Transcript 5.3 - Chemistry

Apparatus for Studying the Relationship Between
Pressure and Volume of a Gas
As P (h) increases
V decreases
1
Boyle’s Law
P a 1/V
P x V = constant
P1 x V1 = P2 x V2
Constant temperature
Constant amount of gas
2
A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg.
What is the pressure of the gas (in mmHg) if the volume is reduced at constant
temperature to 154 mL?
P x V = constant
P1 x V1 = P2 x V2
P2 =
P1 = 726 mmHg
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1
V2
726 mmHg x 946 mL
=
= 4460 mmHg
154 mL
3
Variation in Gas Volume with Temperature at Constant Pressure
As T increases
V increases
4
Variation of Gas Volume with Temperature
at Constant Pressure
Charles’ &
Gay-Lussac’s
Law
VaT
V = constant x T
V1/T1 = V2 /T2
Temperature must be
in Kelvin
T (K) = t (0C) + 273.15
5
A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature
will the gas occupy a volume of 1.54 L if the pressure remains constant?
V1 /T1 = V2 /T2
V1 = 3.20 L
V2 = 1.54 L
T1 = 398.15 K
T2 = ?
T1 = 125 (0C) + 273.15 (K) = 398.15 K
T2 =
V2 x T1
V1
=
1.54 L x 398.15 K
3.20 L
= 192 K
6
Avogadro’s Law
V a number of moles (n)
Constant temperature
Constant pressure
V = constant x n
V1 / n1 = V2 / n2
7
Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many
volumes of NO are obtained from one volume of ammonia at the same temperature
and pressure?
4NH3 + 5O2
1 mole NH3
4NO + 6H2O
1 mole NO
At constant T and P
1 volume NH3
1 volume NO
8
Summary of Gas Laws
Boyle’s Law
9
Charles Law
10
Avogadro’s Law
11
Ideal Gas Equation
Boyle’s law: P a 1 (at constant n and T)
V
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
Va
nT
P
V = constant x
nT
P
=R
nT
P
R is the gas constant
PV = nRT
12
The conditions 0 0C and 1 atm are called standard temperature and pressure
(STP).
Experiments show that at STP, 1 mole of an ideal gas occupies
22.414 L.
PV = nRT
(1 atm)(22.414L)
PV
R=
=
nT
(1 mol)(273.15 K)
R = 0.082057 L • atm / (mol • K)
13
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
T = 0 0C = 273.15 K
P = 1 atm
PV = nRT
nRT
V=
P
n = 49.8 g x
= 1.37 mol
36.45 g HCl
1.37 mol x 0.0821
V=
1 mol HCl
L•atm
mol•K
x 273.15 K
1 atm
V = 30.7 L
14
Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A
certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at
constant volume. What is the final pressure of argon in the lightbulb (in atm)?
PV = nRT
n, V and R are constant
nR
P
=
= constant
T
V
P1 = 1.20 atm
P2 = ?
T1 = 291 K
T2 = 358 K
P1
P2
=
T1
T2
T2
= 1.20 atm x 358 K = 1.48 atm
P2 = P1 x
291 K
T1
15
Sample Problem 5.2
PROBLEM:
Applying the Volume-Pressure Relationship
Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8 cm3
at 1.12 atm. By adding mercury to the tube, he increases the pressure on
the trapped air to 2.64 atm. Assuming constant temperature, what is the
new volume of air (in L)?
PLAN:
P and T are constant
SOLUTION:
V1 in cm3
1cm3=1mL
V1 in mL
unit
conversion
103 mL=1L
V1 in L
xP1/P2
P1 = 1.12 atm
P2 = 2.64 atm
V1 = 24.8 cm3
V2 = unknown
24.8 cm3
1 mL
1 cm3
gas law
calculation
P1V1
V2 in L
n1T1
V2 =
P1V1
P2
=
L
= 0.0248 L
103 mL
P2V2
P1V1 = P2V2
n2T2
= 0.0248 L
1.12 atm
2.46 atm
= 0.0105 L
Sample Problem 5.3
Applying the Pressure-Temperature Relationship
A steel tank used for fuel delivery is fitted with a safety valve that
opens when the internal pressure exceeds 1.00x103 torr. It is filled
with methane at 230C and 0.991 atm and placed in boiling water at
exactly 1000C. Will the safety valve open?
PROBLEM:
PLAN:
SOLUTION:
T1 and T2(0C)
P1(atm)
1atm=760torr
P1(torr)
K=0C+273.15
T1 and T2(K)
x T2/T1
P2 = unknown
T1 = 230C
T2 = 1000C
P1V1
n1T1
P2(torr)
0.991 atm
P2 =
P1 = 0.991atm
P1
n2T2
760 torr = 753 torr
1 atm
T2
T1
=
P2V2
= 753 torr
373K
296K
= 949 torr
P1
T1
=
P2
T2
Sample Problem 5.4
A scale model of a blimp rises when it is filled with helium to a volume of
55 dm3. When 1.10 mol of He is added to the blimp, the volume is 26.2
dm3. How many more grams of He must be added to make it rise?
Assume constant T and P.
PROBLEM:
PLAN:
Applying the Volume-Amount Relationship
We are given initial n1 and V1 as well as the final V2. We have to find n2
and convert it from moles to grams.
n1(mol) of He
P and T are constant
SOLUTION:
x V2/V1
n1 = 1.10 mol
n2 = unknown
n2(mol) of He
V1 = 26.2 dm3
V2 = 55.0 dm3
subtract n1
mol to be added
V1
n1
=
V2
n2
xM
g to be added
n2 = 1.10 mol
n2 = n 1
55.0 dm3
26.2
dm3
P1V1
n1 T1
=
P2V2
n2T2
V2
V1
4.003 g He
= 2.31 mol
mol He
= 9.24 g He
Sample Problem 5.5
Conditions
A steel tank has a volume of 438 L and is filled with 0.885 kg of O2.
Calculate the pressure of O2 at 210C.
PROBLEM:
PLAN:
V, T and mass, which can be converted to moles (n), are given. We use the
ideal gas law to find P.
SOLUTION:
0.885kg
V = 438 L
T = 210C (convert to K)
n = 0.885 kg (convert to mol)
P = unknown
103 g
mol O2
kg
32.00 g O2
24.7 mol
P=
Solving for an Unknown Gas Variable at Fixed
nRT
V
=
= 27.7 mol O2
x 0.0821
438 L
atm*L
mol*K
210C + 273.15 = 294.15K
x 294.15K
= 1.53 atm
Sample Problem 5.6
Using Gas Laws to Determine a Balanced Equation
The piston-cylinders below depict a gaseous reaction carried out at
constant pressure. Before the reaction, the temperature is 150K;
when it is complete, the temperature is 300K.
PROBLEM:
New figures go here.
Which of the following balanced equations describes the reaction?
(1) A2 + B2
(3) A + B2
PLAN:
2AB
AB2
(2) 2AB + B2
(4) 2AB2
2AB2
A2 + 2B2
We know P, T, and V, initial and final, from the pictures. Note that the
volume doesn’t change even though the temperature is doubled. With a
doubling of T then, the number of moles of gas must have been halved in
order to maintain the volume.
SOLUTION:
Looking at the relationships, the equation that shows a
decrease in the number of moles of gas from 2 to 1 is
equation (3).