Elements that exist as gases at 250C and 1 atmosphere

Pressure

Remember:• Pressure = Force / area or

mass x gravity / area• Force = mass x gravity

Units of Pressure

1 pascal (Pa) = 1 N/m2

1 atm = 760 mmHg = 760 torr

1 atm = 101,325 Pa

Barometer

Pressure = ForceArea

Measuring PressureDevices used:• Barometer-Invented by Torricelli• Pressure Gauge• Tire Gauge• Sphygmomanometer

Units of Pressure

• Standard atmospheric pressure (1 atm) = 760 mm Hg or 760 torr (1 mm Hg = 1 torr)

• In the United States we use inches in atmospheric pressure.

• Pascal- used in science as a unit of pressure (Pa).

• Kilopascals are used as well (KPa).

PressureColumn height measures

Pressure of atmosphere• 1 standard atmosphere

(atm) *

= 760 mm Hg (or torr) *

= 29.92 inches Hg *

= 14.7 pounds/in2 (psi) *HD only

= 101.3 kPa (SI unit is PASCAL) * HD only

= about 34 feet of water!

* Memorize these!

Practice

• How many torrs are in 4.5 atm?

• The barometer reads 1,760 torr, how many atm?

• I have a reading of 3700 torr, convert to atm.

• You have 3 atm, how many torr?

Boyle’s Law

• Shows the relationship between Pressure and Volume.

• The Law assumes that the temperature does not change.

• Proposed by Robert Boyle.

Boyle’s LawP1V1 = P2V2

• P1= Pressure 1• V1= Volume 1• P2= Pressure 2• V2= Volume 2• This formula works with any unit of Pressure (torr, Pa)• Any unit of volume works (mL, L).• Make sure you have the same units on both sides! If not,

change one to match the other.• If one side has mL and the other has L, convert one to

match the other.

Boyle’s Law• A gas measured a volume of 100 mL under

pressure of 740 torr. What would be the volume under a pressure of 780 torr with constant temperature?

Step 1: Figure out what you have.• P1 = 740 torr• V1= 100 mL• P2= 780 torr• V2= X

Boyle’s Law

• Step 2: Plug the numbers into the formula

P1V1 = P2V2

(740 torr)(100 mL) = (780 torr)(X)

• Step 3: Solve for x

74000 = 780X

X= 94.87 mL

Boyle’s Law• So the new volume in this case is 94.87 mL• Since we increased the pressure, the volume is decreased. • The formula is a proportion. If something increases, something

else will decrease!

Practice

1. A sample of gas is confined to a 100 mL flask under pressure of 740 torr. If the same gas were transferred to a 50 mL flask, what’s the new pressure?

P1 =

V1 =

P2 =

V2 =

P1V1 = P2V2

740 torr

100 ml

X

50 ml

P2 = P1V1/V2

P2 = (740)(100)/50

P2 = 74000/50

P2 = 1480 torr

Practice

2. You are given a gas that you measure under a pressure of 720 Pa. When the pressure is changed to 760 Pa, the volume became 580 mL. What is the first volume?

P1V1 = P2V2

P1 =

V1 =

P2 =

V2 =

720 Pa

X

760 Pa

580 ml

V1 = P2V2/P1

V1 = (760)(580)/720

V1 = 440800/720

V1 = 612.2 ml

Practice3. A pressure on 134 mL of air is changed to

1200 torr at a constant temperature, if the new volume is 45 mL what is the original pressure?

P1V1 = P2V2

P1 =

V1 =

P2 =

V2 =

X

134 ml

1200 torr

45 ml

P1 = P2V2/V1

P1 = (1200)(45)/134

P1 = 54000/134

P1 = 402.9 torr

Practice4. An amount of Oxygen occupies 2 L when

under pressure of 680 torr. If the volume is increased to 3 L what is the new pressure?

P1V1 = P2V2

P1 =

V1 =

P2 =

V2 =

680 torr

2 L

X

3L

P2 = P1V1/V2

P2 = (740)(100)/50

P2 = 74000/50

P2 = 1480 torr

Reminders• Check your units• Temperature must remain constant in

Boyle’s Law.• P= Pressure (torr, Pa, KPa).• V= Volume (mL, L).

“If you don’t exhale on your way up Cookie, your lungs will explode!

Robert De Niro as Chief Saturday

When the temperature increases, the volume increases.

As T increases V increases

Charles Law

• Shows a relationship between Volume and Temperature. One thing increase, so does the other. Both behave the same way.

• Pressure remains constant.

• V= Volume

• T= Temperature

Few things first…

• ALL TEMPERATURES MUST BE CONVERTED TO KELVIN!

• The Law only works with Kelvin

• If you have Celsius, make sure you convert it to Kelvin.

• Remember Kelvin = Celsius + 273

Temperature Conversion Practice

1. Convert 85°C to K. ____________________

2. Convert 376K to °C. ____________________

3. Convert 154K to °C. ____________________

4. Convert -65°C to K. ____________________

5. Convert 0°C to K. _____________________

6. Convert 0K to °C. _____________________

85 + 273 =

376 - 273 =

154 - 273 =

-65 + 273 =

0 + 273 =

0 - 273 =

358K

103 °C

-119 °C

208K

273K

-273 °C

Few Practice Problems

You heat 100 mL of a gas at 25 C to 80 C. What is the new volume of the gas?

V1 =

T1 =

V2 =

T2 =

V1/T1 = V2/T2

100 ml

25ºC

X

80ºC

+ 273 =

+ 273 =

298K

353K

V2 = V1T2/T1

V2 = (100)(353)/298

V2 = 35300/298

V2 = 118.5 ml

Few Practice Problems

You have 200 mL of a gas at 55 C and you freeze it to 0 C. What is the new volume?

V1/T1 = V2/T2

V1 =

T1 =

V2 =

T2 =

200 ml

55ºC

X

0ºC

+ 273 =

+ 273 =

328K

273K

V2 = V1T2/T1

V2 = (200)(273)/328

V2 = 54600/328

V2 = 166.5 ml

Few Practice Problems

A 150 mL sample of gas is at 125 K. After cooling the new volume is 80 mL. What is the new temperature?

V1/T1 = V2/T2

V1 =

T1 =

V2 =

T2 =

150 ml

125K

80 ml

X

T2 = V2T1/V2

T2 = (80)(125)/150

T2 = 10000/150

T2 = 66.67K

Gay-Lussac’s LawP1/T1 = P2/T2

• Shows relationship between pressure (P) and temperature (T). If Temperature increases, so does the pressure.

• Based on Charles’ Law.• Volume remains constant.• Like Charles’ Law, all temperature units must be in

KELVIN.• Used in pressure cookers and autoclaves.

Practice

A soda bottle with a temperature of 25ºC and 3 atm of pressure was put into a freezer with a temperature of –1°C. What is the pressure on the bottle inside the freezer?

P1 =

T1 =

P2 =

T2 =

3 atm

25ºC

X

-1ºC

+ 273 =

+ 273 =

298K

272K

P1/T1 = P2/T2

P2 = P1T2/T1

P2 = (3)(272)/298

P2 = 816/298

P2 = 2.74 atm

Combined Gas Law

• Combines Boyle’s Law, Gay-Lussac’s Law, and Charles’ Law.

• Shows the relationship of Pressure, Volume and Temperature.

Combined Gas Law

• Derived from Boyle’s, Gay-Lussac, and Charles Law.• Used when both temperature and pressure changes.• Can be used to find a constant.• If two things increase, one thing will decrease.• Math was used to derive the law.

Combined Gas Law (P, V, T)

Combined Gas Law: nR = PV

T

1. A gas has a volume of 800.0 mL at -23.00 °C and 300.0 torr. What would the volume of the gas be at 227.0 °C and 600.0 torr of pressure?

P1 =

V1 =

T1 =

P2 =

V2 =

T2 =

300 torr

800 ml

-23ºC

600 torr

X

227.0ºC

+ 273 =

+ 273 =

250K

500K

P1V1 = P2V2

T1 T2

V2 = P1V1T2

T1 P2

V2 = (300)(800)(500)

(250)(600)

V2 = 800 ml

Remember this…

• When you are working with Charles’ Law and Combined Gas Law, make sure you have the proper units!

• The laws help us predict the behavior of gases under different circumstances.

How is this possible?

KC-135 Jet

Rail Car

Avogadro’s LawV1/n1 = V2/n2

• Allows us to calculate the number of moles (n) of a gas with in a volume (V).

• Pressure and Temperature remains the same.• If volume increases, so does the number of moles.

Practice

In 150 mL sample, there are 2.5 grams of Cl2 gas, if I double the volume to 300 mL how much Cl2 gas in moles, is in the new sample?

V1 =

n1 =

V2 =

n2 =

150 ml

2.5g

300 ml

x

÷ 17g/m = 0.147 m

V1/n1 = V2/n2

n2 = V2 n1/ V1

n2 = (300)(0.147)/150

n2 = 44.1/150

n2 = 0.294 m

Ideal gas law

• Formula: PV= nRT

• P= Pressure in atmospheres• V= Volume in Liters• n= number of moles of a gas• R= constant, 0.0821 L • atm / mol • K• T= Temperature in Kelvin

Ideal Gas Law

• An ideal gas is a gas that obeys Boyle’s and Charles’ Law.

• Based on the Kinetic Theory.• Shows the relationship between Pressure,

Volume, number of moles and Temperature.• Make sure everything has the proper unit.• Can be used to figure out density of gas or

determine the behavior of a tire.

Practice

How many moles of gas are contained in 890.0 mL at 21.0 °C and 750.0 mm Hg pressure?

PV = nRT

P =

V =

n =

R = 0.0821 Latm/molK

T =

750.0 mm Hg

890.0 ml

X

21.0ºC + 273 = 294 K

n = PV/RT

n = (0.987)(0.89)/(0.0821)(294)

n = 0.878/24.1

n = 0.036 m

÷ 760 mmHg/atm = 0.987 atm

÷ 1000 ml/L = 0.89 L

And now, we pause for this commercial message from STP

OK, so it’s really not THIS kind of STP…

STP in chemistry stands for Standard Temperature and Pressure

Standard Pressure = 1 atm (or an equivalent)

Standard Temperature = 0 deg C (273 K)

STP allows us to compare amounts of gases between different pressures and temperatures

STP allows us to compare amounts of gases between different pressures and temperatures

The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).

PV = nRT

R = PV

nT=

(1 atm)(22.414L)

(1 mol)(273.15 K)

R = 0.082057 L • atm / (mol • K)

Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

PV = nRT

V = nRT

P

T = 0 0C = 273.15 K

P = 1 atm

n = 49.8 g x 1 mol HCl36.45 g HCl

= 1.37 mol

V =1 atm

1.37 mol x 0.0821 x 273.15 KL•atmmol•K

V = 30.6 L

GAS DENSITYGAS DENSITY

High density

Low density

22.4 L of ANY gas AT STP = 1 mole

Density (d) Calculations

d = mV

=PMRT

m is the mass of the gas in g

M is the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance

dRT

PM = d is the density of the gas in g/L

Gas Stoichiometry

What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)

g C6H12O6 mol C6H12O6 mol CO2 V CO2

5.60 g C6H12O6

1 mol C6H12O6

180 g C6H12O6

x6 mol CO2

1 mol C6H12O6

x = 0.187 mol CO2

V = nRT

P

0.187 mol x 0.0821 x 310.15 KL•atmmol•K

1.00 atm= = 4.76 L

Gases and Stoichiometry

2 H2O2 (l) ---> 2 H2O (g) + O2 (g)

Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP?

Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

Deviations from Ideal Gas Law

• Real molecules have volume.The ideal gas consumes the entire

amount of available volume. It does not account for the volume of the molecules themselves.

• There are intermolecular forces.

An ideal gas assumes there are no attractions between molecules. Attractions slow down the molecules and reduce the amount of collisions.– Otherwise a gas could

not condense to become a liquid.

Gases in the Air

The % of gases in air Partial pressure (STP)

78.08% N2 593.4 mm Hg

20.95% O2 159.2 mm Hg

0.94% Ar 7.1 mm Hg

0.03% CO2 0.2 mm Hg

PAIR = PN + PO + PAr + PCO = 760 mm Hg

Total Pressure 760 mm Hg

2 22

Dalton’s Law of Partial Pressures

V and T are constant

P1 P2 Ptotal = P1 + P2

Dalton’s Law

John Dalton1766-1844

Consider a case in which two gases, A and B, are in a container of volume V.

PA = nART

V

PB = nBRT

V

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB XA = nA

nA + nB

XB = nB

nA + nB

PA = XA PT PB = XB PT

Pi = Xi PT

A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?

Pi = Xi PT

Xpropane = 0.116

8.24 + 0.421 + 0.116

PT = 1.37 atm

= 0.0132

Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

2KClO3 (s) 2KCl (s) + 3O2 (g)

Bottle full of oxygen gas and water vapor

PT = PO + PH O2 2

Collecting a gas “over

water”

• Gases, since they mix with other gases readily, must be collected in an environment where mixing can not occur. The easiest way to do this is under water because water displaces the air. So when a gas is collected “over water”, that means the container is filled with water and the gas is bubbled through the water into the container. Thus, the pressure inside the container is from the gas AND the water vapor. This is where Dalton’s Law of Partial Pressures becomes useful.

Table of Vapor Pressures for Water

Solve This!

A student collects some hydrogen gas over water at 20 degrees C and 768 torr. What is the pressure of the gas?

768 torr – 17.5 torr = 750.5 torr

GAS DIFFUSION AND EFFUSION

• diffusion is the gradual mixing of molecules of different gases.

• effusion is the movement of molecules through a small hole into an empty container.

GAS DIFFUSION AND EFFUSION

Graham’s law governs effusion and diffusion of gas molecules.

Thomas Graham, 1805-1869. Professor in Glasgow and London.

Rate of effusion is inversely proportional to its molar mass.

Rate of effusion is inversely proportional to its molar mass.

M of AM of B

Rate for B

Rate for A

GAS DIFFUSION AND EFFUSION

Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is

• proportional to T• inversely proportional to M.

Therefore, He effuses more rapidly than O2 at same T.

He

Gas Diffusionrelation of mass to rate of diffusion

Gas Diffusionrelation of mass to rate of diffusion

• HCl and NH3 diffuse from opposite ends of tube.

• Gases meet to form NH4Cl

• HCl heavier than NH3

• Therefore, NH4Cl forms closer to HCl end of tube.

• HCl and NH3 diffuse from opposite ends of tube.

• Gases meet to form NH4Cl

• HCl heavier than NH3

• Therefore, NH4Cl forms closer to HCl end of tube.