February 5, 2008
Go over Charles’s Law and Avogadro’s Law Homework
Introduce Combined Gas Law
Introduce Ideal Gas Law
Work Sample Problems
HOMEWORK: 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
He
CH4
H2O
H2
-273.15 oC
All gases extrapolate to zero volume at a temperature
corresponding to –273.15 oC (absolute zero).
Temp
Kelvin Temperature Scale
Charles
• 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
Combined Gas Law
• This brings us to the
combined gas law:
P1V1
T1
P2 V2

T2
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.
P1V1
T1
P2 V2

T2
P1 = 1.10 atm
V1 = 1.50 L
T1 = 25 °C = 298 K
P2 = 1.50 atm
V1 = ???? L
T1 = 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?
PV
PV
1 1
 2 2
T1
T2
P1 =
T1 =
V1 =
P2 =
T2 =
V2 =
(101.3 kPa) x (500 cm3) = (93.3 kPa) x (V2)
101.3 kPa
273 K
500 cm3
93.3 kPa
0 oC + 273 = 273 K
X cm3
273 K
273 K
V2 = 542.9 cm3
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
Volume
Pressure
PV=nRT
No. of moles
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
Universal Gas Constant
Temperature
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: PV = nRT
Step 3) Solve for variable
V =
nRT
P
Step 4) Substitute in numbers and solve
(500 g)(0.0821 atm . L / mol . K)(300oC)
V =
740 mm Hg
V=
What MISTAKES did we make in this problem?
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 Kelvin
T = 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
Step 3) Solve for variable
V =
nRT
P
Step 4) Substitute in numbers and solve
(1.9685 mol)(0.0821 atm . L / mol . K)(573 K)
V =
0.9737 atm
V = 95.1 L I2