25°C, 740 t = 1.2 g/L

UH2 MMI2

UI2 MMH2

=√ =RateH2

timeI2

RateI2 timeH2

=

52 s 253.8gtimeH2 2.02 g

=√

amount H2/timeH2

amount I2/timeI2

timeH2= 4.6 s

derivationthe way it really is. the way we choose to view it a.p. chem. ch. 5 boyle’s law pv=k...

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