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Gas Laws
Gas Laws
1. Pressure and Volume (Boyle’s Law)
2.
3.
Pressure and VolumeExperiment Pressure Volume P x V
(atm) (L) (atm x L)
1 8.0 2.0 16
2 4.0 4.0 _____
3 2.0 8.0 _____
4 1.0 16 _____
Boyle's Law P x V = k (constant) when
T remains constant
P1V1= 8.0 atm x 2.0 L = 16 atm L
P2V2= 4.0 atm x 4.0 L = 16 atm L
P1V1 = P2V2 = k
Use this equation to calculate how a volume changes when pressure changes, or how pressure changes when volume changes.
new vol. old vol. x Pfactor new P old P x Vfactor
V2 = V1 x P1 P2 = P1 x V1
P2 V2
P and V Changes
P1
P2
V1 V2
Boyle's Law
The pressure of a gas is inversely related to the volume when T does not change
Then the PV product remains constant
P1V1 = P2V2
P1V1= 8.0 atm x 2.0 L = 16 atm L
P2V2= 4.0 atm x 4.0 L = 16 atm L
Boyle’s Law Problem
Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 1.6 L sample of Freon gas initially at 50 mm Hg after its pressure is changed to 200 mm Hg at constant T?
PV Calculation
Prepare a data table
DATA TABLE
Initial conditions Final conditions
P1 = 50 mm Hg P2 = 200 mm Hg
V1 = 1.6 L V2 = ?
Find New Volume (V2)
Solve for V2: P1V1 = P2V2
V2 = V1 x P1 /P2
V2 = 1.6 L x 50 mm Hg = 0.4 L
200 mm Hg
Check your understanding
A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? (T constant) Explain.
1) 3.2 L
2) 6.4 L
3) 12.8 L
Solution
A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? (T constant)
6.4 L x 0.70 atm = 3.2 L (1)
1.40 atm
Volume must decrease to cause an increase in the pressure
Check your understanding
A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain.
1) 200. mmHg
2) 400. mmHg
3) 1200 mmHg
Solution
A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain.
600. mm Hg x 12.0 L = 200. mmHg (1)
36.0 L
Pressure decrease when volume increases.
Gas Laws
1. Pressure and Volume (Boyle’s Law)
2. Temperature and Volume (Charles’ Law)
3.
Charles’ Law
V = 125 mL V = 250 mL
T = 273 K T = 546 K
Observe the V and T of the balloons. How does volume change with temperature?
Charles’ Law: V and T
At constant pressure, the volume of a gas is
directly related to its absolute (K) temperature
V1 = V2
T1 T2
Understanding Charles’ Law
Use Charles’ Law to complete the statements
below:
1. If final T is higher than initial T, final V
is (greater, or less) than the initial V.
2. If final V is less than initial V, final T is
(higher, or lower) than the initial T.
Solution
V1 = V2
T1 T2
1. If final T is higher than initial T, final V
is (greater) than the initial V.
2. If final V is less than initial V, final T is (lower) than the initial T.
V and T Problem
A balloon has a volume of 785 mL on a Fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?
VT Calculation
Complete the following setup:
Initial conditions Final conditions
V1 = 785 mL V2 = ?
T1 = 21°C = 294 K T2 = 0°C = 273 K
V2 = _785__ mL x _273 K = _______ mL
V1 294 K
Check your answer: If temperature decreases,
V should decrease.
VT Calculation
Complete the following setup:
Initial conditions Final conditions
V1 = 785 mL V2 = ?
T1 = 21°C = 294 K T2 = 0°C = 273 K
V2 = _785__ mL x _273 K = _729___ mL
V1 294 K
Check your answer: If temperature decreases,
V should decrease.
Understanding Check
A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. What temperature (in °C) is needed to change the volume to 640 mL?
1) 443°C 2) 170°C 3) - 82°C
Solution
A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. What temperature (in °C) is needed to change the volume to 640 mL?
2) 170°C
T2 = 291 K x 640 mL = 443 K
420 mL
= 443 K - 273 K = 170°C
Gas Laws
1. Pressure and Volume (Boyle’s Law)
2. Temperature and Volume (Charles’ Law)
3. Temperature and Pressure(Gay-Lussac’s Law)
Gay-Lussac’s Law: P and T
The pressure exerted by a confined gas
is directly related to the temperature
(Kelvin) at constant volume.
P (mm Hg) T (°C)
936 100
761 25
691 0
Understanding Gas Laws
Use Gay-Lussac’s law to complete the statements below:
1. When temperature decreases, the
pressure of a gas (decreases or increases).
2. When temperature increases, the pressure
of a gas (decreases or increases).
Solution
1. When temperature decreases, the
pressure of a gas (decreases).
2. When temperature increases, the
pressure of a gas (increases).
PT Problem
A gas has a pressure at 2.0 atm at 18°C. What will be the new pressure if the temperature rises to 62°C? (V constant)
T = 18°C T = 62°C
PT Calculation
P1 = 2.0 atm T1 = 18°C + 273 = 291 K
P2 = ? T2 = 62°C + 273 = 335 K
What happens to P when T increases?
P increases (directly related to T)
P2 = P1 x T2
T1
P2 = 2.0 atm x 335 K = 2.30 atm
291 K
?
Understanding Gas Laws
Complete with 1) Increases 2) Decreases
3) Does not change
A. Pressure _____, when V decreases
B. When T decreases, V _____.
C. Pressure _____ when V changes from 12.0 L to 24.0 L (constant n and T)
D. Volume _____when T changes from 15.0 °C to 45.0°C (constant P and n)
Solution
A. Pressure 1) Increases, when V decreases
B. When T decreases, V 2) Decreases
C. Pressure 2) Decreases when V changes
from 12.0 L to 24.0 L (constant n and T)
D. Volume 1) Increases when T changes from 15.0 °C to 45.0°C (constant P and n)