Chapter 10 Notes
The kinetic molecular theory states that
1)Gases consist of large numbers of tiny particles that are far apart relative to their size
2)Collisions between gas particles and between particles and container walls are elastic collisions
3)Gas particles are in continuous, rapid, random motion
4)There are no forces of attraction or repulsion between gas particles.
5)The average kinetic energy of gas particles depends on the temperature of the gas
**The kinetic molecular theory applies only to ideal gases
Diffusion-a spontaneous mixing of particles of two
substances caused by their random motion.
Effusion-process by which gas particles pass through a tiny
opening
To describe a gas, you need to know:
1)volume
2)temperature
3)number of molecules
4)pressure
Pressure-force per unit area (The SI unit of force is the Newton-N)
Standard Pressure Standard Temperature
1 atm 0ºC
101.3 kPa
29.92 inches Hg
760mm Hg
760 torr
**Standard temperature and pressure is abbreviated STP.
Temperature Conversion
K=273.15+ºC
Conversion ProblemsI. Convert the following into K
Q:1)0ºC 2)27ºC 3)-50ºC 4)-273ºC
A:1)273K 2)300K 3)223K 4)0K
II. Convert the following into ºC
Q:1)273K 2)350.K 3)100.K 4)20.K
A:1)0ºC 2)77ºC 3)-173ºC 4)-253ºC
III. Convert each of the following into the unit specified
- 125mmHg into atm
125mmHg x 1 atm = 0.164 atm
760 mmHg
- 5.38kPa into torr
5.38kPa x 760 torr = 40.4 torr
101.3kPa
IV. A chemical reaction produces 0.0680mol of oxygen gas. What volume in liters is occupied by this gas at STP.
0.0680mol O2 x 22.4L O2= 1.52L O2
1 mol O2
V. Calculate the density of oxygen at STP.
D=M D=32g = 1.4g/L O2
V 22.4L
VI. Combined Gas Law
P1V1 = P2V2 T has to be in Kelvin, Omit what is constant
T1 T2
Examples
1) A sample of oxygen gas has a volume of 150.mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm is the temperature remains constant?
G: V1=150. mL, P1=0.947atm, P2=0.987atm
S: V2 in mL
CF: P1V1=P2V2
SU: (0.947atm)(150.mL)=(0.987atm)V2
142=0.987V2
V2=144mL
2) A sample of neon gas occupies a volume of 752mL at 25ºC. What volume will the gas occupy at 50ºC if the pressure remains constant?
G: V1=752mL T1=25ºC+273=298K T2=50ºC+273=323K
S: V2=? mL
CF: T1V2=T2V1
SU: (298)V2=323(752)
298V2=242896
V1=815mL
3) A helium-filled balloon has a volume of 50.0L at 25ºC and 1.08 atm. What volume will it have at 0.855atm and 10.ºC?
G: V1=50.0L T1=25ºC+273=298K P1=1.08atm
P2=0.855atm T2=10ºC+273=283K
S: V2=? L
CF: P1V1=P2V2
T1 T2
SU: (1.08)(50.0)=(0.855)V2
298 283
.181=0.855V2
283
51.3=0.855V2 V2=60.0L
VII. Dalton’s Law of Partial Pressures
PT=P1 + P2 + P3…..
Three of the primary components of air are carbon dioxide, nitrogen, and oxygen. In a sample containing a mixture of only these gases at exactly one atmosphere of pressure, the partial pressures of carbon dioxide and nitrogen are given at PCO2=0.285torr and PN2=593.525torr. What is the partial pressure of oxygen?
PT=PCO2+PN2+PO2 The total pressure is 1atm, but the other units are in torr so we have
to convert atm to torr, which is 760torr.
760torr=0.285torr + 593.525torr + P02
760=593.81+ PO2
P O2=166torr
Ideal Gas Law
PV=nRT
P=atm V=liters n=moles T=Kelvin R=0.082L·atm
mol· K
Problem
1. What is the pressure in atmospheres exerted by a 0.500mol sample of nitrogen gas in a 10.0L container at 298K
G: n=0.500mol V=10.0L T=298K R=0.082L·atm
mol·K
S: P=? in atm
CF: PV=nRT
SU: P(10.0)=(0.500)(0.082)(298)
10.0P=12.2
P=1.22atm
2. What is the volume, in liters, of 0.250mol of oxygen gas at 20.0ºC and 0.974atm pressure?
G: n=0.250mol P=0.974atm T=293K R=0.082L·atm
mol·K
S: V=? in L
CF: PV=nRT
SU: (0.974)V=(0.250)(0.082)(293)
0.974V=6.01
V=6.17L
3. What mass of chlorine gas in grams is contained in a 10.0L tank at 27ºC and 3.50atm of pressure?
G: V=10.0L P=3.50atm T=300K R=0.082L·atm
mol·K
S: mass in grams (need n first)
CF: PV=nRT
SU: (3.50)(10.0)=n(0.082)(300)
35=24.6n
n=1.42mol
To find the mass: 1.42mol Cl2 x 70g Cl2= 99.4g Cl2
1 mol Cl2