Transcript Bell Ringer

Bell Ringer
298 K
A sample of nitrogen occupies 10.0 liters at 25ºC
and 98.7 kPa. What would be the volume at 20ºC
and 102.7 kPa?
293 K
T2
P1 V1
P2
T1
A 7.87 L
V2 =
B 9.45 L
C 10.2 L
D 10.6 L
V2 =
=
P2 V2
T2
T2
P2
P1 V1 T2
T1 P2
(98.7 kPa) (10.0 L) (293 K)
(298 K) (102.7 kPa)
Gas Laws Quiz
For Next Class:
• Homework:
– Gas Laws Packet #2, problems 1-10
• Quiz next class on Ideal Gas Law, Partial
Pressures, and Density
– 5 questions; 22 points total
• 2 short answer/FITB (2 points each)
• 3 math problems (6 points each)
The Ideal Gas Law & Co.
Ms. Besal
3/14/2006
A Reminder…
We
that we live in an
world where:
• Gas particles have no mass
• Gas particles have no volume
• Gas particles have elastic collisions
These assumptions are used when trying to
calculate the AMOUNT of a gas we have!
Why are these assumptions
important?
PV = nRT
Image source: thefreedictionary.com
PV = nRT
P RESSURE
V OLUME
n MOLES OF GAS
R GAS CONSTANT
TEMPERATURE
Image source: popartuk.com
The MysteRious R
• R is a constant (doesn’t change).
• Number value of R depends on other units.
• Units of R are a combination of many units.
62.4 mmHg · L
mol · K
8.31 kPa · L
mol · K
0.0821 atm · L
mol · K
Image source: toysrusemail.com
Ummm… What?
PV = nRT
Solve for R:
P
V
R =
nT
Plug in units:
(kPa)
(atm)
Hg) (L)
R = (mm
(mol) (K)
Gas Laws, Gas Laws Everywhere!
V1
T1
=
V2
T2
P1 x V1 = P2 x V2
P1 V1
P2 V2
=
T1
T2
P VCONDITIONS
=nRT
Used with CHANGING
Used with only ONE SET OF CONDITIONS
When to Use PV = nRT
• Calculating amount of gas in moles
• Calculating P, V, or T if moles of gas are
known.
– IMPORTANT! We must have 3 out of 4 pieces
of information:
•P
•V
•n
•T
Practice with the Ideal Gas Law
1. A gas sample occupies 2.62 L at 285ºC and 3.42 atm.
How many moles are present in this sample?
PV = nRT
P
V
n =
RT
n =
(3.42 atm) (2.62 L)
= 0.196 mol
0.0821 L · atm (558 K)
mol · K
But Let’s Be Practical…
We don’t usually measure in moles!
We usually measure quantities in GRAMS!
PV = nRT
PVM = gRT
PVM = gRT
P RESSURE
V OLUME
M OLAR MASS OF GAS (g/mol)
g RAMS OF GAS
R GAS CONSTANT
TEMPERATURE
Image source: popartuk.com
Practice with the Ideal Gas Law
A balloon is filled with 0.2494 g of helium to a pressure
of 1.26 atm. If the desired volume of the balloon is
1.250 L, what must the temperature be in ºC?
PVM = gRT
T =
PVM
gR
(1.26 atm) (1.250 L) 4.00 g
mol
T =
0.0821 L · atm
mol · K
(0.2494 g)
=
308 K
- 273
35 ºC
PV=nRT vs. PVM=gRT
• Use PV=nRT when:
– You are given moles in the problem.
– You are searching for moles as an answer.
• Use PVM=gRT when:
– You are given grams in the problem.
– You are searching for grams as an answer.
What Else Happens Under
Unchanging Conditions?
At constant V and T, pressure is easy to calculate!
“The sum of the individual pressures is equal to the total pressure.”
Total Pressure = Pressure of gas 1 + Pressure of gas 2 +
Pressure of gas 3 + Pressure of gas 4 …
Ptotal = P1 + P2 + P3 + …
Partial Pressures Practice
A sample of hydrogen gas is collected over water at
25ºC. The vapor pressure of water at 25ºC is 23.8
mmHg. If the total pressure is 523.8 mmHg, what is
the partial pressure of the hydrogen?
Ptotal = PH2 + PH2O
523.8 mm Hg
P H2
=
=
PH2 + 23.8 mm Hg
500.0 mm Hg
Source: 2003 EOC Chemistry Exam
What do Changing Conditions
Affect?
We have learned that we can change 3 variables:
Temperature, Volume, and Pressure.
If MASS remains constant…
…But VOLUME changes…
Then DENSITY CHANGES!
D = M
V
Two Types of Density Problems:
At STP:
Not at STP:
• molar volume of any
gas at STP =
• Determine new volume (V2 )
using Combined Gas Law
P1 V1
P2 V2
=
T1
T2
• Density at STP =
• Density at non-STP =
molar mass
molar mass
molar
volume
22.4 Liters
V2
Practice with Density Problems:
Determine the density of
ethane (C2H6) at STP:
molar mass
D (at STP) =
molar volume
molar mass = 30.08 g
molar volume = 22.4 L
D =
30.08 g
22.4 L
Determine the density of
C2H6 at 3.0 atm and 41ºC.
V2P1=V8.6
1 L P2 V2
T1
=
T2
D =V molar
mass
=
P1 V1 T2
2
V2 T P
1
2
P1 = 1.0 atm
P2 = 3.0 atm
V2 D
= (1.0
atm) (22.4
(314 K)
= 30.08
g =L) 3.5
g/L
V1 = 22.48.6
L L V2 = ?
(273 K) (3.0 atm)
= 1.34 g/L
T1 = 273 K
V2 = 8.6 L
T2 = 314 K
For Next Class:
• Homework:
– Gas Laws Packet #2, problems 1-10
• Quiz next class on Ideal Gas Law, Partial
Pressures, and Density
– 5 questions; 22 points total
• 2 short answer/FITB (2 points each)
• 3 math problems (6 points each)
What do Changing Conditions
Affect?
• Density
• Stoichiometry problems
So Far:
Now…
Mass-Volume at Non-STP
Two parts to solving these problems:
• Use Stoichiometry
• Use Gas Law