Chapter 5 Gases
I.
Pressure
A.
Properties of Gases
•
Expand to completely fill their container
•
Take the Shape of their container
•
Low Density, much less than solid or liquid state
•
Compressible
•
Mixtures of gases are always homogeneous
•
Fluid
B. Pressure = total force applied to a certain area
1. larger force = larger pressure
2. smaller area = larger pressure
3. Gas pressure caused by gas molecules colliding with container or surface
4. More forceful or frequent collisions mean higher gas pressure
F
P
A
C. Measuring Pressure
1. Atmospheric Pressure is constantly present
a. Decreases with altitude because of less air
b. Varies with weather conditions
2. Measuring pressure using a barometer
a. Column of mercury supported by air pressure
b. Longer mercury column = higher pressure
c. Force of the air on the surface of the mercury is balanced by the pull of
gravity on the column of mercury
3. Units of Gas Pressure
a. atmosphere (atm) = 29.92 in Hg = 760 mm Hg
b. Torr = mm Hg 1 atm = 760 torr
c. Pascal (Pa) = 1 N/m2 1 atm = 101,325 Pa
4. Example: Convert 49 torr to atm and Pa
49torr 
1atm
 760torr

  0.064atm

49torr  101,325Pa   6500Pa
 760torr 
5. Measuring the Pressure of a Trapped Gas Sample
a. Use an manometer to compare to atmospheric pressure
b. Open-end manometer
i. if gas end lower than open end, Pgas = Pair + h
ii. if gas end higher than open end, Pgas = Pair – h
II.
Simple Gas Laws
A.
Boyle’s Law = Pressure is inversely proportional to Volume (constant T, n)
1. PV = k or P1V1 = P2V2
(k = Boyle’s Law Constant)
2. As Pressure on a gas increases, the Volume decreases
V = k(1/P)
3.
Example: What is new V of 1.53L of SO2 at 5600Pa when changed to
15,000Pa?
P1V1 5,600Pa1.53L
P1V1  P2V2  V2 

 0.57 L
15,000Pa
P2
B. Charles’ Law = V is directly proportional to T (constant P, n)
V1
V2
1. V = bT
b
T2
2. Another way to write Charles Law T1
3. As the temperature decreases, the volume of a gas decreases as well
4. Absolute Zero
a. Theoretical temperature at which a gas would have zero V and P
b. 0 K = -273.2 °C = -459 °F (K = oC + 273)
c. All gas law problems use Kelvin temperature scale!
5. Example: V = ? If 2.58L of gas at 15 oC is heated to 38 oC at same P?
V1 V2
T2V1 38  273K 2.58L 311K 2.58L

 V2 


 2.79L
15  273K 
288K 
T1 T2
T1
C. Avogadro’s Law = V directly proportional to moles of gas (Constant T, P)
1. V = an
V1
V2
a
2. Another way to write Avagodro’s Law:
n1
n2
3. More gas molecules = larger volume
4. Count number of gas molecules by moles = n
5. One mole of any gas occupies 22.414 L (at 1 atm, 0 oC) = molar volume
6. Equal volumes of gases contain equal numbers of molecules
7. Example: V of O3 = ? If we convert 11.2L (0.50 mol) O2 to O3?
a. 3O2(g) -----> 2O3(g)
b.
c.
 2 mol O3
0.50 mol O 2 
 3 mol O 2

  0.33 mol O3


V1 V2
n V 0.33mol 11.2 L 
  V2  2 1 
 7.4 L
0.50mol 
n1 n2
n1
III. Ideal Gas Law
A. By combining the constants from the 3 gas laws we can write a general
equation
1. Each simple gas law holds something constant
2. To consider changes in P, V, T, n at the same time, we combine constants
k
 Tn 
 Tn 





V     bT P,n  anT, P  kba 
 R

 P  T, n
 P 
 P 
PV  nRT
3. PV = nRT is called the Ideal Gas Law
4. R is called the gas constant
a. The value of R depends on the units of P and V
b. Generally use R = 0.08206Latm/Kmol when P in atm and V in L
B. Only “Ideal gases” obey this law exactly
1. Most gases obey when P is low (< 1 atm) and T is high (> 0°C)
2. An ideal gas is only a hypothetical substance
3. Constant conditions drop out of the equation to give simpler laws
a. If you hold n constant
b.
P1V1 P2 V2
PV
PV  nRT 
 nR 

T
T1
T2
C. Ideal Gas Law Problems
1. Example: How many moles of H2 gas if: 8.56L, 0 oC, 1.5atm?
PV  nRT  n 
1.5atm8.56L  0.57mol
PV

RT  0.08206Latm 

273K 
Kmol


2. Example: P = ? if 7.0 ml gas at 1.68 atm is compressed to 2.7 ml?
PV  nRT  constant  Reduces to Boyle's Law
P1V1 1.68atm7.0ml
P1V1  P2 V2  P2 

 4.4atm
2.7ml
V2
3. Example: V = ? If 345 torr, -15 oC, 3.48L changes to 36 oC, 468torr?
P1V1 P2 V2
PV
PV  nRT 
 nR  constant  Reduces to

T
T1
T2
P1V1 P2 V2
T2 P1V1 309K 345torr 3.48L

 V2 

 3.07 L
468torr 258K 
T1
T2
P2 T1
4. Example: DV = ? If 0.35mol, 13oC, 568torr changes to 56oC, 897torr
n 2 R 2 T 2 n1R1T1
ΔV  V2 - V1 

P2
P1
nRT
PV  nRT  V 
P
ΔV 
0.35mol  0.08206Latm  329K  0.35mol  0.08206Latm  286K 
Kmol
897torr  1atm
 760torr




Kmol
568torr  1atm
 760torr



 8L  11L  3L
D. Gas Stoichiometry
1. Molar volume = volume one mole of any gas occupies
2. Standard Temperature and Pressure (STP) are 0 oC and 1atm
PV  nRT  V 
nRT (1mol)(0.08206 Latm / Kmol )(273.2K )

 22.42 L
P
1.000atm
3. Real gases differ only slightly from the ideal molar volume
4. Example: n = ? For 1.75L of N2 at STP?
1.75L
1mol 
  0.0781mol
 22.42 L 
5. Example: V(CO2) = ? If 152g CaCO3 decomposes to CaO+CO2?
a. CaCO3(s) -------> CaO(s) + CO2(g)
b. Assume ideal gas behavior for CO2
 1mol CaCO 3 
 1 mol CO 2


152 g CaCO 3 
 1.52 mol CaCO 3 

 1mol CaCO
3
 100.09g CaCO 3 

1.52 mol CO 2  22.42L   34.1 L CO 2
 mol 

  1.52 mol CO 2


6. Example:
a. 2.80 L of CH4 at 25 oC and 1.65atm
b. 35.0L of O2 at 31 oC and 1.25 atm
c. What is the volume of CO2 formed at 2.50 atm and 125 oC?
d. CH4(g) + 2O2(g) -------> CO2(g) + 2H2O(g)
e. Find Limiting reagent
n(CH 4 ) 
1.65atm2.80L
PV

 0.189mol
RT 0.08206Latm / Kmol 298K 
n(O2 ) 
1.25atm35.0L
 1.75mol
0.08206Latm / Kmol 304K 
f. Calculate the amount of CO2 product based on CH4 as limiting reagent
 1 molCO2
0.189mol CH 4 
 1 mol CH
4


  0.189 mol CO 2


g. Use the ideal gas law to find the volume of this amount of CO2
PV  nRT  V 
nRT (0.189mol)(0.08206Latm / Kmol )(398K )

 2.47 L
P
2.50atm
7. Calculating Molar Mass of a Gas from density Example:
m
grams
a.
Molar Mass 
d
V
mol
b. Example: d = 1.95g/L at 1.50atm and 27 oC. Molar Mass = ?
c. Assume we have 1.0L of this gas
 1.95 g 
m
1L   1.95 g
 m  dV  
V
 L 
PV
(1.50atm)1L
PV  nRT  n 

 0.0609mol
RT (0.08206Latm / Kmol )300K 
d
d. Calculate Molar Mass from its definition
Molar Mass 
grams
1.95 g

 32.0 g / mol
mol
0.0609mol