Chapter 14

Expansion of Gases

Younes Sina

For a gas, molecules are free to move and bounce repeatedly against eachother as well as their container's walls. In each collision, a gas moleculetransfers some momentum to its container's walls. Gas pressure is the resultof such momentum transfers. The faster they move, the greater impulse percollision they impart to the container's walls causing a higher pressure. For afixed volume, if the temperature of a gas increases (by heating), its pressureincreases as well. This is simply because of increased kinetic energy of gasmolecules that cause more number of collisions per second and thereforeincreased pressure.

The thermal expansion of a gas involves 3 variables:VolumeTemperaturePressure

Pressure is the result of the collision of its molecules on the walls of that container.

Temperature is a result of the vibrations of its atoms and molecules. It is an indicator of kinetic energy of the gas molecules.

The average K.E. of gas molecules is a function of temperature only.

(K.E.)avg = (3/2) kT

absolute temperature in Kelvin

Boltzmann's constant (k = 1.38x10-23 J /K)

K.E. = (1/2)MV2

average speed of gas molecules

Mass of gas molecules

Example :a) Calculate the average K.E. of air molecules at 27.0 oC.b) Calculate the average speed of its constituents (oxygen molecules and nitrogen molecules).1 mole of O2 = 32.0 grams= 6.02x1023 molecules of O2

1 mole of N2 = 28.0 grams= 6.02x1023 molecules of N2

Solution:K.E. = (3/2) k TK.E. = (3/2) (1.38x10-23 J/K)(27+273)K = 6.21x10-21 J/molecule(Every gas molecule at this temperature, has 6.21x10-21 J whether it is a single O2 molecule or a single N2 molecule).For every O2 molecule: K.E. = (1/2)MV2

6.21x10-21 J = (1/2) [ 32.0x10-3 kg / 6.02x1023]V2 → V = 483m/sFor every N2 molecule: K.E. = (1/2)MV2

6.21x10-21 J = (1/2) [ 28.0x10-3 kg / 6.02x1023 ]V2 → V = 517m/sEach N2 molecule is lighter; therefore, its average speed is higher. Each O2 molecule is heavier; therefore, its average speed is lower at the same temperature.

Expansion of GasesPerfect Gas Law:

PV = nRTgas absolute pressure

V is its volume (the volume of its container)

Universal gas constant = 8.314 [J/(mole K)]

gas absolute temperature in Kelvin

The two conditions for a gas to be ideal or obey this equation are:1) The gas pressure should not exceed about 8 atmospheres.2) The gas must be superheated (gas temperature sufficiently

above its boiling point) at the operating pressure and volume.

number of moles of gas

Equation of State

The Unit of " PV ":Note that the product " PV " has dimensionally the unit of "energy “.In SI, the unit of "P" is [ N / m2 ] and the unit of volume " V " is [ m3 ].On this basis, the unit of the product " PV " becomes [ Nm ] or [ Joule ].The " Joule " that appears in R = 8.314 J /(mole K) is for this reason.

Example :A 0.400 m3 tank contains nitrogen at 27 oC. The pressure gauge on it reads3.75 atmosphere. Find(a) the number of moles of gas in the tank(b) its mass in kg.

Solution:

Pabs. = Pgauge + 1 atm = 4.75 atm

Tabs. = 27oC + 273 = 300K

PV = nRT → n = (PV) / [RT]n = (4.75x101000Pa)(0.400m3) / [(8.314 J / (mole K))300K](a) n = 76.9 moles(b) M = (76.9 moles)(28.0 grams /mole) = 2150 grams = 2.15 kg

Example :

A 0.770 m3 hydrogen tank contains 0.446 kg of hydrogen at 127 oC.The pressure gage on it is not working. What pressure should the gauge show?Each mole of H2 is 2.00 grams.

Solution:n = (0.446x103 grams) / (2.00 grams / mole) = 223 molesPV = nRT → P = (nRT) / VP = (223 moles)[ 8.314 J/(mole K)] (127 + 273)K / (0.770 m3)Pabs = 963,000 PaPgauge = Pabs - 1atm = 963,000 Pa - 101,000Pa = 862,000Pa (≈ 8.6 atm)

For a fixed mass of a gas:

P1V1 = nRT1→ n= P1V1/ RT1

P2V2 = nRT2→ n= P2V2/ RT2

(P1V1) / (P2V2) = T1 / T2

P1V1/ RT1= P2V2/ RT2

(P1V1)/ T1= (P2V2)/ T2

Example :1632 grams of oxygen is at 2.80 atm. of gauge pressure and a temperature of 127 oC. Find a) its volumeIt is then compressed to 6.60 atm. of gauge pressure while cooled downto 27 oC. Find(b) its new volume

Solution:n = (1632 / 32.0) moles = 51.0 moles(a) PV = nRT → V = nRT/pV = (51.0moles)[(8.314 J/(mole K)](127+273)K /(3.80x101,000)Pa = 0.442m3

(b) (P2V2)/(P1V1) =T2 /T1

(7.6atm)(V2) /[(3.8atm)(0.442m3)] = 300K/400KV2 = 0.166m3

Constant Pressure (Isobar) Processes:

A process in which the pressure of an ideal gas does not change is called an" isobar" process.Constant pressure means P2=P1

(P2V2) / (P1V1) = T2 / T1 → V2 / V1 = T2 / T1

Example :A piston-cylinder mechanism as shown below may be used to keep a constant pressure. The pressure on the gas under the piston is 0 gauge plus the extra pressure that the weight generates. Let the piston's radius be 10.0 cm and the weight 475 N, and suppose that the position of the piston at 77 oC is 25.0 cm from the bottom of the cylinder. Find its position when the system is heated and the temperature is 127 oC.

Solution:V1 = πr2h1 = π(10.0cm)2(25.0 cm) = 2500π cm3

V2 = πr2h2 = π(10.0cm)2( h2 ) = (100π)h2 cm3

P2= P1= ConstantT1 = 77 oC + 273 oC = 350 KT2 = 127oC + 273oC = 400 KV2 / V1 = T2 / T1

(100πh2)/(2500π) = 400/350

h2 = 28.6cm

Constant Temperature (Isothermal) Processes:A process in which the temperature of a gas does not change is called an "isothermal process“.

Constant temperature means T2=T1Equation (P2V2) / (P1V1) = T2 / T1 becomes:(P2V2) /(P1V1) = 1

P2V2 = P1V1

Example :A piston cylinder system has an initial volume of 420 cm3 and the air in it is at apressure of 3.00 atmospheres as its gauge shows. The gas is compressed to avolume of 140 cm3 by pushing the piston. The generated heat is removed by enough cooling such that the temperature remains constant.Find the final pressure of the gas.

140 cm3420 cm3

3.00 atm ? atm

T2=T1

Solution:P2V2 = P1V1

P2(140 cm3) = (4.00 atm)(420 cm3)(P2)abs. = 12.0 atm(P2)gauge = 11.0 atm

Constant Volume (Isometric) Processes:A process in which the volume of an ideal gas does not change is called an" isometric process“.

V2=V1Equation (P2V2) / (P1V1) = T2 / T1 becomes:

P2 / P1 = T2 / T1Rigid gas cylinders have constant volumes

Example :

A 15.0 liter gas cylinder contains helium at 7 oC and 11.0 atm of gauge pressure. It is warmed up to 147 oC. Find its new gauge pressure.

Solution:

P2 / P1 = T2 / T1

P2 = P1 (T2 / T1 )P1= 11 atm +1 atm = 12 atmP2= 18.0 atm(P2)gauge = 17.0 atm

Homework:Problems 4, 5, 6