The Ideal Gas: Boyle’s LawThe Ideal Gas: Boyle’s LawThe Ideal Gas: Boyle’s LawThe Ideal Gas: Boyle’s Law

Presented by: Kimmy AtoloyePresented by: Kimmy Atoloye

Dec 20th, 2002Dec 20th, 2002

SC441L-8771SC441L-8771

Dr. Roman KezerashviliDr. Roman Kezerashvili

ObjectivesObjectives

To study and verify Boyle’s Law.To study and verify Boyle’s Law. Illustrate that pressure, P is directly Illustrate that pressure, P is directly

proportional to the inverse height of air, proportional to the inverse height of air, 1/h.1/h.

EquipmentsEquipments

Boyle’s law apparatusBoyle’s law apparatus BarometerBarometer

TheoryTheory

There are three major concepts in measuring There are three major concepts in measuring the behavior of various gases.the behavior of various gases.

The volume V is proportional to the number of The volume V is proportional to the number of moles, n when the temperature and pressure moles, n when the temperature and pressure are constant. If we are to double the number are constant. If we are to double the number of moles, keeping the temperature and of moles, keeping the temperature and pressure constant, the volumes will double. pressure constant, the volumes will double.

The volume changes inversely with the The volume changes inversely with the pressure when the temperature and quantity pressure when the temperature and quantity of air are constant.of air are constant.

Theory (cont’d)Theory (cont’d)

If once again, we double the pressure while If once again, we double the pressure while the temperature and number of moles are the temperature and number of moles are constant, the air is compressed to one half constant, the air is compressed to one half of its initial volume. of its initial volume.

The pressure is proportional to absolute The pressure is proportional to absolute temperature for the given amount of gas, temperature for the given amount of gas, when volume is constant. If the absolute when volume is constant. If the absolute temperature is doubled, keeping the temperature is doubled, keeping the volume and number of moles constant the volume and number of moles constant the pressure doubles.pressure doubles.

Theory (cont’d)Theory (cont’d)

These three details are all connected These three details are all connected into a single equation which is called into a single equation which is called the ideal gas equation:the ideal gas equation:

pV = nRTpV = nRT (1)(1)

where R is the proportionality constant where R is the proportionality constant or universal gas constant which in the SI or universal gas constant which in the SI system of units, R = 8.31 J/mole.Ksystem of units, R = 8.31 J/mole.K

Theory (cont’d)Theory (cont’d)

If a constant number of moles or a If a constant number of moles or a constant mass of gas of an ideal gas the constant mass of gas of an ideal gas the product nR is constant, so the quantity product nR is constant, so the quantity pV/T is also constant. The subscripts 1 pV/T is also constant. The subscripts 1 and 2 refer to two states of the same and 2 refer to two states of the same mass of gas, but different pressures, mass of gas, but different pressures, volumes and absolute temperatures.volumes and absolute temperatures.

pp11VV11 = = pp22VV22 (2)(2) T T1 1 T T22

Theory (cont’d)Theory (cont’d)

This equation is known as the ideal gas This equation is known as the ideal gas law which states that the product of law which states that the product of pressure and volume of the given mass pressure and volume of the given mass of gas divided by the absolute of gas divided by the absolute temperature of the gas is a constanttemperature of the gas is a constant

If the temperature TIf the temperature T11 and T and T2 2 are the are the same, thensame, then

pp11VV1 1 = = pp22VV22 = const, T = const, m = const (3)= const, T = const, m = const (3) oror

pV = const, T = const, m = const (4)pV = const, T = const, m = const (4)

Theory (cont’d)Theory (cont’d)

Therefore Therefore the product of the pressure and the product of the pressure and volume of a given mass of a gas at constant volume of a given mass of a gas at constant temperature is equal to a constanttemperature is equal to a constant, which has , which has became to be known as the Boyle’s law, in became to be known as the Boyle’s law, in honor of the British physicist and chemist honor of the British physicist and chemist Robert Boyle. It is also true that the pressure Robert Boyle. It is also true that the pressure of the given mass of a gas is inversely of the given mass of a gas is inversely proportional to the volume at a constant proportional to the volume at a constant temperature.temperature.

pp is is inversely proportional toinversely proportional to 11 (5) (5)

VV

Theory (cont’d)Theory (cont’d)

Meaning, if we were to measure the Meaning, if we were to measure the volume of gas for each pressure and volume of gas for each pressure and graph the pressure of the gas as a graph the pressure of the gas as a reciprocal of its volume, we will derive a reciprocal of its volume, we will derive a straight line.straight line.

pp11hh11 = = pp22hh22, , T = const, m = constT = const, m = const

(6)(6)

ph = constant, T = const, m = constph = constant, T = const, m = const (7) (7)

ProcedureProcedure

Fill the ¾ of the larger cylinder with water. Fill the ¾ of the larger cylinder with water. Using the barometer determine the current Using the barometer determine the current atmospheric pressure.atmospheric pressure.

Record the length of the small cylinder, LRecord the length of the small cylinder, L0 0 in in the data table.the data table.

Invert the smaller cylinder and immerse it Invert the smaller cylinder and immerse it into the water column to a 100mm depth into the water column to a 100mm depth and measure the immersion depth (D) and and measure the immersion depth (D) and height of the water (H) using the scale height of the water (H) using the scale provided on the small column.provided on the small column.

Procedure (cont’d)Procedure (cont’d)

Record the data in the data table.Record the data in the data table. Then repeat step 4 for every 100 mm Then repeat step 4 for every 100 mm

until the small cylinder is at the bottom until the small cylinder is at the bottom of the large cylinder.of the large cylinder.

Data AnalysisData Analysis

Atmospheric pressure, PAtmospheric pressure, P0 0 = = 765 mm Hg765 mm Hg

Height of large cylinder, LHeight of large cylinder, L00 = = 920 mm920 mm

Impression Depth, D (mm) Height of water, H (mm) Height of air, h (mm)

100 7 913200 15 905300 22 898400 30 890500 39 881600 46 874700 54 866750 60 860

Pressure, P (mm Hg) 1/h (mm^ -1) Ph

772 0.0011 704711779 0.0011 704681786 0.0011 705394792 0.0011 705153799 0.0011 703939806 0.0011 704344813 0.0012 703777816 0.0012 701693

Data Analysis (cont’d)Data Analysis (cont’d)

Pressure vs Height of air

y = 474953x-0.942

760

770

780

790

800

810

820

850 860 870 880 890 900 910 920

height of air, h (mm)

pre

ss

ure

, P

(m

m H

g)

Series1

Power(Series1)Power(Series1)

Data Analysis (cont’d)Data Analysis (cont’d)

Pressure vs Inverse height

y = 2E-06x - 7E-05

0.0011

0.0011

0.0011

0.0011

0.0011

0.0011

0.0012

0.0012

0.0012

760 770 780 790 800 810 820

inverse height, 1/h (mm)

pre

ss

ure

, P (

mm

Hg

)

Series1

Linear(Series1)

CalculationsCalculations

Calculate the volume of the compressed Calculate the volume of the compressed gas at the depth x, h = Lgas at the depth x, h = L0 0 - H- H

hh11 = 920 mm Hg - 7 mm = 913 mm = 920 mm Hg - 7 mm = 913 mm

hh22 = 920 mm Hg - 15 mm = 905 mm = 920 mm Hg - 15 mm = 905 mm

hh33 = 920 mm Hg - 22 mm = 898 mm = 920 mm Hg - 22 mm = 898 mm

hh44 = 920 mm Hg - 30 mm = 890 mm = 920 mm Hg - 30 mm = 890 mm

hh55 = 920 mm Hg - 39 mm = 881 mm = 920 mm Hg - 39 mm = 881 mm

hh66 = 920 mm Hg - 46 mm = 874 mm = 920 mm Hg - 46 mm = 874 mm

hh77 = 920 mm Hg - 54 mm = 866 mm = 920 mm Hg - 54 mm = 866 mm

hh8 8 = 920 mm Hg - 60 mm = 860 mm= 920 mm Hg - 60 mm = 860 mm

Calculations (cont’d)Calculations (cont’d)

Calculate the absolute pressure using the Calculate the absolute pressure using the equation P = Pequation P = P00 + + D-H D-H

13.5513.55

where 13.55 is the density of mercury at room where 13.55 is the density of mercury at room temperature.temperature.

P = 765 mm Hg + (100 mm - 7 mm) = 772 mm HgP = 765 mm Hg + (100 mm - 7 mm) = 772 mm Hg

P = 765 mm Hg + (200 mm - 15 mm) = 779 mm HgP = 765 mm Hg + (200 mm - 15 mm) = 779 mm Hg

P = 765 mm Hg + (300 mm - 22 mm) = 786 mm HgP = 765 mm Hg + (300 mm - 22 mm) = 786 mm Hg

P = 765 mm Hg + (400 mm - 30 mm) = 792 mm HgP = 765 mm Hg + (400 mm - 30 mm) = 792 mm Hg

P = 765 mm Hg + (500 mm - 39 mm) = 799 mm HgP = 765 mm Hg + (500 mm - 39 mm) = 799 mm Hg

P = 765 mm Hg + (600 mm - 46 mm) = 806 mm HgP = 765 mm Hg + (600 mm - 46 mm) = 806 mm Hg

Calculations (cont’d)Calculations (cont’d)

P = 765 mm Hg + (700 mm - 54 mm) = 813 P = 765 mm Hg + (700 mm - 54 mm) = 813 mm Hgmm Hg

P = 765 mm Hg + (750 mm - 60 mm) = 816 P = 765 mm Hg + (750 mm - 60 mm) = 816 mm Hgmm Hg

QuestionsQuestions

Boyle’s Law assumes that the gas is at Boyle’s Law assumes that the gas is at constant temperature. What change in constant temperature. What change in pressure might be expected if the gas pressure might be expected if the gas temperature increased? temperature increased?

If the gas temperature increases, the pressure will If the gas temperature increases, the pressure will also increase.also increase.

If 1 liter of gas at a pressure of 20 mm Hg is If 1 liter of gas at a pressure of 20 mm Hg is compressed to a volume of 10cm^3, what compressed to a volume of 10cm^3, what will the resulting pressure be? What would will the resulting pressure be? What would the volume be if the desired pressure is 760 the volume be if the desired pressure is 760 mm Hg ( 1 atmosphere )?mm Hg ( 1 atmosphere )?

Questions(cont’d)Questions(cont’d)

PV = constantPV = constant

10 mm = 1 cm10 mm = 1 cm

(20 mm Hg)(100 mm^3) = 2000 mm Hg(20 mm Hg)(100 mm^3) = 2000 mm Hg

when V = 760 mm Hgwhen V = 760 mm Hg

(760 mm Hg)(100 mm^3) = 76000 mm Hg(760 mm Hg)(100 mm^3) = 76000 mm Hg

ConclusionConclusion

In closing, our main goal for this laboratory In closing, our main goal for this laboratory exercise was achieved. The graph of pressure, exercise was achieved. The graph of pressure, P versus inverse height of air, 1/h illustrates P versus inverse height of air, 1/h illustrates that the pressure of a fixed mass of air is that the pressure of a fixed mass of air is directly proportional to the inverse height of directly proportional to the inverse height of air. Boyle’s law stated that the product of the air. Boyle’s law stated that the product of the pressure and volume of a given mass of air at pressure and volume of a given mass of air at constant temperature is equal to a constant. constant temperature is equal to a constant. Our values for Ph was almost constant but we Our values for Ph was almost constant but we were are slightly off due to experimental were are slightly off due to experimental errors.errors.

The EndThe End

Thanks for listening