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