Review of the Gas Laws

PV = nRT

PV = nRT

• Boyle’s Law (isothermal & fixed amount)

• Charles’s Law (isobaric & fixed amount)

• Avogadro’s Law (isothermal & isobaric)

• ????? Law (isochoric & fixed amount)

• ????? Law (isothermal & isochoric)

• ????? Law (isobaric & isochoric)

Boyle’s Law

atmLVP

atmLVP

KmolK

atmLmolePV

nRTPV

4.22

4.22

)273(0821.0)1(

• Pressure and volume are inversely proportional.• As pressure increases, volume decreases.• If pressure increases by 2x, volume cuts in half.

Charles’s Law

atmmolKatmL

mole

T

V

P

nR

T

V

1

0821.0)1(

• Temperature and volume are directly proportional.

• As temperature increases, volume also increases.

• If temperature increases by 2x, volume also doubles.

• Temperature must be measured in Kelvin.

Avogadro’s Law

mol

L

n

V

atm

KmolKatmL

n

V

P

RT

n

V

4.22

)1(

)273(0821.0

• Moles of gas and volume are directly proportional.• As the number of moles increases, the volume also

increases.• If the number of moles increases by 2x, the volume also

doubles.

????? Lawisothermal & isochoric

mol

atm

n

P

L

KmolKatmL

n

P

V

RT

n

P

1

4.22

)273(0821.0

• Moles of gas and pressure are directly proportional.

• As the moles of gas increase, the pressure also increases.

• If the number of moles of gas increases by 2x, the pressure also doubles.

????? Lawisobaric & isochoric

• Moles of gas and temperature are inversely proportional.

• As the number of moles of gas increase, the temperature decreases.

molKTn

molKatmLLatm

nT

R

PVnT

273

0821.0

)4.22)(1(

Units of Pressure

• 1 atm = 760 torr = 760 mmHg

• 1 atm = 101.325 kPa

• 1 bar = 105 Pa = 100 kPa

• 1 Pa = mN

2

1

How does 1 atm = 101.325 kPa?Let the area of the base of a cylinder = 1 m2

Volume = area x height = 1 m2 x 0.76 m = 0.76 m3

Convert volume to cubic centimeters.

363

33 1076.0

1

10076.0 cmx

m

cmxm

Use the density of mercury and the acceleration due to gravity to calculate the weight of mercury in the column.

Nxs

mx

g

kgx

cm

gxcmx 3

2336 103.101

8.9

1000

16.131076.0

ContinueHow does 1 atm = 101.325 kPa?

Pressure is force (or weight) per unit area. Divide the weight of mercury by the area it is resting on.

kPam

Nx

m

Nx

area

Nx101103.101

1

103.101103.1012

32

33

Let the area of the cylinder = 1cm2

kPa

m

Nx

m

cmx

cm

N

A

N

Nors

mkg

s

mx

g

kgx

cm

gxcm

cmAxhV

cmA

cmh

101103.1011

100

1

13.1013.10

13.1013.108.9

1000

16.1376

76

1

76

23

2

2

2

2233

3

2

Barometric Formula

As elevation increases, the height of the atmosphere decreases and its pressure decreases.

hhgP

Check units.

22

2

223 m

N

msm

kg

sm

kgmx

s

mx

m

kg

Continue Derivation of Barometric Formula

Write in differential form. gdhdP

density V

Mmoles

volume

mass W

Rewrite PV = nRT asRT

P

V

n

Therefore,RT

PMW

Continue Derivation of Barometric Formula

Substitute the expression for density into the differential eqn.

dhRT

gPMdP W

Divide both sides of the above equation by P and integrate.

dhRT

gM

P

dP W

Continue Derivation of Barometric Formula

Integration of the left side and moving the constants outside the integral on the right side of the differential equation gives,

hRT

gMdh

RT

gMP WWln

Continue Derivation of Barometric Formula

Evaluating the integral between the limits of Pi at zero height and Pf at height h, gives

RT

ghM

P

PW

i

f

ln

Sample Problem Using the Barometric Formula

torrP

P

P

P

P

KmolKJ

msm

molkg

x

P

P

RT

ghM

P

P

f

i

f

i

f

i

f

W

i

f

464

6099.0

4944.0ln

)293(314.8

42678.9108.28ln

ln

23

Dalton’s Law of Partial Pressures

Pressure is additive.baT PPP

Write each pressure as, V

nRTP

V

RTn

V

RTn

V

RTn baT

ContinueDalton’s Law of Partial Pressures

V

RTn

V

RTn

V

RTn baT

Multiply through by V (the combined volume of the gases) and divide by R T.

baT nnn

Moles are indeed additive.

Mole Fraction & Partial Pressure

T

bb

T

aa n

nXand

n

nX

T

T

T

ba

T

b

T

aba n

nor

n

nn

n

n

n

nXX

Therefore, 1 ba XX

ContinueMole Fraction & Partial Pressure

Show that TbbTaa PXPPXP &

T

bb

T

aa P

PX

P

PX &

1

T

T

T

ba

T

b

T

aba P

P

P

PP

P

P

P

PXX