Lecture 26 © slgCHM 151

Topics:

1. Ideal Gas Law calculations2. Density, Molar Mass of gases3. Stoichiometry involving gases

Combined Gas Law

We can combine the three factors which describe a “confined gas” (constant n, number of particles)as follows:

P 1 / V, P T :

P T P =Cc T V V

PV = Cc AND P1V1 = Cc =P2V2

T T1 T2

If we hold the one factor constant for this “confinedgas”, then :

P1V1 = P2V2

T1 T2

V1 = V2

T1 T2

Constant T

P1V1 = P2V2

Constant P

Constant V

P1 = P2 T1 T2

Units Utilized in Gas Law Problems:

PRESSURE*: generally done in reference to a column of mercury immersed in a dish of mercury open to atmospheric pressure. (CD ROM)

Pressure as a FORCE PER UNIT AREA:

English SI

= 14.7 lbs/ft2(psi) = 101.3 kilopascal (k Pa)

760 mm Hg = 760 Torr = 1 atmosphere (atm)

1 Pa, pascal = 1 Newton / m2 1 Newton = 1 kg . m / s2 (SI unit of force)

Volume*: Liters, L (most general)

Temperature: Absolute or Kelvin Scale; K= oC +273.15

In all cases, K, Kelvin scale must be used! Where P, V = 0, T=0, absolute zero, - 273.15 oC

* Any V or P unit OK for combined gas law, if used consistently

Units Utilized in Gas Law Problems, continued:

Since gas law problems always include lots of data, itis a standard practice to set up a table of all given databefore proceeding: For the combined gas law, anappropriate table is below:

V1 =

P1 =

T1 =

n is consta nt

INITIAL

V2 =

P2 =

T2 =

FINAL

A gas exerts a pressure of 735 mm Hg in volume of 2.50 L at a temperature of 73o C. What pressure would itexert if the temperature were increased to 110o C and the volume increased to 3.00 L?

V1 = 2.50 L

P1 = 735 mm Hg

T1 = 73 oC +273

= 346 K

n is consta nt

V2 = 3.00L

P2 = ?

T2 = 110oC + 273 = 383 K

INITIAL FINAL

data

formula

solve

P1V1

T1

P2V2

T2

=P2 = P1V1T2

T1 V2

P2 =

V1 = 2.50 L

P1 = 735 mm Hg

T1 = 73 oC +273

= 346 K

V2 = 3.00L

P2 = ?

T2 = 110oC + 273 = 383K

INITIALFINAL

735 mm Hg X 2.50 L X 383K

346 K X 3.00 L

= 678 mm Hg

A sample of CO2 gas has a pressure of 56.5 mm Hg in a 125 mL flask. The gas is transferred to a new flask where it has a pressure of 62.3 mm Hg at the sametemperature. What is the volume of the new flask?

V1 =125 ml = .125 L

P1 = 56.5 mm Hg

T1 = same

V2 = ?

P2 = 62.3 mm Hg

T2 = same

INITIALFINAL

P1V1

T1

P2V2

T2

=V2 = P1V1

P2

V2 =

V1 =125 ml = .125 L

P1 = 56.5 mm Hg

T1 = s ame

V2 = ?

P2 = 62.3 mm Hg

T2 = s ame

INITIALFINAL

56.5 mm Hg X .125 L

62.3 mm Hg

= .113 L = 113 mL

GROUP WORK

A sample of gas occupies 12.0 liters at 240. oC under a pressure of 80.0 kPa. At what temperature would the gas occupy 15.0 liters if the pressure were increased to 107 kPa?

V1 = P1 =

T1 =

n is consta nt

V2 =

P2 =

T2 =

INITIAL FINAL

KEY:

P1V1

T1

P2V2=

T2 =

V1 =12.0 L

P1 = 80.0 kPa

T1 = 240. oC +273 = 513 K

V2 = 15.0 L

P2 = 107 kPa

T2 = ?

INITIALFINAL

= K -273 = oC

T2

= T2 =

P1V1

T1

P2V2=

T2 =

V1 =12.0 L

P1 = 80.0 kPa

T1 = 240. oC +273 = 513 K

V2 = 15.0 L

P2 = 107 kPa

T2 = ?

INITIALFINAL

107 kPa X 15.0 L X 513 K

80.0 kPa X 12.0 L

= 858 K -273 = 585 oC

T2

P2V2T1P1V1T2 = T2 =P2V2T1

P1V1

KEY:

VARIATION OF P with n at constant V, T

If we keep the volume of our container constant and also the temperature, the introduction of more particlesinto the container will mean more collision with thewall of the container.

Accordingly, we can say:

P n P = Cn n

P1 P2

= Cn =n1 n2

Relating P to n, V, T

P 1 / V, P T , P n :

P n T P =R nT V V

where R = gas constant used to interrelate the fourfactors. Rearranging:

PV = nRT the “ideal gas equation”

(No such thing as an “ideal gas” but if very low T’s andvery high P’s are avoided, most gases can be describedfairly accurately using the “ideal gas law”.)

Evaluation of value of R

If P, V, n, and T for a gas are experimentally determined, the value for this constant can readily be calculated, which is appropriate for use with any gas (if the conditionsare not too extreme): PV / nT =R

We can calculate R for one mole of any gas using what isdescribed as “Standard Conditions”, a conventionalP and T adopted as the norm for gases (“STP”):

Standard Pressure = 1 atmosphere or 760 Torr (mm Hg)Standard Temperature = 0oC = 273 K

It was determined by Avogadro (1811) that equal volumesof any gas, at the same temperature and pressure,contain the same number of molecules.

It has since been experimentally determined that at STP, 1.00 mole of any “ideal” gas occupies 22.414 L, called thestandard molar volume.

1.00 mole gas, STP = 22.4 L

Calculation of R, STP:

Let us use experimentally determined molar volume of 1.00 mole of gas at STP to evaluate R. Note that ourtable of values changes form when using PV= nRT :

P = 760 Torr

V= 22.414 L

T= 273 K

n= 1.00 mole

R= ?

PV = n R T

PV

nT= R

R = 760 Torr X 22.414 L

1.00 mol X 273 K

= 62.4 Torr L mol-1 K-1

Calculation of R, changing pressure unit to atmospheres:

P = 1.00 atm

V= 22.414 L

T= 273 K

n= 1.00 mole

R= ?

PV = n R T

PV

nT= R

R = 1.00 atm X 22.414 L

1.00 mol X 273 K

= 0.0821 atm L mol-1 K-1

Useful Gas Law Knowledge, to date:

P1V1 = P2V2

T1 T2

V1 = V2

T1 T2

Constant T

P1V1 = P2V2

Constant P

Constant V

P1 = P2 T1 T2

Combined Gas Law, constant n:

PV = nRT the “ideal gas equation”

1.00 mole gas, STP = 22.4 L

STP:Standard Pressure = 1 atmosphere or 760 Torr (mm Hg)Standard Temperature = 0oC = 273 k

R = 62.4 Torr L / K mol = 62.4 Torr L K-1 mol-1

R = .0821 atm L / K mol = .0821 atm L K-1 mol-1

n = ? 25.0 g N2 gas = ? moles 28.0 g N2 = 1 mole N2 25.0 g N2 1 mol = .893 mol N2

28.0 g

25.0 g = .893 mol 28.0 g mol-1

More convenientform, gas laws

n = grams X 1 mole

# g= # moles

mass, g

molar massn = = # moles

Problems using the ideal gas equation:

A hot air balloon holds 30.0 kg of helium. What is the volume of the balloon if the final pressure is 1.20 atm and the temperature 22oC?

P = 1.20 atm

V= ?

T= 22oC +273 = 295 K

g= 30.0 kg He

M = 4.00 g/mol

n =

R=

30.0 kg X 1000 g1 kg

X 1 mol He

4.00 g He

= 7500 mol= 7.50 X 103 mol.0821 atm L /K mol

V = 151,371.875 L = 1.51 X 105 L

P = 1.20 atm

V= ?

T= 22oC +273 = 295 K

g= 30.0 kg He

M = 4.00 g/mol

n =

R=

30.0 kg X 1000 g1 kg

X 1 mol He

4.00 g He

= 7500 mol= 7.50 X 103 mol.0821 atm L / K mol

PV = n R T

V= nRT

P =

7.50 X 103 mol X .0821 L atm K-1 mol-1 X 295 K

1.20 atm

Convert formula first!

Group Work

The nitrogen gas in an air bag , with a volume of 65 L, exerts a pressure of 829 mm Hg at 25o C. How many moles and how many g of N2 are in the air bag? (Solvefor moles using ideal gas equation, then solve for g)

PV =nRT

n=

P = 829 Torr

V= 65 L

T= 25o+ 273= 298 K

n= ?

R= 62.4 Torr L K-1 mol -1

PV = n R T

PV = nRT

n =829 Torr X 65 L

62.4 Torr L k-1 mol -1 X 298 k

= mol2.9

2.9 mol N2 X 28.0 g N2

1 mol N2

= 81 g

P = 829 Torr

V= 65 L

T= 25o+ 273= 298 K

n= ?

R= 62.4 Torr L K-1 mol -1

Another handy “formula” to utilize:

The ideal gas law can be used to calculate molar mass if grams of gas, and P,V, T are known:

1. Calculate moles of gas, n = PV / RT

2. Use n, moles and mass, g to calculate M

n, moles = mass, g molar mass , M = mass, g molar mass, M n, moles

Widely used

Gas Laws : Density, Molar Mass

Chloroform is a common liquid which vaporizes readily. If the pressure of the vaporized liquid is 195mm Hg at25 oC, and the density of the gas is 1.25 g/L, what is themolar mass of the chloroform?

V= 1.00 Lmass= 1.25 gP= 195 mm HgT= 25 oC = 298 KR = 62.4 L mm Hg K-1 mol-1

M, molar mass = ? g/mol

Procedure:

1) solve for n, # mol

2) M = #g / #mol

density

V= 1.00 Lmass= 1.25 gP= 195 mm HgT= 25 oC = 298 KR = 62.4 L mm Hg K-1 mol-1

M, molar mass = ? g/mol

CHCl3 = 12 + 1 + 3(35.4) = 119.2 g/mol

n = PV

RT=

195 mm Hg x 1.00 L

62.4 l mmHg K-1 mol-1 x 298 K

= .0105 mol

M = #grams

# mol=

1.25 g

.0105 mol= 119 g/mol

Group Work

What is the molar mass of a gas which has a density of 1.83 g/L measured at 27.0 oC and 0.538 atm?R = .0821 L atm K-1mol-1

V= mass= P= T= R = n = ? M, molar mass = ? g/mol

M = #grams

# mol=

1.83 g

.0218 mol= 83.9 g/mol

n = PV

RT=

0.538 atm x 1.00 L

.0821 L atm K-1 mol-1 x 300.1 K

= .0218 mol

V= 1.00 Lmass= 1.83 gP= 0.538 atmT= 27.0 oC = 300.1 KR = .0821 L atm K-1 mol-1

n = ?M, molar mass = ? g/mol

Gas laws, Stoichiometry

Problem 49, text:

A self contained breathing apparatus uses canisters containing potassium superoxide. The superoxide consumes the CO2 exhaled by the person and replacesit with O2.

What mass of solid KO2, in grams, is required to react with 8.90 L of CO2 at 22.0 oC and 767 mm Hg?

4 KO2(s) + 2 CO2(g) 2 K2CO3(s) + 3 O2(g)

When gases are included in a balanced equation for a reaction, the number of moles or grams of thegas used can be computed by the Ideal gasequation, PV = nRT.

Use Ideal gas equation in equation situation to compute moles of gas, then proceed normally to do problem.

Word of Warning: never use grams of solid or liquidin gas law equation: it won’t give a correct answer...

4 KO2(s) + 2 CO2(g) 2 K2CO3(s) + 3 O2(g)

? grams

V= 8.90 LT= 22.0 oC = 296.1 KP= 767 mm HgR= 62.4 L mm Hg K-1 mol-1

n =?

n = PV

RT=

767 mm Hg x 8.90 L

62.4 L mm Hg K-1 mol-1 x 296.1 K

= .36945 mol = .369 mol

CO 2

FIRST

4 KO2(s) + 2 CO2(g) 2 K2CO3(s) + 3 O2(g)

? grams

1 K =39.12 O= 32.0 71.1 g/mol

n = .369 moles

.369 mol CO2 = ? g KO2

.369 mol CO2

2 mol CO2

4 mol KO2

1 mol KO2

71.1 g KO2= 52.5 g KO2

Group Work

What volume of O2, collected at 22.0 oC and 728mm Hg would be produced by the decomposition of8.15 g KClO3, M = 122.5 g/mol?

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

1. Calculate moles of O2 from equation

2. Calculate V of O2 from ideal gas law

8.15 g KClO3

122.5 g KClO3

1 mol KClO3

2 mol KClO3

3 molO2

= .0998 mol O2

V = nRT

P=

.0998 mol x 62.4 L mm Hg K-1 mol-1 x 296.1 K

728 mm Hg

= 2.53 L

V= ? LT= 22.0 oC = 296.1 K P= 728 mm Hg R= 62.4 L mm Hg K-1 mol-1

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

8.15 g

122.5 g/mol

V, L =?