The Ideal Gas LawBringing It All Together

Objectives

When you complete this presentation, you will be able to

state the ideal gas lawderive the ideal gas law constant and discuss its unitsuse the ideal gas law to calculate pressure, volume, temperature, or amount of gas in a systemcalculate molar mass or density of a gas

Introduction

When we use the combined gas lawwe allow P, V, and T to vary.we keep the amount of gas constant.

If we vary the amount of gas as well as the pressure, volume, and temperature

we will use the Ideal Gas Law

Introduction

When we use the phrase “Ideal Gas Law,”we are talking about an ideal gaswe are not talking about a real gas

An ideal gas obeys all of the assumptions of the kinetic theory of gases -

small particlesno attractionmoving rapidlyperfectly elastic collisions

Introduction

Under most common circumstances, real gases act like ideal gases.

Only under conditions of -low temperature

orhigh pressure

will real gases deviate from ideality.

Application

Applying the kinetic theory, as we add an amount of gas to a container of gas, we are introducing additional particles to collide with the walls of the container.

Pressure goes up to keep V & T constantVolume goes up to keep P & T constantTemperature goes down to keep P & V constant

Application

This means that the amount (number of mols) of material varies -

directly with pressure and volume inversely with temperature

We will use the equation -PV = nRTP is pressure, V is volume, n is mols of gas, T is temperature, and R is the gas constant.

Application

PV = nRTP: measured in atm, kPa, or mm HgV: measured in Ln: measured in molT: measured in KR: a constant whose value depends on the units of P, V, n, and T

The Value of R

R is a constant whose value depends on the units of P, V, n, and T (mostly on the units of pressure)

We can find R by solving the ideal gas law for R and entering the appropriate values.

We remember from our studies of stoichiometry that 1.00 mol of a gas has a volume of 22.4 L at STP (standard temperature and pressure).

This gives us enough information to find R for each unit of pressure.

The Value of R

Where P is in atm:STP is 1.00 atm at 273 KpV = nRT ⇒ R = PV/nTR = [(1.00 atm)(22.4 L)]/[(1.00 mol)(273 K)]R = 0.0821 L-atm/mol-K

The Value of R

Where P is in mm Hg:STP is 760 mm Hg at 273 KpV = nRT ⇒ R = PV/nTR = [(760 mm Hg)(22.4 L)]/[(1.00 mol)(273 K)]R = 62.4 L-mm Hg/mol-K

The Value of R

Where P is in kPa:STP is 101.3 kPa at 273 KpV = nRT ⇒ R = PV/nTR = [(101.3 kPa)(22.4 L)]/[(1.00 mol)(273 K)]R = 8.314 L-mm Hg/mol-K

The Value of R

PV = nRTIf P is measured in atm:

R = 0.0821 atm-L/mol-KIf P is measured in kPa:

R = 8.314 kPa-L/mol-KIf P is measured in mm Hg:R = 62.4 mm Hg-L/mol-K

The Value of R

PV = nRT

Remember:The value of R is dependent on the units of pressure.Always use the correct value of R.

All appropriate values for R will be given to you for any test or quiz.

Example 1 – Finding PWhat is the pressure, in atm, of 0.125 mols of helium in a 4.00 L container at a temperature of 430 K?

P = ? atmV = 4.00 Ln = 0.125 molR = 0.0821 L-atm/mol-KT = 430 K

PV = nRT ⇒ P =

nRTV

=(0.125)(0.0821)

(430)(4.00) atm

P = 1.10321875 atm

= 1.10 atm

This is the value we use when pressure is

expressed in atm.

Practice Problems – Finding P1. What is the pressure, in atm, of 3.50 mols of N2

gas held in 45.0 L at a temperature of 310 K?

2. What is the pressure, in mm Hg, of 0.0400 mols of CO2 gas held in 10.0 L at a temperature of 350 K?

3. What is the pressure, in kPa, of 172 mols of He gas held in 675 L at a temperature of 273 K?

4. What is the pressure, in atm, of 0.00250 mols of Cl2 gas held in 0.100 L at a temperature of 455 K?

P = 0.198 atm

P = 87.4 mm Hg

P = 578 kPa

P = 0.934 atm

Example 2 – Finding VWhat is the volume of 2.50 mols of oxygen at a pressure of 85.0 kPa and a temperature of 315 K?

P = 85.0 kPaV = ? Ln = 2.50 molR = 8.31 L-kPa/mol-KT = 315 K

PV = nRT ⇒ V =

nRTP

=(2.50)(8.31)(315)

(85.0) L

P = 76.98970588 L

= 77.0 L

This is the value we use when pressure is

expressed in kPa.

Practice Problems – Finding V1. What is the volume of 1.00 mols of F2 gas at a

pressure of 0.450 atm and a temperature of 298 K?

2. What is the volume of 40.2 mols of UF6 gas at a pressure of 645 mm Hg and a temperature of 655 K?

3. What is the volume of 0.0424 mols of Ne gas at a pressure of 4.53 kPa and a temperature of 242 K?

4. What is the volume of 3.22 mols of O2 gas at a pressure of 4.67 atm a temperature of 273 K?

V = 54.4 L

V = 2,550 L

V = 18.8 L

V = 15.5 L

Example 3 – Finding nHow many mols of nitrogen is contained in a volume of 22.4 L at a pressure of 760 mm Hg and a temperature of 273 K?

P = 760 mm HgV = 22.4 Ln = ? molR = 62.4 L-mm Hg/mol-KT = 273 K

PV = nRT ⇒ n =

PVRT

=(760)(22.4)(62.4)(273)

mol

P = 0.999342538 mol

= 0.999 mol

This is the value we use when pressure is

expressed in mm Hg.

Practice Problems – Finding n1. How many mols of CO2 gas is in 75.0 L at a

pressure of 2.75 atm and a temperature of 298 K?

2. How many mols of SF6 gas is in 0.500 L at a pressure of 950 mm Hg and a temperature of 350 K?

3. How many mols of Ar gas is in 22.4 L at a pressure of 95.9 kPa and a temperature of 298 K?

4. How many mols of CH4 gas is in 0.0782 L at a pressure of 32.5 atm a temperature of 653 K?

n = 8.43 mol

n = 0.0217 mol

n = 0.867 mol

n = 0.0474 mol

Example 4 – Finding TWhat is the temperature of 1.60 mols of neon contained in a volume of 15.0 L at a pressure of 1.20 atm?

P = 1.20 atmV = 15.0 Ln = 1.60 molR = 0.0821 L-atm/mol-KT = ? K

PV = nRT ⇒ T =

PVnR

=(1.20)(15.0)

(1.60)(0.0821)

K

P = 137.0280146 K

= 137 K

This is the value we use when pressure is

expressed in atm.

Practice Problems – Finding T1. What is the temperature of 3.00 mol of CO2 gas

in 75.0 L and at a pressure of 1.00 atm?

2. What is the temperature of 0.755 mol of He gas in 4.25 L and at a pressure of 2,320 mm Hg?

3. What is the temperature of 35.0 mol of CH4 gas in 33.5 L and at a pressure of 2,520 kPa?

4. What is the temperature of 1.25 mol of CO2 gas in 25.0 L and at a pressure of 1.70 atm?

T = 305 K

T = 209 K

T = 290 K

T = 414 K

Application

The ability to measure the amount of the gas from pressure, volume, and temperature measurements is a powerful tool for exploring other properties of gases.

If we can also measure the mass of the gas, we can determine

the molar mass of the gasthe density of the gas

The Molar Mass of a Gas

We can determine the number of mols, n, of a gas by using pressure, volume, and temperature measurements and the ideal gas law.

n = PV/RT

The molar mass, M, is the mass, m, divided by the number of mols.

M =m/n

Putting the two equations togetherM = m/(PV/RT) = mRT/PV

Example 5 – Finding MAt 301 K and 0.974 atm, 1.00 L of a gas has a mass of 5.16 g. What is the molar mass of this gas?

P = 0.974 atmV = 1.00 LR = 0.0821 L-atm/mol-KT = 301 Km = 5.16 g

M =

mRTPV

=(5.16)(0.0821)

(301)(0.974)(1.00) g/mol

M = 130.9183121 g/mol

= 131 g/mol

This is the value we use when pressure is

expressed in atm.

Practice Problems – Finding M1. At 302 K and 1.05 atm, 1.81 L of a gas has a mass

of 5.42 g. What is the molar mass of this gas?

2. At 260 K and 695 mm Hg, 5.41 L of a gas has a mass of 10.2 g. What is the molar mass of this gas?

3. At 285 K and 97.2 kPa, 95.6 L of a gas has a mass of 329 g. What is the molar mass of this gas?

4. At 310 K and 4.15 atm, 0.350 L of a gas has a mass of 3.32 g. What is the molar mass of this gas?

M = 70.0 g/mol

M = 44.0 g/mol

M = 83.9 g/mol

M = 58.2 g/mol

The Density of a Gas

The density of a gas is the mass of the gas divided by its volume

ρ = m/V

If we measure the density of a gas and know its pressure and temperature, we can find the molar mass, M .

M =mRTPV

m RTV P=

m RTV P=

ρRTP=

We are just rearranging V and P in the denominator.

We separate out the m/V

term.

We replace m/V with

ρ .

Example 6 – Finding MWhat is the molar mass of a gas with a density of 3.42 g/L at 293 K and 0.980 atm?

ρ = 3.42 g/LR = 0.0821 L-atm/mol-KT = 293 KP = 0.980 atm

M =

ρRTP

=(3.42)(0.0821)

(293)(0.980) g/mol

M = 83.94808776 g/mol

= 83.9 g/mol

This is the value we use when pressure is

expressed in atm.

Practice Problems – Finding M1. What is the molar mass of a gas with a

density of 3.25 g/L at 150 K and 10.0 atm?

2. What is the molar mass of a gas with a density of 5.76 g/L at 273 K and 672 mm Hg?

3. What is the molar mass of a gas with a density of 5.17 g/L at 452 K and 672 kPa?

4. What is the molar mass of a gas with a density of 0.508 g/L at 345 K and 0.450 atm?

M = 4.00 g/mol

M = 146 g/mol

M = 352 g/mol

M = 32.0 g/mol

The Density of a Gas

If we know the molar mass of a gas, then we can calculate the density of that gas under specific conditions of pressure and temperature.

M =

ρ =

M PRT

ρRTP

If we rearrange and solve

for ρ …

We used this equation to solve for molar mass when we knew

density.

Example 7 – Finding ρWhat is the density of a sample of ammonia gas, NH3, M = 17.04 g/mol, at 0.928 atm and 336 K?

M = 17.04 g/molP = 0.928 atmR = 0.0821 L-atm/mol-KT = 336 K

ρ =M PRT

=(17.04)(0.928)(0.0821)(336)

g/L

ρ = 0.5732382112 g/L

= 0.573 g/L

This is the value we use when pressure is

expressed in atm.

Practice Problems – Finding r1. What is the density of a sample of CH4, M =

16.05 g/mol, at 1.25 atm and 280 K?

2. What is the density of a sample of H2, M = 2.02 g/mol, at 672 mm Hg and 261 K?

3. What is the density of a sample of CO2, M = 44.01 g/mol, at 175 kPa and 310 K?

4. What is the density of a sample of UF6, M = 352.0 g/mol, at 10.2 atm and 397 K?

r = 1.63 g/L

r = 0.0833 g/L

r = 3.00 g/L

r = 110. g/L

Summary

The ideal gas law allows us to calculate the pressures, volumes, amounts, and temperatures of gases.

PV = nRT

The value of the ideal gas constant, R, depends on the units of pressure.

For units of atm, R = 0.0821 L-atm/mol-K

For units of kPa, R = 8.314 L-kPa/mol-K

For units of mm Hg, R = 62.4 L-mm Hg/mol-K

We can also use the ideal gas law to calculate the molar masses and densities of gases.