February 5, 2008 Go over Charles’s Law and Avogadro’s Law Homework Introduce Combined Gas Law ...
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Transcript of February 5, 2008 Go over Charles’s Law and Avogadro’s Law Homework Introduce Combined Gas Law ...
February 5, 2008Go over Charles’s Law and Avogadro’s Law HomeworkIntroduce Combined Gas LawIntroduce Ideal Gas LawWork Sample ProblemsHOMEWORK: Pg. 480 -- #21, 23ac, 24ac, 25, 28, 29
Equation of State of an Ideal Gas
• Robert Boyle (1662) found that at fixed temperature– Pressure and volume of a gas is inversely proportional
PV = constant Boyle’s Law
• J. Charles found that at fixed pressure– Volume of gas is proportional to change in temperature
Volume
Temp-273.15 oC
All gases extrapolate to zero volume at a temperature corresponding to –273.15 oC (absolute zero).
He CH4
H2O
H2
Kelvin Temperature Scale
• Kelvin temperature (K) is given by
K = oC + 273.15
where K is the temperature in Kelvins, oC is temperature in Celsius
• Using the ABSOLUTE scale, it is now possible to write Charles’ Law as
V / T = constant Charles’ Law• Combining Boyle’s law, Charles’ law, and another law
called Gay-Lussac’s Law (relating pressure and temperature) we can mathematically prove that
P V / T = constant
Charles
Combined Gas Law
• This brings us to the combined gas law:
2
22
1
11
TVP
TVP
Practice Problem
• A 1.50 L sample of neon gas at 1.10 atm and 25 °C is heated to 45 °C. The neon gas is the subjected to a pressure of 1.50 atm. Determine the new volume of the neon gas.
2
22
1
11
TVP
TVP
P1 = 1.10 atmV1 = 1.50 LT1 = 25 °C = 298 K
P2 = 1.50 atmV1 = ???? LT1 = 45 °C = 318 K
V2 = 1.17 L
The Combined Gas Law
When measured at STP, a quantity of gas has a volume of 500 cm3. What volume will it occupy at 0 oC and 93.3 kPa?
P1 = 101.3 kPaT1 = 273 KV1 = 500 cm3
P2 = 93.3 kPaT2 = 0 oC + 273 = 273 KV2 = X cm3
(101.3 kPa) x (500 cm3) = (93.3 kPa) x (V2)
273 K 273 K
V2 = 542.9 cm3
1 1 2 2
1 2
PV PV
T T
Ideal vs. Real Gases
No gas is ideal.
Most gases behave ideally (almost) at pressures of approximately 1 atm or lower, when the temperature is approximately 0 °C or higher.
When we do calculations, we will assume our gases are behaving as ideal gases
Ideal Gas Equation
P V = n R T
Universal Gas ConstantVolume
No. of moles
Temperature
Pressure
R = 0.0821 atm L / mol K
R = 8.314 kPa L / mol K
Kelter, Carr, Scott, Chemistry A Wolrd of Choices 1999, page 366
Ideal Gas Law
What is the volume that 500 g of iodine will occupy under the conditions: Temp = 300oC and Pressure = 740 mm Hg?
Step 1) Write down given information. mass = 500 g iodine T = 300oC P = 740 mm Hg R = 0.0821 atm . L / mol . K
Step 2) Equation:
V =nRT
P
V (500 g)(0.0821 atm . L / mol . K)(300oC)
740 mm Hg=
Step 3) Solve for variable
Step 4) Substitute in numbers and solve
V =
What MISTAKES did we make in this problem?
PV = nRT
What mistakes did we make in this problem?
What is the volume that 500 g of iodine will occupy under the conditions: Temp = 300oC and Pressure = 740 mm Hg?
Step 1) Write down given information. mass = 500 g iodine Convert mass to moles;
recall iodine is diatomic (I2)500 g I(1 mole I2/254 g I2)
n = 1.9685 mol I2
T = 300oC Temperature must be converted to KelvinT = 300oC + 273
T = 573 K
P = 740 mm Hg Pressure needs to have same unit as R;therefore, convert pressure from mm Hg to atm.x atm = 740 mm Hg (1 atm / 760 mm Hg)
P = 0.8 atm
R = 0.0821 atm . L / mol . K
Ideal Gas Law
What is the volume that 500 g of iodine will occupy under the conditions: Temp = 300oC and Pressure = 740 mm Hg?
Step 1) Write down given information. mass = 500 g iodine
n = 1.9685 mol I2 T = 573 K (300oC) P = 0.9737 atm (740 mm Hg) R = 0.0821 atm . L / mol . K V = ? L
Step 2) Equation: PV = nRT
V =nRT
P
V (1.9685 mol)(0.0821 atm . L / mol . K)(573 K)
0.9737 atm=
Step 3) Solve for variable
Step 4) Substitute in numbers and solve
V = 95.1 L I2