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Ideal gases Assumptions: There are a huge number N of molecules, each of PowerPoint Presentation

Ideal gases Assumptions: There are a huge number N of molecules, each of

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- Ideal gases
- Assumptions:
- There are a huge number N of molecules, each of
- mass m, moving in random directions at various speeds.
- On average, the molecules are large distances from each other.
- --> The average separation is much greater than the size
- The molecules only interact when they collide.
- Collisions with each other and the wall are perfectly elastic,
- like perfectly elastic pool balls.

- These assumptions are usually valid when a gas is at:
- low density; and
- relatively higher temps (away from condensation point)

Ideal gas law: PV = nRT (memorize)

Usually,

P = pressure in Pa = Nm-2

V = volume in m3

n = the number of moles of gas

R = the universal gas constant =8.31 J mol-1 K-1 (given)

T = the absolute (kelvin) temperature

Also useful to know:

Standard temperature and pressure (STP):

1. T = 273 K = 0 0C

2. Pressure = 1 atm = 1.01 x 105 N m-2 (Pa)

Review: What is a mole (mol)?

1 mol = the amount of substance that contains as many

atoms or molecules as there are in 12.00 g of carbon-12

1 mol= the number of grams of a substance numerically equal to the atomic (or molecular) mass

--> The “grams per mol” is called its “molar mass”

Ex 1. What is the mass in grams (g) of 1 mol of He gas?

Ex 2. How many moles are in 0.700 g of He gas?

Ex 3. What is the mass of 1 mole of O2 gas?

Ex 4. How many moles are in 12.8 g of oxygen gas?

Note: 1 mol contains Avagadro’s number NA of particles

NA = 6.02 x 1023 particles per mole (given)

= 6.02 x 1023 mol-1

Ex 5. How many atoms are in 0.700 g of helium gas?

Ex 6. How many molecules are in 12.8 g of oxygen gas?

Ex 7. What is the volume occupied by 5.60 g of nitrogen

(molecular mass = 14.0) gas is at a temperature of 40.0 0C

and a pressure of 0.860 atm?

T = _______ K; P = _________ Pa; n = _______moles

Use: P V = n R T

How many liters is this? (1000 L = 1 m3)

- Ex. 8. A gas occupies a volume of 5.0 liters at a pressure of
- 2.0 atm. If the gas compressed to a volume of 3.0 liters
- at constant temperature, what is its new pressure (in atm)?
- Constant T: PV = nRT --> solve for T = PV/nR = constant
- before = after
- P1V1/nR = P2V2/nR
- P1V1 = P2V2 (n also constant)
- With the equation in this form, any P and V units are OK
- as long as you are consistent. So substitute values in the last
- equation and solve for P2:

The previous example uses pressure of Boyle’s Law:

PV = nRT = constant b/c T is constant

This is like: xy = constant

--> P and V are inversely proportional

Graph: P

V

Explain Boyle’s law using particle collisions.

- Ex. 9. Heat is added to a gas initially at a temp of 37 pressure of 0C and
- a pressure of 1.33 atm at constant volume. What will be the
- new pressure (in atm) if the temperature rises to 57 0C?
- Constant V: PV = nRT
- Solve for: V = nRT/P = constant
- before = after
- nRT1/P1 = nRT2/P2
- T1/P1 = T2/P2 (n is constant)
- With the equation in this form, any P units are OK as long
- as you are consistent, but T must be in kelvins. Substitute in:

This is an example of pressure of Guy Lussac’s Law:

PV = nRT --> V = nRT/P = constant

--> T/P = constant

--> T = constant x P --> T and P are directly proportional

Graph: P

T

Explain Guy-Lussac’s law using particle collisions:

Ex. 10. A gas occupies a volume of 2.00 liters at a temp of

of 25 0C. What will be its new volume (in liters) when it is at

a new temp of 50. 0C at constant pressure?

Constant P: PV = nRT

--> solve for P = nRT/V = constant

before = after

nRT1/V1 = nRT2/V2

T1/V1 = T2/V2 (n constant)

Make sure T is in kelvins, Then substitute values in:

This is an example of ofCharles’ Law:

PV = nRT --> P = nRT/V = constant

--> T/V = constant

--> T = constant x V --> T and V are directly proportional

Graph: V

T

Explain Charles’ law using particle collisions. Especially,

how can pressure remain constant when T increases?

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