gas laws and relationships between p, v, and t
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Gas Laws and Relationships between P, V, and T. Boyle’s Law Charles’s Law Gay-Lusaac’s Law How to use each. Gases and their Variables. Four Variables when exploring gases: P , V , T , and n P : Pressure in atm’s V : Volume in L T : Temperature in K - PowerPoint PPT PresentationTRANSCRIPT
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Gas Laws and Relationships between P, V, and T
Boyle’s Law
Charles’s Law
Gay-Lusaac’s Law
How to use each
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Gases and their Variables• Four Variables when exploring gases:
P, V, T, and n
• P: Pressure in atm’s
• V: Volume in L
• T: Temperature in K
• n: quantity of matter in moles
How do they relate to one another? Let’s put on our PVT Cards and find out!
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Gas Properties can be modeled using MATH!
And again, Scientists have done all of the work for us!
http://www.youtube.com/watch?v=13WUqWd_Yk8&feature=related
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Boyle’s LawStates: The volume of a sample of gas is inversely proportional to its pressure, if temperature remains constant.
Translation: At constant temperature and n, 1 α P
VRobert Boyle
EX. Volume Decreases, Pressure Increases Volume Increases, Pressure Decreases
In an inverse relationship, the product of the two quantities is a constant.
P1 x V1 = P2 x V2
P1 V1 = P2 V2
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Boyle’s Law
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Example 1 - A sample of gas collected in a 350 cm3 container exerts a pressure of 103 kPa. What would be the volume of this gas at 150 kPa of pressure? (Assume that the temperature remains constant.)
Solving: If temperature remains constant - use Boyle’s Law.
Write the original formula: P1V1 = P2V2
Then list what is given and what is unknown.
P1 = 103 kPa V1= 350 cm3 P2 = 150 kPa V2 = ?
Then PLUG AND CHUG! Let’s work it together.
103 kPa (350 cm3) = 150 kPa V2
150 kPa 150 kPa240 cm3= V2
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Charles’s LawStates: The volume of a sample of gas is directly proportional to its Kelvin temperature, if pressure remains constant.
Translation: At constant pressure and n, V α T **Temperature ALWAYS in Kelvin**
Jacques Charles
EX. Volume Increases, Temperature Increases Volume Decreases, Temperature Decreases
In an direct relationship, the quotient of the two quantities is a constant.
V1 / T1 = V2 / T2 V1 = V2
T1 T2
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Charles’s Law
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Ex. If a gas occupies 733 cm3 at 10.0 oC, at what temperature will it occupy 950 cm3? Assume that pressure remains constant.
Solving: If pressure remains constant - use Charles’s Law.
Write the original formula:V1 = V2
T1 T2
Then list what is given and what is unknown.
V1 = 733 cm3 T1= 10.0 oC V2 = 950 cm3 T2 = ?
Then PLUG AND CHUG! Let’s work it together. First convert oC to Kelvin: K = 10.0 oC + 273 = 283 K
733 cm3 = 950 cm3 283 K T2
366.7 or 370 K= T2
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How does a hot air balloon work?
TIP: Think about Charles’s Law.
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Gay-Lussac’s Law
States: The Kelvin temperature of a sample of gas is directly proportional to pressure, if volume remains constant.
Translation: At constant volume and n, T α P **Temperature ALWAYS in Kelvin**
Joseph Gay-Lussac
EX. Temperature Increases, Pressure Increases Temperature Decreases, Pressure Decreases
In an direct relationship, the quotient of the two quantities is a constant.
P1 / T1 = P2 / T2 P1 = P2
T1 T2
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Gay-Lussac’s Law
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EX. If a gas is cooled from 323.0 K to 273.15 K and the volume is kept constant what final pressure would result if the original pressure was 750.0 mm Hg?
Solving: If volume remains constant - use Gay-Lussac’s Law.
Write the original formula:P1 = P2
T1 T2
Then list what is given and what is unknown.
P1 = 750.0 mm Hg T1= 323.0 K P2 = ? T2 = 273.15 K
Then PLUG AND CHUG! Let’s work it together.
750.0 mm Hg = P2 323.0 K 273.15 K
634.2 mm Hg= P2
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ONE MORE FUN FACT! Standard Temperature and
Pressure (STP)• At STP:
– Temperature = 273 K or 0 oC – Pressure = 1 atm = 760 mm Hg
Gas Laws Song
http://www.youtube.com/watch?v=13WUqWd_Yk8
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Let’s Practice!1. A sample of neon has a volume of 239 cm3 at 2.00 atm of pressure. What would the pressure have to be in order for the gas to have a volume of 5.00 x 102 cm3?
2. A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. The pressure inside the container at 20.0 °C was at 3.00 atm. What is the pressure of the nitrogen after its temperature is increased?
3. A sample of gas at 3.00 x 103 mm Hg inside a steel tank is cooled from 500.0 °C to 0.00 °C. What is the final pressure of the gas in the steel tank?
HOMEWORK = Practice problems with each law!FINISH LAB TO TURN IN AT END OF CLASS!