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1As P (h) increases V decreases
Apparatus for Studying the Relationship BetweenPressure and Volume of a Gas
2
P 1/V
P x V = constant
P1 x V1 = P2 x V2
Boyle’s Law
Constant temperatureConstant amount of gas
3
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?
P1 x V1 = P2 x V2
P1 = 726 mmHg
V1 = 946 mL
P2 = ?
V2 = 154 mL
P2 = P1 x V1
V2
726 mmHg x 946 mL154 mL
= = 4460 mmHg
P x V = constant
4As T increases V increases
Variation in Gas Volume with Temperature at Constant Pressure
5
Variation of Gas Volume with Temperatureat Constant Pressure
V T
V = constant x T
V1/T1 = V2 /T2T (K) = t (0C) + 273.15
Charles’ & Gay-Lussac’s Law
Temperature must bein Kelvin
6
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 = 3.20 L
T1 = 398.15 K
V2 = 1.54 L
T2 = ?
T2 = V2 x T1
V1
1.54 L x 398.15 K3.20 L
= = 192 K
V1 /T1 = V2 /T2
T1 = 125 (0C) + 273.15 (K) = 398.15 K
7
Avogadro’s Law
V number of moles (n)
V = constant x n
V1 / n1 = V2 / n2
Constant temperatureConstant pressure
8
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 4NO + 6H2O
1 mole NH3 1 mole NO
At constant T and P
1 volume NH3 1 volume NO
9
Summary of Gas Laws
Boyle’s Law
10
Charles Law
11
Avogadro’s Law
12
Ideal Gas Equation
Charles’ law: V T(at constant n and P)
Avogadro’s law: V n(at constant P and T)
Boyle’s law: P (at constant n and T)1V
V nT
P
V = constant x = RnT
P
nT
PR is the gas constant
PV = nRT
13
The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).
PV = nRT
R = PVnT
=(1 atm)(22.414L)
(1 mol)(273.15 K)
R = 0.082057 L • atm / (mol • K)
Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.
14
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
PV = nRT
V = nRTP
T = 0 0C = 273.15 K
P = 1 atm
n = 49.8 g x 1 mol HCl
36.45 g HCl= 1.37 mol
V =1 atm
1.37 mol x 0.0821 x 273.15 KL•atmmol•K
V = 30.7 L
15
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
nRV
= PT
= constant
P1
T1
P2
T2
=
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K
P2 = P1 x T2
T1
= 1.20 atm x 358 K291 K
= 1.48 atm
Sample Problem 5.2 Applying the Volume-Pressure Relationship
PROBLEM: 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: SOLUTION:
V1 in cm3
V1 in mL
V1 in L
V2 in L
unit conversion
gas law calculation
P1 = 1.12 atm P2 = 2.64 atm
V1 = 24.8 cm3 V2 = unknown
P and T are constant
24.8 cm3 1 mL
1 cm3
L
103 mL= 0.0248 L
P1V1
n1T1
P2V2
n2T2
=P1V1 = P2V2
P1V1
P2
V2 = = 0.0248 L1.12 atm
2.46 atm= 0.0105 L
1cm3=1mL
103 mL=1L
xP1/P2
Sample Problem 5.3 Applying the Pressure-Temperature Relationship
PROBLEM: 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?
PLAN: SOLUTION:
P1(atm) T1 and T2(0C)
P1(torr) T1 and T2(K)
P1 = 0.991atm P2 = unknown
T1 = 230C T2 = 1000C
P2(torr)
1atm=760torr
x T2/T1
K=0C+273.15
P1V1
n1T1
P2V2
n2T2
=P1
T1
P2
T2
=
0.991 atm1 atm
760 torr = 753 torr
P2 = P1 T2
T1 = 753 torr
373K
296K= 949 torr
Sample Problem 5.4 Applying the Volume-Amount Relationship
PROBLEM: 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.
PLAN:
SOLUTION:
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
n2(mol) of He
mol to be added
g to be added
x V2/V1
x M
subtract n1
n1 = 1.10 mol n2 = unknown
V1 = 26.2 dm3 V2 = 55.0 dm3
P and T are constant
P1V1
n1T1
P2V2
n2T2
=
V1
n1
V2
n2
= n2 = n1 V2
V1
n2 = 1.10 mol55.0 dm3
26.2 dm3= 2.31 mol
4.003 g He
mol He= 9.24 g He
Sample Problem 5.5 Solving for an Unknown Gas Variable at Fixed Conditions
PROBLEM: 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.
PLAN:
SOLUTION:
V, T and mass, which can be converted to moles (n), are given. We use the ideal gas law to find P.
V = 438 L T = 210C (convert to K)
n = 0.885 kg (convert to mol) P = unknown
210C + 273.15 = 294.15K0.885kg103 g
kg
mol O2
32.00 g O2
= 27.7 mol O2
P = nRT
V=
24.7 mol 294.15Katm*L
mol*K0.0821x x
438 L= 1.53 atm
Sample Problem 5.6 Using Gas Laws to Determine a Balanced Equation
PROBLEM: 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.
PLAN:
SOLUTION:
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.
Which of the following balanced equations describes the reaction?
(1) A2 + B2 2AB (2) 2AB + B2 2AB2
(4) 2AB2 A2 + 2B2(3) A + B2 AB2
Looking at the relationships, the equation that shows a decrease in the number of moles of gas from 2 to 1 is equation (3).
New figures go here.