february 5, 2008 go over charles’s law and avogadro’s law homework introduce combined gas law ...
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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