Gases and Moles

Gases and MolesGases and MolesThe Ideal Gas EquationThe Ideal Gas Equation

What factors affect the What factors affect the pressure of a confined gas?pressure of a confined gas?

1.1. Number of moleculesNumber of molecules

2.2. TemperatureTemperature

3.3. Volume of the Volume of the containercontainer

Think in terms of the number of collisions.Think in terms of the number of collisions.

Number of moleculesNumber of moleculesIncreasing the number of Increasing the number of

molecules increases the number molecules increases the number of collisions …of collisions …

P P nn Where n is the Where n is the number of moles number of moles of moleculesof molecules

… … which increases the pressure.which increases the pressure.

TemperatureTemperatureIncreasing the temperature makes the Increasing the temperature makes the

molecules move faster, increasing molecules move faster, increasing the number of collisions …the number of collisions …

P P TT Where T is the Where T is the absolute absolute temperaturetemperature

… … which increases the pressure.which increases the pressure.

VolumeVolumeIncreasing the volume of the Increasing the volume of the

container decreases the number container decreases the number of collisions …of collisions …

P P Where V is Where V is the volumethe volume

11VV

… … which decreases the pressure.which decreases the pressure.

Sooooo…Sooooo…

P P nn P P TT

P P 11VV

P P nnVVTT

nn TTRR

Make it into an equationMake it into an equation

P P nnVVTT

P P VV

The Ideal Gas EquationThe Ideal Gas Equation

nn TTRRP P VV

nn TTRRPPVV… … is usually written as …is usually written as …


nn TTRRP P VV

R = R = 0.08210.0821 L atmL atmmol Kmol K

R is the “gas constant”R is the “gas constant”


nn TTRRP P VVCan R be in units Can R be in units

other thanother than L atmL atmmol Kmol K

?


nn TTRRP P VVR = 0.0821 L atm/mol KR = 0.0821 L atm/mol KR = 8.314 L kPa/mol KR = 8.314 L kPa/mol KR = 62.4 L torr/mol KR = 62.4 L torr/mol K


nn TTRRP P VVR = 0.0821 L R = 0.0821 L atmatm/mol K/mol KR = 8.314 L kPa/mol KR = 8.314 L kPa/mol KR = 62.4 L torr/mol KR = 62.4 L torr/mol K


nn TTRRP P VVR = 0.0821 L atm/mol KR = 0.0821 L atm/mol KR = 8.314 L R = 8.314 L kPakPa/mol K/mol KR = 62.4 L torr/mol KR = 62.4 L torr/mol K


nn TTRRP P VVR = 0.0821 L atm/mol KR = 0.0821 L atm/mol KR = 8.314 L kPa/mol KR = 8.314 L kPa/mol KR = 62.4 L R = 62.4 L torrtorr/mol K/mol K

Ideal Gas EquationIdeal Gas Equation

The ideal gas equation relates The ideal gas equation relates pressure, volume, temperature pressure, volume, temperature and the number of moles of a and the number of moles of a quantity of gas.quantity of gas.

PV = nRTPV = nRT

Ideal Gas EquationIdeal Gas Equation

Use the ideal gas equation Use the ideal gas equation whenever the problem gives you whenever the problem gives you mass or moles, or asks for a mass mass or moles, or asks for a mass or a number of moles.or a number of moles.

PV = nRTPV = nRT

Ideal gas equation problem:Ideal gas equation problem:

Some ammonia gas (NHSome ammonia gas (NH33) is ) is

contained in a 2.50 L flask at contained in a 2.50 L flask at a temperature of 20.0 C. If a temperature of 20.0 C. If there are 0.0931 moles of the there are 0.0931 moles of the gas, what is its pressure?gas, what is its pressure?

SolutionSolution

PV = nRTPV = nRT P = (nRT)/VP = (nRT)/V

= (0.0931 = (0.0931 molmol))

P = P = 0.896 atm0.896 atm

2.50 L2.50 L

)(293 )(293 KK))(0.0821(0.0821L atmL atm

mol Kmol K

Here’s another oneHere’s another one

Find the volume of 1.00 Find the volume of 1.00 mole of nitrogen gas (Nmole of nitrogen gas (N22) )

at 0.0 C and 1.00 atm of at 0.0 C and 1.00 atm of pressure.pressure.

SolutionSolution

V = (1.00 V = (1.00 molmol))

1.00 atm1.00 atm

)(273 )(273 KK))(0.0821(0.0821L atmL atm

mol Kmol K

PV = nRTPV = nRT V = (nRT)/PV = (nRT)/P

V = V = 22.4 L22.4 L

Ideal gas equation problem:Ideal gas equation problem:

How many grams of sulfur How many grams of sulfur trioxide are in an 855 mL trioxide are in an 855 mL container at a pressure of container at a pressure of 1585 torr and a temperature 1585 torr and a temperature of 434 C?of 434 C?

The answer isThe answer is 2.46 g SO2.46 g SO33

The Ideal Gas EquationThe Ideal Gas Equationcan be used tocan be used to

derive thederive theCombined Gas LawCombined Gas Law

The Combined Gas LawThe Combined Gas Law

Start with the ideal gas equation:Start with the ideal gas equation:

PV = nRTPV = nRT


PP11VV11 = nRT = nRT11

PP22VV22 = nRT = nRT22 andand

Suppose the volume, pressure and Suppose the volume, pressure and temperature change to give a new temperature change to give a new pressure, volume and temperature.pressure, volume and temperature.


Now, solve for what doesn’t Now, solve for what doesn’t change, the constants n and R:change, the constants n and R:

PP11VV11 = nRT = nRT11

PP22VV22 = nRT = nRT22 andand


PP11VV11 = nR= nRTT11


Now, solve for what doesn’t Now, solve for what doesn’t change, the constants n and R:change, the constants n and R:

andand


Since both are equal to nR, Since both are equal to nR, we can make a new equation.we can make a new equation.

andand




Since both are equal to nR, Since both are equal to nR, we can make a new equation.we can make a new equation.

==PP11VV11

TT11

PP22VV22

TT22


This is the This is the Combined Gas LawCombined Gas Law

==PP11VV11

TT11

PP22VV22

TT22


It can be derived from the laws of It can be derived from the laws of Boyle, Amonton and Charles, orBoyle, Amonton and Charles, or

the Ideal Gas Equationthe Ideal Gas Equation

==PP11VV11

TT11

PP22VV22

TT22

The Ideal Gas EquationThe Ideal Gas Equationand and

DensityDensity

Density calculationsDensity calculations

And an equation for “moles”:And an equation for “moles”:

Start with the equation Start with the equation for density:for density:

D = D = mmVV

n =n = mmMM

Where m = massWhere m = massand and MM = molar mass = molar mass


into the ideal gas equation …into the ideal gas equation …

PV = PV = nnRTRT

Now substituteNow substitute n =n = mmMM

PV =PV = mmRTRTMM and getand get


to getto get

Now rearrangeNow rearrange PV =PV = mRTmRTMM

P =P = mRTmRTVVMM


Recall that Recall that D = D = mmVV

P =P = DDRTRT MM

P =P = mRTmRTVVMM


Solving for density,Solving for density,

P =P = DDRTRT MM

becomes: becomes:

DD = = PPMMRTRT


The density of a The density of a gas depends on gas depends on the molar mass the molar mass and the pressure and the pressure and temperature.and temperature.

D = D = PPMMRTRT

Density ProblemDensity Problem

(a) at STP(a) at STP(b) at a pressure of 695 torr and (b) at a pressure of 695 torr and

a temperature of 40.0 C.a temperature of 40.0 C.

1. Determine the density 1. Determine the density of nitrogen, Nof nitrogen, N22, gas, gas

The answers areThe answers are 1.25 g/L,1.25 g/L, and 0.996 g/L.and 0.996 g/L.

2. Determine the molar mass 2. Determine the molar mass of a gas which has a density of a gas which has a density of 8.53 g/L at a pressure of of 8.53 g/L at a pressure of 2.50 atm and a temperature 2.50 atm and a temperature of 500.0 K?of 500.0 K?

Another ProblemAnother Problem

The answer isThe answer is 140. g/mol140. g/mol

Gases and Moles

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