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Combined Gas Law. The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! P 1 V 1 P 2 V 2 =

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combined gas law
Combined Gas Law
  • The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION!

P1 V1 P2 V2

=

T1 T2

No, it’s not related to R2D2

combined gas law1
Combined Gas Law

If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law!

=

P1

V1

P2

Boyle’s Law

Charles’ Law

Gay-Lussac’s Law

V2

T1

T2

combined gas law problem
Combined Gas Law Problem

A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

Set up Data Table

P1 = 0.800 atm V1 = 180 mL T1 = 302 K

P2 = 3.20 atm V2= 90 mL T2 = ??

*Make volume units the same!

calculation
Calculation

P1 = 0.800 atm V1 = 180 mL T1 = 302 K

P2 = 3.20 atm V2= 90 mL T2 = ??

P1 V1 P2 V2

=

T1 T2

T2 = P2 V2 T1

P1 V1

T2 = 3.20 atm x 90.0 mL x 302 K 0.800 atm x 180.0 mL

T2 = 604 K - 273 = 331 °C

= 604 K

learning check
Learning Check

A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

Note: Volumes must be the same unit and pressures must be the same unit!

slide7

1. List variables:

  • V1 = 675 mL = 0.675 L
  • T1 = 35°C = 308 K
  • P1 = 0.850 atm
  • V2 = .315 L
  • P2 = 802 mmHg = 1.06 atm
    • Decide on the appropriate gas law:
      • Everything’s changing, so Combined!
      • 3. Rearrange to solve for unknown:
      • T2 =

(P2) (V2) (T1)

= 179.2 K =

(178.4K)

(P1) (V1)

slide8

PV

=nRT

Ideal gas law

  • IF WE COMBINE ALL OF THE LAWS we’ve looked at TOGETHER - INCLUDING AVOGADRO’S LAW - WE GET:
slide9

Ideal gas constant(R)

  • R IS A CONSTANT THAT CONNECTS THE 4 VARIABLES
  • R IS DEPENDENT ON THE UNITS OF THE VARIABLE FOR PRESSURE
    • TEMP IS ALWAYS IN KELVIN
    • VOLUME IS ALWAYS IN LITERS
    • PRESSURE IS IN EITHER atm OR mmHg OR kPa
slide10

L•atm

L•kPa

R=.0821

R=8.314

L•mmHg

R=62.4

mol•K

mol•K

mol•K

  • Because of the different pressure units we use there are 3 different values for “R””
  • IF PRESSURE IS GIVEN IN atm
  • IF PRESSURE IS GIVEN IN mmHg
  • IF PRESSURE IS GIVEN IN kPa
learning check1
Learning Check

Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (in atm) in the tank in the dentist office?

slide12

Using Ideal gas law

L•atm

0.0821

mol•K

(2.86 mol)(.0821)(296K)

PV = nRT

20.0 L

1. List variables:

n = 2.86 moles

  • R
  • P
  • ?
  • T
  • V
  • 20.0 L
  • 296K

2. Rearrange to solve for unknown:

P = nRT

V

3, Plug & Chug:

  • =3.48 atm
gas diffusion and effusion
GAS DIFFUSION AND EFFUSION
  • Diffusion is the movement of molecules to fill a container
  • Effusion is the movement of molecules through a small hole into an empty container.
graham s law
Graham’s Law

Rates of effusion of gases are inversely proportional to the square root of their molar masses, at constant temp. & pressure.

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

M = molar mass & Gas B is the heavier gas!

graham s law1
Graham’s Law

Molecules effuse thru holes in a rubber balloon – that’s the main reason they get ‘whimpy’ after awhile!

They do this at a rate that is inversely proportional to molar mass.

Therefore, He (4 g/mol) effuses more rapidly than O2 (32 g/mol) at same T. (It’s lighter!)

He

Lighter gases effuse faster than heavier ones

graham s law2
Graham’s Law
  • We can use the entire equation to calculate the actual speed of gas particles, however…
  • We will just use the square root side to COMPARE rates of effusion (speeds)
  • Ex. Compare the rates of effusion of oxygen gas & hydrogen gas.
  • 1st – find their molar masses!
    • O2 = 32.0 g/molH2 = 2.0 g/mol

2nd -Put the heavier gas (Gas B) in the numerator!

4

32 g

You’re not done yet!

2.0 g

graham s law3
Graham’s Law
  • The number 4 is not much of a “comparison”!
  • You must put your answer in sentence form!
  • Try this:
  • “_______ gas travels (or effuses) at a rate

___ times faster than _________ gas.”

So, the answer is…

  • “Hydrogen gas travels (or effuses) at a rate

4 times faster than oxygen gas.”

(Lighter gas)

(#)

(heavier)

graham s law4
Graham’s Law
  • You try it!
  • Compare the rates of effusion of Ar and nitrogen gas (N2)

39.9 g/mol

28.0 g/mol

39.9

28.0

1.19

“Nitrogen gas travels (or effuses) at a rate 1.19 times faster than argon gas.”

slide19

All of these gas laws work just ducky

assuming the gases are ‘ideal”

(Points with “no volume” & “no mutual

attraction”

Most of the time, gases conform to

ideal conditions.

So, when are gases not “ideal”?

Under conditions of low temperature

& high pressure

(force molecules close enough to

affect each other!)

deviations from ideal gas law
Deviations from Ideal Gas Law
  • Real Molecules have volumes and attractive forces between them.
  • Gases are not “Ideal” under conditions of high pressure & low temperature which bring particles close enough together to affect each other!