gases
DESCRIPTION
GASES. General Properties of Gases. There is a lot of “free” space in a gas. Gases can be expanded infinitely. Gases fill containers uniformly and completely. Gases diffuse and mix rapidly. Properties of Gases. Gas properties can be modeled using math. Model depends on — - PowerPoint PPT PresentationTRANSCRIPT
GASES
General Properties of Gases
• There is a lot of “free” space in a gas.• Gases can be expanded infinitely.• Gases fill containers uniformly and completely.• Gases diffuse and mix rapidly.
Properties of Gases• Gas properties can be modeled using math. Model depends on—
• V = volume of the gas (L, mL)
• T = temperature (K)• ALL temperatures in the entire chapter MUST be in Kelvin!!! No Exceptions!
• n = amount (moles)
• P = pressure (atmospheres, mmHg, torr, kPa)
Pressure• Column height measures Pressure of atmosphere• 1 standard atmosphere (atm) *• = 760 mm Hg (or torr) *• = 101.3 kPa (SI unit is PASCAL)
Pressure conversions• A.) What is 475 mm Hg expressed in atm?
• 475 mmHg 1 atm = 0.625 atm
760 mm Hg
• B.) The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg?
• 29.4 psi 760 mmHg = 1.52 x 103 mmHg
14.7 psi
Your Turn: Learning Check for Pressure Conversions • A.) What is 2 atm expressed in torr?
• B.) The pressure of a tire is measured as 32.0 psi. What is this pressure in kPa?
Boyle’s Law• This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and
temperature are constant (do not change). For example, P goes up as V goes down.
• P1V1 = P2 V2
• V1 is the original volume • V2 is the new volume • P1 is original pressure • P2 is the new pressure
• Sample Problem• Suppose you have a gas with 45.0 ml of volume and has a pressure of
760.mmHg. If the pressure is increased to 800mmHg and the temperature remains constant then according to Boyle's Law the new volume is 42.8 ml.
• (760mmHg)(45.0ml) = (800mmHg)(V2) •
V2 = 42.8ml
Robert Boyle
Charles’s Law• V and T are directly proportional. If one temperature goes up, the volume goes up!• V1 V2
T1 = T2
• V1 is the initial volume T1 is the initial temperature • V2 is the final volume T2 is the final temperature
• Sample Problem• You have a gas that has a volume of 2.5 liters and a temperature of 250 K. What
would be the final temperature if the gas has a volume of 4.5 liters?• V1 / T1 = V2 / T2 • V1 = 2.5 liters• T1 = 250 K• V2 = 4.5 liters• T2 = ?• Solving for T2, the final temperature equals 450 K.
• Important: Charles's Law only works when the pressure is constant. • Note: Charles's Law is fairly accurate but gases tend to deviate from it at very high
and low pressures.
Jacques Charles
Gay-Lussac’s Law• If n and V are constant,
then P α T• P and T are directly proportional.• P1 P2
T1 T2
• If one temperature goes up, the pressure goes up!
• Sample problem
• The pressure inside a container is 770 mmHg at a temperature of 57 C. What would the pressure be at 75 C?• P1= 770 mmHg• T1 = 57°C• T2= 75°C• P2 = ?
=
Combined Gas Law• Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION!
P1 V1 P2 V2
T1 T2=
Combined Gas Law Problem• A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?
• Set up Data Table• P1 = 0.800 atm V1 = 180 mL T1 = 302 K• P2 = 3.20 atm V2= 90 mL T2 = ??
Calculations• P1 = 0.800 atm V1 = 180 mL T1 = 302 K• P2 = 3.20 atm V2= 90 mL T2 = ??
• P1 V1 P2 V2
T1 = T2 P1 V1 T2 = P2 V2 T1
• T2 = P2 V2 T1
P1 V1
• T2 = 3.20 atm x 90.0 mL x 302 K
0.800 atm x 180.0 mL
• T2 = 604 K - 273 = 331 °C