Combined and ideal gas lawsCombined and ideal gas laws

Gases Have MassGases DiffuseGases Expand To Fill

ContainersGases Exert PressureGases Are CompressiblePressure & Temperature

Are Dependent

Gases Have MassGases DiffuseGases Expand To Fill

ContainersGases Exert PressureGases Are CompressiblePressure & Temperature

Are Dependent

Gas propertiesGas properties

VOLUME (V)– UNITS OF VOLUME (L)

AMOUNT (n)– UNITS OF AMOUNT (MOLES)

TEMPERATURE (T)– UNITS OF TEMPERATURE (K)

PRESSURE (P)–UNITS OF PRESSURE (mmHg)–UNITS OF PRESSURE (kPa)–UNITS OF PRESSURE (atm)–UNITS OF PRESSURE (torr)

VOLUME (V)– UNITS OF VOLUME (L)

AMOUNT (n)– UNITS OF AMOUNT (MOLES)

TEMPERATURE (T)– UNITS OF TEMPERATURE (K)

PRESSURE (P)–UNITS OF PRESSURE (mmHg)–UNITS OF PRESSURE (kPa)–UNITS OF PRESSURE (atm)–UNITS OF PRESSURE (torr)

Gas variablesGas variables

P1V1 = P2V2

P1V1 = P2V2

BOYLE’S LAW–PRESSURE & VOLUME–AS P THEN V–AT CONSTANT T, n

BOYLE’S LAW–PRESSURE & VOLUME–AS P THEN V–AT CONSTANT T, n

A little reviewA little review

A Little reviewA Little reviewCHARLES’ LAW:

–TEMPERATURE & VOLUME

–AS T THEN V–AT CONSTANT P, n

CHARLES’ LAW: –TEMPERATURE & VOLUME

–AS T THEN V–AT CONSTANT P, n

V1V1

T2T2

==T1T1

V2V2

A Little reviewA Little reviewGAY-LUSSAC’S LAW:

–TEMPERATURE & PRESSURE

–AS P THEN T–AT CONSTANT V, n

GAY-LUSSAC’S LAW: –TEMPERATURE & PRESSURE

–AS P THEN T–AT CONSTANT V, n

P1P1

T2T2

==T1T1

P2P2

Another step up…Another step up…

PV=k1PV=k1 V/T=k2V/T=k2 P/T=k3P/T=k3

If we combine all of the relationships from the 3 laws covered thus far (Boyle’s, Charles’s, and Gay-Lussac’s) we can develop a mathematical equation that can solve for a situation where 3 variables change :

If we combine all of the relationships from the 3 laws covered thus far (Boyle’s, Charles’s, and Gay-Lussac’s) we can develop a mathematical equation that can solve for a situation where 3 variables change :

Combined gas law

Combined gas lawAMOUNT IS HELD

CONSTANTIS USED WHEN YOU HAVE A CHANGE IN VOLUME, PRESSURE, OR TEMPERATURE

AMOUNT IS HELD CONSTANT

IS USED WHEN YOU HAVE A CHANGE IN VOLUME, PRESSURE, OR TEMPERATURE P1V1P1V1

T1T1

= k= kP2V2P2V2

T2T2

= k= k

Combined gas law

Combined gas lawAMOUNT IS HELD CONSTANT

IS USED WHEN YOU HAVE A CHANGE IN VOLUME, PRESSURE, OR TEMPERATURE

AMOUNT IS HELD CONSTANT IS USED WHEN YOU HAVE A

CHANGE IN VOLUME, PRESSURE, OR TEMPERATURE

P1V1P1V1

T1T1

P2V2P2V2

T2T2

==

P1V1T2P1V1T2 P2V2T1P2V2T1==

A GAS WITH A VOLUME OF 4.0L AT STP. WHAT IS ITS VOLUME

AT 2.0ATM AND AT 30°C?

A GAS WITH A VOLUME OF 4.0L AT STP. WHAT IS ITS VOLUME

AT 2.0ATM AND AT 30°C?

Example problemExample problem

P1 P1 V1 V1 T1 T1

P2P2V2 V2 T2 T2

1atm1atm4.0 L4.0 L273K273K

2.0 atm

2.0 atm??

30°C + 273

30°C + 273=

303K= 303K

PLUG & CHUGPLUG & CHUG

P1V1P1V1

T1T1

P2V2P2V2

T2T2

==

2.22L = V22.22L = V2

SO FAR WE’VE COMPARED ALL THE VARIABLES EXCEPT THE AMOUNT OF A GAS (n).

There is a lesser known law called avogadro’s law which relates v & n.

It turns out that they are directly related to each other.

As # of moles increases then v increases.

SO FAR WE’VE COMPARED ALL THE VARIABLES EXCEPT THE AMOUNT OF A GAS (n).

There is a lesser known law called avogadro’s law which relates v & n.

It turns out that they are directly related to each other.

As # of moles increases then v increases.

V/n = k

V/n = k

ideal gas lawideal gas lawWHICH LEADS US TO THE

IDEAL GAS LAW – SO FAR WE HAVE ALWAYS

HELD AT LEAST 1 OF THE VARIABLES CONSTANT.

WE CAN SET UP A MUCH MORE POWERFUL EQN, WHICH CAN BE DERIVED BY COMBINING THE PROPORTIONS EXPRESSED BY THE PREVIOUS LAWS.

WHICH LEADS US TO THE IDEAL GAS LAW –

SO FAR WE HAVE ALWAYS HELD AT LEAST 1 OF THE VARIABLES CONSTANT.

WE CAN SET UP A MUCH MORE POWERFUL EQN, WHICH CAN BE DERIVED BY COMBINING THE PROPORTIONS EXPRESSED BY THE PREVIOUS LAWS.

IF WE COMBINE ALL OF THE LAWS TOGETHER INCLUDING AVOGADRO’S LAW MENTIONED EARLIER WE GET:

IF WE COMBINE ALL OF THE LAWS TOGETHER INCLUDING AVOGADRO’S LAW MENTIONED EARLIER WE GET:

PVPVTTnn

= R= R

WHERE R IS THE

UNIVERSAL GAS

CONSTANT

WHERE R IS THE

UNIVERSAL GAS

CONSTANTNORMALLYWRITTEN

AS

NORMALLYWRITTEN

ASPVPV=nRT=nRT

Ideal gas lawIdeal gas law

Ideal gas constant(R)

Ideal gas constant(R)R IS A CONSTANT THAT

CONNECTS THE 4 VARIABLES R IS DEPENDENT ON THE

UNITS OF THE VARIABLES FOR P, V, & T–TEMP IS ALWAYS IN KELVIN–VOLUME IS IN LITERS–PRESSURE IS IN EITHER atm OR mmHg OR kPa

R IS A CONSTANT THAT CONNECTS THE 4 VARIABLES

R IS DEPENDENT ON THE UNITS OF THE VARIABLES FOR P, V, & T–TEMP IS ALWAYS IN KELVIN–VOLUME IS IN LITERS–PRESSURE IS IN EITHER atm OR mmHg OR kPa

BECAUSE OF THE DIFFERENT PRESSURE UNITS THERE ARE 3 POSSIBILITIES FOR OUR R

BECAUSE OF THE DIFFERENT PRESSURE UNITS THERE ARE 3 POSSIBILITIES FOR OUR R

R=.0821R=.0821L•atmL•atmmol•Kmol•K

– IF PRESSURE IS GIVEN IN mmHg

– IF PRESSURE IS GIVEN IN mmHg

R=62.4R=62.4L•mmHgL•mmHgmol•Kmol•K

– IF PRESSURE IS GIVEN IN kPa

– IF PRESSURE IS GIVEN IN kPa

R=8.314R=8.314L•kPaL•kPamol•Kmol•K

– IF PRESSURE IS GIVEN IN atm

– IF PRESSURE IS GIVEN IN atm

Using Ideal gas law

Using Ideal gas lawEG #1: WHAT VOL DOES 9.45g

OF C2H2 OCCUPY AT STP?EG #1: WHAT VOL DOES 9.45g

OF C2H2 OCCUPY AT STP?

P P

V V T T

1atm1atm

?? 273K273K

R R

n n =.3635 mol

=.3635 mol

.0821 .0821L•atmL•atmmol•Kmol•K

9.45g9.45g

26g26g

PV = nRTPV = nRT(1.0at

m)(1.0at

m)(V)(V)

(.3635mol)(.3635mol) (273K)(273K)

V = 8.15LV = 8.15L

==(.0821 )(.0821 )L•atm

mol•KL•atmmol•K

(1.0atm)

(1.0atm)

(V)(V) (8.147L•atm

)

(8.147L•atm

)

==

Using Ideal gas law

Using Ideal gas lawEG #2: A CAMPING STOVE

PROPANE TANK HOLDS 3000g OF C3H8. HOW LARGE A CONTAINER

WOULD BE NEEDED TO HOLD THE SAME AMOUNT OF PROPANE AS A GAS AT 25°C AND A PRESSURE OF

303kPa?

EG #2: A CAMPING STOVE PROPANE TANK HOLDS 3000g OF C3H8. HOW LARGE A CONTAINER

WOULD BE NEEDED TO HOLD THE SAME AMOUNT OF PROPANE AS A GAS AT 25°C AND A PRESSURE OF

303kPa?

Using Ideal gas law

Using Ideal gas law

P P

V V T T

303kPa

303kPa?? 298K298K

R R

n n =68.2 mol

=68.2 mol

8.314 8.314L•kPaL•kPa

mol•Kmol•K

3000g3000g44g44g

PV = nRTPV = nRT

(303kPa)

(303kPa)

(V)(V)(68.2mol)(68.2mol) (298K)(298K)

==(8.314 )(8.314 )L•kPa

mol•KL•kPamol•K

(303kPa)(303kPa) (V)(V) (168,970.4L•kPa)(168,970.4L•kPa)==

V = 557.7LV = 557.7L