Gas Law'sm1

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    EH 1

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    Air is the mostimportant gas toliving things on the

    Earth. The atmosphere of

    the Earth is amixture of

    nitrogen, oxygen,water vapor, argon,and a few tracegases.

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    Gases assume the volume and shape oftheir containers.

    Gases are the most compressible state of

    matter.

    Gases will mix evenly and completely whenconfined to the same container.

    Gases have much lower densities thanliquids and solids.

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    Boyles Law is one of the laws in physics thatconcerns the behaviour of gases

    When a gas is under pressure it takes up lessspace:

    The higher the pressure, the smaller the volume Boyles Law tells us about the relationship between

    the volume of a gas and its pressure at a constanttemperature

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    The volume of a gas is inversely proportionalto the absolute pressure applied to it whenthe temperature is kept constant. Pis theabsolute pressure.

    P

    V1

    contPV

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    P1V1 = P2V2

    Final pressure (N/m2

    )

    Final volume (m3

    )Original volume (m3

    )

    Original pressure

    (N/m2)

    Robert Boyle (1627-1691). Son of Earl of

    Cork, Ireland.

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    A deep sea diver isworking at a depth wherethe pressure is 3.0atmospheres. He isbreathing out air bubbles.

    The volume of each airbubble is 2.0 cm2. At thesurface the pressure is1.0 atmosphere. What isthe volume of each

    bubble when it reachesthe surface?(assume thetemperature of the wateris constant)

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    Real gases DO experience inter-molecular attractions

    Real gases DO have volume

    Real gases DO NOT have elastic collisions

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    The volume of a gas is directly

    proportional to absolutetemperature when the

    pressure is kept constant

    2

    2

    1

    1

    T

    V

    T

    V

    Temperature MUST be in KELVIN

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    At constant volume theabsolute pressure of agas is directly

    proportional to theabsolute Temperature. TP

    kT

    P

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    You basicallytake BoylesCharles and

    Gay-LussacsLaw andcombine themtogether.

    Moles areconstant

    T

    PV

    T

    VP

    T

    PVk

    T

    VPk

    kTVPTVP

    o

    oo

    o

    oo

    oooooo

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    Pure helium gas isadmitted into a leakproof cylindercontaining a movablepiston. The initialvolume, pressure,

    and temperature ofthe gas are 15 L, 2.0atm, and 300 K. Ifthe volume isdecreased to 12 Land the pressureincreased to 3.5 atm,find the finaltemperature of thegas.

    )2)(15(

    )300)(5.3)(12(

    T

    VP

    PVT

    TT

    PV

    T

    VP

    oo

    o

    o

    oo

    T= 420 K

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    Combining Boyles and Charles laws allowsfor developing a single equation:

    P*V = n*R*T

    P = pressure

    V = volume

    n = number of molesR = universal gas constant

    T = temperature

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    nRTPV

    KmolatmLR

    R

    nTPV

    /0821.0

    KmolJ8.31ConstantGasUniversalR

    alityproportionofconstant

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    PV = nRT

    The conditions 0 0C and 1 atm are calledstandard temperature and pressure (STP).

    Experiments show that at STP, 1mole of an ideal gas occupies

    22.414 L.

    R =

    PV

    nT =

    (1 atm)(22.414L)

    (1 mol)(273.15 K)

    R= 0.082057 L atm / (mol K)

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    R can also be expressed asR = 8.314 J/mol K

    The value of the constantdepends on the units used todefine the other variables.

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    P*V = n*R*T Learn it!

    Use it!

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    PV = nRT

    V =nRT

    P

    T= 0 0C = 273.15 K

    P = 1atm

    n= 49.8 g x1 mol HCl

    36.45 g HCl= 1.37 mol

    V= 1 atm

    1.37 x 0.0821 x

    273.15 K

    V= 30.7 L

    [R= 0.082057 L atm /(mol K)]

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    PV = nRT

    V =nRT

    P

    V= 1 atm x1.01 x105

    1.37 x 8.314 x 273.15 K

    V= 0 .0307 m3

    T= 0 0C =273.15 K

    P = 1 atm =1.01 x105 Pa

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    How many particles are in 49.8 g of HCl atSTP?

    How many particles are in 1 mole of HCl?

    n= 49.8 g x1 mol HCl

    36.45 g HCl= 1.37 mol

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    When measured at STP, a quantity of gas hasa volume of 500 dm3. What volume will itoccupy at

    0 oC and 93.3 kPa?

    1 1 2 2

    1 2

    PV PV T T

    P1 = 101.3 kPa

    T1 = 273 KV1 = 500 dm

    3

    P2 = 93.3 kPaT2 = 0

    oC + 273 = 273 K

    V2 = X dm3

    (101.3 kPa) x (500 dm

    3

    ) = (93.3kPa) x (V2)273 K 273 K

    V2 = 542.9 dm3

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    When a scuba diver is severalhundred feet under water, thehigh pressures cause N2 from thetank air to dissolve in the blood.If the diver rises too fast, thedissolved N2 will form bubbles inthe blood, a dangerous andpainful condition called "thebends". Helium, which is inert,less dense, and does not dissolvein the blood, is mixed with O2 inscuba tanks used for deepdescents.