Ideal Gas Ideal Gas

Review Review Q = Energy transfer in form of heat Q = Energy transfer in form of heat

SI unit : J ( Joule)SI unit : J ( Joule)

What is a Joule ?What is a Joule ?

Joule <- base SI unit named after a dead guy Joule <- base SI unit named after a dead guy

English physicist James Prescott Joule (1818–1889English physicist James Prescott Joule (1818–1889

J= (Kg * mJ= (Kg * m22)/ s)/ s2 2 = N *m = Pa *m= N *m = Pa *m3 3

Ideal Gas Ideal Gas Objective:Objective:

Relationship among pressure, heat, volume and the Relationship among pressure, heat, volume and the amount of gasamount of gas

Red Bull StratosRed Bull Stratos

Felix Baumgartner.Felix Baumgartner.

On 14 October 2012On 14 October 2012

Ideal Gas LawIdeal Gas LawPV = nRTPV = nRT

Macroscopic Microscopic

P- PressureV- Volume T- Temperature

Atoms and MoleculesNumber of moles

MassVelocity (KE) – Kinetic Energy

Kinetic theory of Kinetic theory of PressurePressure

Change in momentumChange in momentum

Δ Δ P = m (VP = m (Vff –V –Vii) )

I = I = PΔ

Ft = PΔ

F = P/ T Δ Δ

Sample gas = Large Sample gas = Large number of particles number of particles

Force is constant in a Force is constant in a time interval time interval

- Particles move - Particles move randomly in all direction randomly in all direction

-pressure same on all the -pressure same on all the wallswalls

Pressure ( P) Pressure ( P) force( F) force( F) per unit surface area (A)per unit surface area (A)

P=F/A P=F/A

Si unit Si unit

1 Pa = 1 N/m1 Pa = 1 N/m22 = 1 Kg/ms = 1 Kg/ms22

Pressure ( P )Pressure ( P )Standard pressure – Standard pressure – average pressure of the average pressure of the atmosphere at sea level atmosphere at sea level 101.3 K Pa 101.3 K Pa

Boyle’s lawBoyle’s law

Boyle’s law – fixed sample Boyle’s law – fixed sample of gas, at constant of gas, at constant temperature---temperature---volume( V )varies inversely volume( V )varies inversely with the pressure( P )with the pressure( P )

When 2 variables are When 2 variables are inversely related – product is inversely related – product is constant ( PV=Constant)constant ( PV=Constant)

A sample gas is held in a 2.6 mA sample gas is held in a 2.6 m33 volume at 226 kPa. The volume at 226 kPa. The temperature is kept constant while the volume is temperature is kept constant while the volume is decreased until pressure is 565 kPa. What is the new decreased until pressure is 565 kPa. What is the new volume of the gas.volume of the gas.

Given :Given :

PPii = 226 kPa = 226 kPa

VVii = 2.6 m = 2.6 m33

PPff = 565 kPa = 565 kPa

Unknown:Unknown:

VVff ? ?

Relationship –Relationship –

PPi i VVii = P = Pf f VVff

A sample gas is held in a 2.6 mA sample gas is held in a 2.6 m33 volume at 226 kPa. The volume at 226 kPa. The temperature is kept constant temperature is kept constant while the volume is while the volume is decreased until pressure is 565 kPa. What is the new decreased until pressure is 565 kPa. What is the new volume of the gasvolume of the gas

Solution :Solution :

PPiiVVii = P = Pff V Vff

VVff = P = PiiVVii/ P/ Pff

(226 kPa ) (2.6 m(226 kPa ) (2.6 m33)/565 kPa)/565 kPa

= 1.0 m= 1.0 m33

Pressured (P) went up – Pressured (P) went up – Volume (V) went down Volume (V) went down

Charles’s LawCharles’s Lawalso known as the law of volumesalso known as the law of volumes

What did Charles What did Charles observed?observed?

Increased Temperature Increased Temperature

0 C° 0 C° 1 C° 1 C°

Volume increased by :Volume increased by :

1/ 273 of original Volume1/ 273 of original Volume

Increase T by 2 C° Increase T by 2 C°

V increased by 2/273V increased by 2/273

If Increased T by 273C° If Increased T by 273C°

V would of doubledV would of doubled

Charles’s LawCharles’s Lawalso known as the law of volumesalso known as the law of volumes

What about when cooling What about when cooling the gas?the gas?

Volume sharked by 1/273 Volume sharked by 1/273 for decreasing each C° for decreasing each C°

Max cooling point -20 C° Max cooling point -20 C°

Extended the graph to its Extended the graph to its lower limit. Extrapolated lower limit. Extrapolated that the lowest -273 C° that the lowest -273 C°

At -273C° volume is zero.At -273C° volume is zero.

Now called Absolute zero Now called Absolute zero ( Kelvin) ( Kelvin)

Charles’s LawCharles’s LawGases tend to expand Gases tend to expand when heated when heated

Volumes of gas depends Volumes of gas depends on temperatureon temperature

Volume of fixed amount of Volume of fixed amount of gas at a constant gas at a constant pressure depends linearly pressure depends linearly on the temperature. on the temperature.

Constant Pressure Constant Pressure

VVii/T/Tii = V = Vf f /T/Tf f = Constant= Constant

VVii/T/Tii = V = Vf f /T/Tf f

A container of .22 mA container of .22 m33 of nitrogen gas at 20.0 ° C of nitrogen gas at 20.0 ° C is heated under is heated under constant pressure constant pressure to 167.0 ° Cto 167.0 ° C What is its new volume? What is its new volume?

Given:Given:

VVii = .22 m = .22 m33

TTii = 20.0 ° C = 20.0 ° C

TTff = 167 ° C = 167 ° C

Unknown: VUnknown: V22

VVii/T/Tii = V = Vf f /T/Tf f

A container of .22 mA container of .22 m33 of nitrogen gas at 20.0 ° C of nitrogen gas at 20.0 ° C is heated under is heated under constant pressure constant pressure to 167.0 ° Cto 167.0 ° C

What is its new volume? What is its new volume?

First switch ° C to K. Use First switch ° C to K. Use Kelvin for gas law Kelvin for gas law

TTii = 20.0 ° C + 273 = 293 K = 20.0 ° C + 273 = 293 K

TTff = 167 ° C + 273 = 440 K = 167 ° C + 273 = 440 K

VVii/T/Tii = V = Vf f /T/Tf f

VVff = ( V = ( Vii T Tf f )/ T)/ Tii

Plug and chug Plug and chug

((.22 m((.22 m33)(440K))/ (293K))(440K))/ (293K)

= .33 m= .33 m33

Boyle’s law Vs. Charles's Law Boyle’s law Vs. Charles's Law

Gay-Lussac's lawGay-Lussac's lawThe combined gas lawThe combined gas law

Boyle’s law and Charles's law combined Boyle’s law and Charles's law combined

For a fixed amount of gas:For a fixed amount of gas:

(P(PiiVVii )/ T )/ Tii = Constant = (P = Constant = (PffVVff)/ T)/ Tff

(P(PiiVVii )/ T )/ Tii = (P = (PffVVff)/ T)/ Tff

A 20.0-L sample of argon gas at 273 K is at atmospheric A 20.0-L sample of argon gas at 273 K is at atmospheric pressure, 101.3 kPa. The temperature is lowered to 77 K, pressure, 101.3 kPa. The temperature is lowered to 77 K, and the pressure is increased to 145 kPa. What is the new and the pressure is increased to 145 kPa. What is the new

volume of the argon samplevolume of the argon sample??

Given: VGiven: Vii = 20.0-L = 20.0-L

PP11= 101.3 kPa= 101.3 kPa

TTii = 273 K = 273 K

PPff = 1.45 kPa = 1.45 kPa

TTff = 77 K = 77 K

Unknown : VUnknown : Vff

(P(PiiVVii )/ T )/ Tii = (P = (PffVVff)/ T)/ Tff

1 Litre (L) = .001 m1 Litre (L) = .001 m33

A 20.0-L sample of argon gas at 273 K is at atmospheric A 20.0-L sample of argon gas at 273 K is at atmospheric pressure, 101.3 kPa. The temperature is lowered to 77 K, pressure, 101.3 kPa. The temperature is lowered to 77 K, and the pressure is increased to 145 kPa. What is the new and the pressure is increased to 145 kPa. What is the new

volume of the argon sample?volume of the argon sample?

Solution Solution

(P(PiiVVii )/ T )/ Tii = (P = (PffVVff)/ T)/ Tff

VVff = (P = (PiiVVii T Tff)/ P)/ PffTTii

Plug and chugPlug and chug

((101.3kPa)( 20.0-L)(77 ((101.3kPa)( 20.0-L)(77 K))/( (145 kPa)(273 K))K))/( (145 kPa)(273 K))

=3.9 L=3.9 L

= 3.9 x10= 3.9 x10-3-3 m m3 3