The Ideal Gas LawThe Ideal Gas Law Section 4.4 Pg. 172-175.

The Ideal Gas Law Section 4.4

Pg. 172-175

IDEAL GAS LAW Before getting too far into this law, it is important to

understand the difference between an ideal gas and a real gas....

IDEAL GAS – does not really exist, it is hypothetical

Follows all gas laws perfectly under all conditions

Does not condense when cooled

Assumes that the particles have no volume and are not attracted to each other

REAL GAS – does not follow gas laws exactly, it deviates at low temperatures and high pressures

Condenses to liquid or sometimes solid when cooled or under pressure

Particles are attracted to each other and have volume

Behaves like an ideal gas at higher temperatures and lower pressures

Ideal vs. Real Gases(a)In an ideal gas, the molecules collide like

perfectly hard spheres and rebound very quickly after collision.

(b) In a real gas, the molecules are “soft” (can be deformed) and intermolecular attractions are important. The process of collision takes a slightly longer time, as a result.

We will be dealing with gases as if they were

ideal

In 1873, Johannes van der Waals

hypothesized the existence of

attractions between gas molecules to

explain deviations from the ideal law.

The general forces of attraction,

called van der Walls forces, include

dipole-dipole and London forces.

Johannes van der Waals (1837-1923)

IDEAL GAS LAW Describes the interrelationship of pressure, temperature, volume

and amount (moles) of matter; the four variables that define a gas system

REMEMBER:

Boyle’s Law: Volume is inversely proportional (α) to pressure V α 1/P

Charles’ Law: Volume is directly proportional to temperature V α T

Avogadro’s Theory: Volume is directly proportional to chemical amount (mol) V α n

From all of these comparisons to volume:

V α 1 x T x n = V α T n P P

V = R T n R is a constant called the universal gas constant P (allows us to change from α to =)

V = nRT PV = nRT (ideal gas law) P

R = universal gas constant

Depends on STP or SATP, atm or kPa

Units: L • kPa/mol • K value = 8.314 L • kPa/mol • K

Units: L • atm/mol • K value = 0.0821 L • atm/mol • K

Make sure you look at the unit for pressure to decide which R value to use

Any idea how we came up with the number??

You substitute SATP or STP conditions for one mole into the ideal gas law and solve for R

R = PV = (101.325 kPa)(22.414L) = 8.314 L • kPa/mol • K . nT (1.0 mol)(273.15K)

Using the Ideal Gas Law When solving for the ideal gas law, start by listing your

variables. If three are known of the four, you can solve for the last one.

Example One: What mass of neon gas should be introduced into an evacuated 0.88L tube to produce a pressure of 90 kPa at 30°C?

P = 90 kPa

V = 0.88L

T = 30°C 303K

R = 8.314 L • kPa/mol • K

m = ?

n = ?

PV = nRT

n = PV RT

n = (90kPa)(0.88L) (8.314 L•kPa/mol•K)(303K)

n = 0.0314 mol x 20.18 g = 0.63 g

1 mol

Using the Ideal Gas Law – Practice 2. A rigid steel cylinder with a volume of 2.00 L is filled with

nitrogen gas to a final pressure of 20.0 atm at 27°C. How many moles of N2 gas does the cylinder contain?

P = 20.0 atm PV = nRT

V = 2.00 L n = PV/RT

T = 27 °C = 300 K n = (20.0 atm)(2.00L)

R = 0.0821 L•atm/mol•K (0.0821 L•atm/mol•K)

(300K)

n = ? n = 1.624 mol

n = 1.62 mol

Using the Ideal Gas Law – Practice

3. Predict the volume occupied by 0.78 g of hydrogen at 22°C and 125 kPa

P = 125 kPa

V = ?

T = 22 °C = 295 K

R = 8.314 L•kPa/mol•K

m = 0.78g

n = ?

n H2 = 0.78 g x 1 mol = 0.386 mol 2.02g

PV = nRT V = nRT P

V = (0.386 mol)(8.314L•kPa/mol•K)(295K) 125 kPa

V = 7.573 L

V = 7.6 L

Homework

Pg. 174 #3-5

Pg .176 # 1,2,6,9,11