lecture 14.2b- gas law equations
DESCRIPTION
Section 14.2 lecture (part B) for Honors & Prep ChemistryTRANSCRIPT
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Bellwork- Gas Variables
What four variables are needed to describe a gas sample?
What units are they measured in?
Do you need to know what kind of gas it is?
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P1V1 P2V2
T1 T2 =
The combined gas law
Uses three gas variables to describe a gas sample at two different times.
If a variable does not change (is constant) it can be removed from the equation.
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Boyle’s Law: Pressure and Volume
If the temperature is constant, as the pressure of a gas increases, the volume decreases.
LAW - P, V
Constants- amount of gas, Temp
P1V1 P2V2
T1 T2 = P1V1 P2V2=
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Boyle’s law states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure.
Only if T and n are held constant
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INDIRECTLY PROPORTIONAL aka INVERSELY PROPORTIONAL
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Boyle ProblemL measures Volume(V)
V1 = 30.0L
All math All math examples examples should be in should be in your notesyour notes
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Boyle ProblemkPa measures Pressure(P)
V1 = 30.0L V2 = ?P1 = 103 kPa P2 = 25.0 kPa
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Boyle ProblemThese two values go together because they describe the gas at the same moment.
V1 = 30.0L V2 = ?P1 = 103 kPa P2 = 25.0 kPa
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Boyle ProblemThis is the unknown
V1 = 30.0L V2 = ?P1 = 103 kPa
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Boyle ProblemThis is a pressure value
V1 = 30.0L V2 = ?P1 = 103 kPa P2 = 25.0 kPa
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Boyle ProblemThese two values go together because they describe the gas at the same moment.
V1 = 30.0L V2 = ?P1 = 103 kPa P2 = 25.0 kPa
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Boyle Problem
Temp is constant and I need an equation that relates pressure and volume.
P1V1 = P2V2
P1V1 = P2V2
T1 T2
V1 = 30.0L V2 = ?P1 = 103 kPa P2 = 25.0 kPa
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Boyle Problem
P1V1 = P2V2
103 kPa x 30L = 25.0 kPa x V2___________ ___________ 25.0 kPa 25.0kPa
= 124 L
P,V GOOD!
V1 = 30.0L V2 = ?P1 = 103 kPa P2 = 25.0 kPa
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for Sample Problem 14.1
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Charles’s Law: Temperature and Volume
As the temperature of an enclosed gas increases, the volume increases, if the pressure is constant.
LAW- T ,V
Constants- moles of gas, Pressure
P1V1 P2V2
T1 T2 =
V1 V2
T1 T2 =
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Charles’s law states that the volume of a fixed mass of gas is directly proportional to its Kelvin temperature
if the pressure is kept constant.
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DIRECTLY PROPORTIONAL
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14.2
T1 = 24C T2 = 58CV1 = 4.00L V2 = ?
T must be in Kelvins!!
+ 273 = 297K + 273 = 331K
4.00L V2
297 K 331 K=
V2 = 4L 297K x 331K = 4.46 L
T V GOOD!
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for Sample Problem 14.2
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Gay-Lussac’s Law: Pressure and Temperature
As the temperature of an enclosed gas increases, the pressure increases, if the volume is constant.
LAW- T , P Constants- moles of gas, Volume
P1V1 P2V2
T1 T2 =
P1 P2
T1 T2 =
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Gay-Lussac’s law states that the pressure of a gas is directly proportional to the Kelvin temperature
if the volume remains constant.
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14.3
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for Sample Problem 14.3
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The combined gas law allows you to do calculations for situations in which only the amount of gas is constant.
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for Sample Problem 14.4
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1. If the volume of a gas in a container were reduced to one fifth the original volume at constant temperature, the pressure of the gas in the new volume would be
a. one and one fifth times the original pressure.
b. one fifth of the original pressure.
c. four fifths of the original pressure.
d. five times the original pressure.
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14.2 Section Quiz.3. At 46°C and 89 kPa pressure, a gas occupies a
volume of 0.600 L. How many liters will it occupy at 0°C and 20.8 kPa?
a. 0.600 L
b. 2.58 L
c. 0.140 L
d. 2.20 L