# Ideal Gas Law PV=nRT.

## Presentation on theme: "Ideal Gas Law PV=nRT."— Presentation transcript:

Ideal Gas Law PV=nRT

Remember… Boyle’s Law Charles’ Law:
Combined Gas Law: (Units MUST Match Temp in Kelvin!!!)

A gas with a volume of 350 ml is collected at 15o C and 120 kPa
A gas with a volume of 350 ml is collected at 15o C and 120 kPa. If the temperature changes to 30o C, what pressure would be required to put this gas in a 300 ml container?

A balloon has a volume of 500 ml at a temperature of 22oC and a pressure of 755 mmHg. If the balloon is cooled to 0o C and a pressure of 145 mmHg, what is its new volume?

Avogadro’s Principle Under similar conditions (same Temp and Pressure) equal volumes of gases contain equal numbers of particles. 10 L of H2 (g) and 10 L of O2 (g) Both at Standard Temperature and Pressure (STP) contain… The same number of particles!

The volume of 1 mole of gas particles
Molar Volume The volume of 1 mole of gas particles at STP is 22.4 L

Ideal Gas Law Animation:

Try this: 1 mole of gas occupies 22.4 L at STP = __________ ml (22400)
= ___________ moles of gas (1 mole) = ___________ particles (6.02 x 1023)

How many particles in 11.2 dm3 of gas at STP?
0.5 moles = 3.01 x 1023 particles 22,400 cm3 of NH3 gas at STP weighs? = 22.4 L = 1 mole = 17 grams (add up MW) 44.8 L of NH3 at STP weighs? = 2 moles = 34 grams _____ grams = 1 mole of nitrogen gas = _____ L at STP? 28.00 22.4

How many N2 molecules are in 22.4 dm3 at STP?
= 1 mole = 6.02 x 1023 What volume will 1.2 x 1024 H2 molecules occupy at STP? = 2 moles = 44.8 L at STP

Ideal Gas Equation Use when NOT at STP!!!
PV= nRT P = Pressure (in kPa) V = Volume (in Liters or dm3) n = number of moles T = Temperature (in Kelvin) R = L• kPa mole • K

Development of R in

1. What volume will 2 moles of NO2 occupy at 300 Kelvin and 90 kPa?

What will be the temp of 2 grams of H2 if 5000 cm3 is at 5 atm?

Finding Molecular Weight of a Gas
Remember: MW = grams / moles Converting grams to moles Divide grams by the molecular weight

PV=nRT 92kPa • 5.0 L= n • 8.31 • 293 K n = 0.19 mol
1) 5.0 L of a gas weighs g at 20o C & 92 kPa. What is the mole weight of the gas? PV=nRT 92kPa • 5.0 L= n • 8.31 • 293 K n = 0.19 mol

2)If the mole weight of a gas is 26 g/mol and 18
2)If the mole weight of a gas is 26 g/mol and g of the gas is 30 L at 21o C, what is the pressure of the gas? PV= nRT P x 30 L=0.69 mol x 8.31 x 294 K P = 56.2kPa

Stoichiometry Solving Steps. Balance the Equation
Change grams to moles Use mole ratio to solve Change moles to volumes

Use mole ratio (coefficients) Use MW on P.T. Use 22.4 L/mol
@ STP Or PV = nRT Use 22.4 L/mol @ STP Or PV = nRT Use MW on P.T.

Mg (c) + HCl (ag)  MgCl2 + H2 (g)
If 2.43 g Mg react what volume of H2 is produced? (at STP) 1:1 0.1 mole 0.1 moles Mg (c) HCl (ag)  MgCl2 + H2 (g) 2 ? L 2.24 L H2 2.43 g Mg

Mg (c) +2 HCl (ag)  MgCl2 + H2 (g)
If 2.43 g Mg react what volume of H2 is produced at 40oC and 85 kPa? 1:1 0.1 mole 0.1 moles Mg (c) HCl (ag)  MgCl2 + H2 (g) 2 ? L 3.06 L H2 2.43 g Mg

Mg (c) + HCl (ag)  MgCl2 + H2 (g)
If 250 ml of H2 is produced at 20o C & 100 kPa, what mass of Mg reacted? 1:1 0.01 mole 0.01 moles Mg (c) HCl (ag)  MgCl2 + H2 (g) 2 0.243 g Mg ? g 250 ml

Volume to Volume THE MOLE RATIO IS THE SAME AS THE VOLUME RATIO.
Liters B Use mole ratio (coefficients) Liters A

What vol. of oxygen is needed to completely burn 1 L of methane?
Burning of methane: What vol. of oxygen is needed to completely burn 1 L of methane? 1:2 1.0 L 2.0 L CH O2  CO2 (g) H2O (l) 2 2

To produce 11.2 L of CO2 requires how many moles of O2 at STP?
Burning of methane: To produce 11.2 L of CO2 requires how many moles of O2 at STP? 2:1 1 mole = 22.4 L 11.2 L CH O2  CO2 (g) H2O (l) 2 2