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Gas Laws
Physical Characteristics of Gases
Physical Characteristics
Typical Units
Volume, V
liters (L)
Pressure, P
Kilopascals (kPa) or
atmosphere (atm)
Temperature, T
Kelvin (K)
Number of atoms or
molecules, n
mole (1 mol = 6.023x1023
atoms or molecules)
8-Boyle’s Law
 Pressure and volume
are inversely related at
constant temperature.
 PV = K
 As one goes up, the other
goes down.
 P1V1 = P2V2
“Father of Modern Chemistry”
Robert Boyle
Chemist & Natural Philosopher
Listmore, Ireland
January 25, 1627 – December 30, 1690
Boyle’s Law: P1V1 = P2V2
9 & 10-Charles’ Law
 Volume of a gas varies
directly with the absolute
temperature at constant
pressure.
 V = KT
 V1 / T1 = V2 / T2
Jacques-Alexandre Charles
Mathematician, Physicist, Inventor
Beaugency, France
November 12, 1746 – April 7, 1823
Charles’ Law: V1/T1 = V2/T2
Charles’ Law: V1/T1 = V2/T2
10- Gay-Lussac Law
 At constant volume,
pressure and absolute
temperature are
directly related.
P=kT
 P1 / T1 = P2 / T2
Joseph-Louis Gay-Lussac
Experimentalist
Limoges, France
December 6, 1778 – May 9, 1850
11-Avogadro’s Law: V1/n1=V2/n2
12- Avogadro’s Hypothesis
•
Equal volumes of gases at the same P and T
all contain the same number of molecules.

Ex: at STP conditions, the following is true:





2H2
2mol H2
44.8L H2
12.04 x 1023
+ O2
-->
1mol O2
22.4L O2
6.02 x 1023
2H2O
2mol H2O
44.8L H2O
12.04 X 1023
That is, VOLUMES of gases at same T and P are
in same ratios as the coefficients (like moles are!)
13- Combined Gas Law
• Combination of ‘minor’ gas laws which
•
enables the solving of problems
involving changes of P, V, T and n all
together
P1V1 = P2V2
n1T1 n2T2
If a variable is held constant, omit it from
the formula
 If a variable is doubled with no initial value
given, just substitute it for x and 2x or ‘10’
and ‘20’

Example
• 2 mol of gas X has a volume of 1L at 25oC. If its
pressure is tripled and temperature raised to
100oC, what would be its new volume?
n1 = n2 = 2 mol (variable can be eliminated)
V1 = 1L
T1 = 25 oC + 273 = 298K
V2 = ?
T2 = 100oC + 273 = 373K
V2 = P1V1T2 = (x)(1L)(373K)
T1P2
(298K)(3x)
P1 = x
P2 = 3x
= 0.417L
14- Molar Volume
• Is the volume of ONE MOLE of any gas
•
•
at a specific T and P
This value changes with any change in
P and T
At STP molar volume = 22.4 L
Ideal Gas vs Real Gas
•
Ideal Gases




Hypothetical gases that
‘obey’ all gas laws under
all conditions of P and T
Don’t exist in real life
Gas laws assume that
gases behave as ideal
gases under any T and P
Assume that at any P and
T ideal gases:


Exhibit no attraction
between particles ie: won’t
liquefy or solidify
Occupy no volume at zero
kelvin
•
Real Gases

In reality, all gases will
eventually liquefy or
solidify at very low
temperatures or high
pressures.


At these conditions ,
particles attract each other
Gas Laws are therefore
not applicable under
these conditions of T and
P
15- Ideal Gas Law
• Involves the collective relationship
•
between the P, V, n and T of ANY gas.
PV = nRT
P = Pressure of gas
 V = Volume of gas
 n = moles of gas
 T = Temperature (Kelvin)
 R = Universal gas constant

16 - Universal Gas
Constant (R)
•
•
•
Value applied to make ideal gas law a valid math
equation
Value changes depending on pressure UNIT
Calculated at STP conditions for one mole of any gas

If P in atm, R calculated by:


R = PV = 1 atm x 22.4L = 0.082 atm L
nT
1mol x 273K
mol K
If P in kPa, R calculated by:

R = PV = 101.3 kPa x 22.4L = 8.31 kPa L
nT
1mol x 273K
mol K
Ideal Gas Law Example
•
n=?
What mass of Ne gas must be put into an
880ml rigid vessel to obtain a gas with a P of
240kPa and temperature of 35oC?
P = 240kPa
V = 880ml (0.88L) T = 35oC + 273 = 308K
n = PV = 240kPa x 0.88L
= 0.083mol Ne gas
RT 8.31 kPa L x 308K
mol K
Mass Ne = 0.083mol x 20g/1mol = 1.66g
17 - Partial Pressure of Gas
• Note: equal VOLUME of gases
(containing same mols of gas) at same
T apply:

THE SAME PRESSURE
Dalton’s Law of Partial
Pressure
•
•
Ptotal= P1 + P2 + P3…
Example #1

2L of gas A, 4L of gas B and 8L of gas C all at the same
temperature are put in a container. The total pressure of the
mixture is 235kPa. What is the partial pressure (PP) of
each?
1)
PPA = 2L x 235kPa = 33.6kPa
14L
2) PPB = 4L x 235kPa = 67.1kPa
14L
3) PPC = 8L x 235kPa = 134.3kPa
14L
Ptotal = 235kPa
Dalton’s Law Ex #2
• 7.84L of CO2 and 0.40 mol H2 at SATP are put
into a 10.5L vessel. The temperature of the
mixture drops to -12.5OC. What is the PP of
each gas? (SATP same as STP BUT at 25OC)
1) nCO2 = PV = 101.3kPa x 7.84L = 0.32mol CO2
RT
8.31 x 298K
2)ntotal = 0.32 mol + 0.40 mol = 0.72 mol
3) Ptotal = nRT = 0.72 mol x 8.31 x 260.5 K = 148.4 kPa
V
10.5L
4) PPCO2 = 0.32/0.72 x 148.4kPa = 65.96 kPa
PPH2 = 0.4/0.72 x 148.4kPa = 82.44kPa
STP vs SATP
•
SATP



•
Standard ambient temperature and pressure
25oC (298k)
101.3kPa
STP



Standard temperature and pressure
273K
101.3kPa
18- How to calculate the
molar volume of a gas at
any T and P
• Ex: What is the molar volume of CO2
gas at 890kPa and 15oC?
•P = 890 kPa
•T= 288 K
•n= 1mol
•R = 8.31 kPa L
mol K
V= nRt
P
V= 1mol x 8.31 x 288
890 kPa
V= 2.69 L
*The volume of a gas is affected by any change in T and P but
19- Density of Gases
•D=m
•
V
If any 2 equivalent variables of a gas
are known, they can be used as a
conversion factor to solve a problem.
Example
•
The density of a gas is 1.8g/L at 25OC and
101.3kPa. What is its molar mass?

Volume for one mol:






V=?
n= 1 mol
T= 298K
P=101.3kPa
V=nRT = 1 x 8.31 x 298 = 24.45L
101.3
Molar Mass = 24.45L x 1.8g = 44g
1L
Example #2
• In the reaction N2 + 3H2 -> 2NH3 , if
4.7mol H2 are reacted, what volume of
NH3 would be produced at a T of 45OC
and P of 156 kPa?
a) 4.7mol H2 x 2mol NH3 = 3.13 mol NH3 produced
3mol H2
b) VNH3 = nRT = 3.13 x 8.31 x 318K = 53.02L
P
156 kPa
Example #3
• For the equation in example 2, if 24.44L NH3
are produced at 1.43 atm and 550C, what mass
of H2 must be reacted?
V= 24.44L
N = PV
P = 1.43atm x 101.3kPa
1 atm
RT
= 144.9kPa x 22.4L
= 144.9 kPa
8.31 x 328K
T = 550C + 273 = 328K
n= ?
= 1.3mol
1.3mol NH3 x 3mol H2 = 1.95molH2 x 2g
2mol NH2
1mol
= 3.9gH2
20- Relative Masses of
Gases
• Remember Avagadro’s Hypothesis!!
• Ex #1

What is the mass of 15.68L of gas X at 350C and
116kPa, if its 1.8 times more dense than O2?
a) 32g/mol x 1.8 = 57.6g/mol Gas X
b) n= PV = 116kPa x 15.68L = 0.71mol Gas X
RT
8.31 x 308K
c) 0.71 mol x 57.6g/mol = 40.9g
Example #2
• At SATP, 7.35 L of gas A weighs 17.5g.
•
•
What is its molar mass?
V= nRT = 1 x 8.31 x 298 = 24.5L
P
101.3kPa
Molar Mass = 24.5L x 17.5g = 58g/mol
mol
7.35L
21- Graham’s Law of
Diffusion
• Diffusion
Process whereby a substance spreads
spontaneously in all directions from regions
of high to low concentration
 At the same T gases diffuse/move at
different rates


The lighter the gas, the faster it diffuses
Graham’s Law
• The relative rate of diffusion of 2 gases
is inversely proportional to the square
root of their molar masses or densities.
______________
RateA = √molar mass B
RateB √molar massA
Example #1
Calculate the RR (Relative Rate) of diffusion of H2 and CO2
__________
√44g/mol = 6.63 = 4.7*
√2g/mol
1.41
• There is no unit for the RR of diffusion
• The 4.7 indicates that H2 diffuses at a rate 4.7 times faster
than
CO2
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