GAS LAWS

CONCEPT MAP

Properties of Gases

Gas Model Gases are composed of tiny, widely-spaced particles. For a typical gas, the average distance between particles is about ten times their diameter. Because of the large distance between the particles, the volume occupied by the particles themselves is negligible (approximately zero). For a typical gas at room temperature and pressure, the gas particles themselves occupy about 0.1% of the total volume. The other 99.9% of the total volume is empty space. This is very different than for a liquid for which about 70% of the volume is occupied by particles.

The particles have rapid and continuous motion. For example, the average velocity of a helium atom, He, at room temperature is over 1000 m/s (or over 2000 mi/hr). The average velocity of the more massive nitrogen molecules, N2, at room temperature is about 500 m/s. Increased temperature means increased average velocity of the particles.

The particles are constantly colliding with the walls of the container and with each other. Because of these collisions, the gas particles are constantly changing their direction of motion and their velocity. In a typical situation, a gas particle moves a very short distance between collisions. O2 molecules at normal temperatures and pressures move an average of 10-7 m between collisions. There is no net loss of energy in the collisions. A collision between two particles may lead to each particle changing its velocity and thus its energy, but the increase in energy by one particle is balanced by an equal decrease in energy by the other particle.

IDEAL GASES The particles are assumed to be point‑masses, that is, particles that have a mass but occupy no volume. There are no attractive or repulsive forces at all between the particles.

Gas Properties and their Units Pressure (P) = Force/Area 1 atm = 101.325 kPa = 760 mmHg = 760 torr Volume (V) unit usually liters (L) 1000 L = 1 m3 Temperature (T) ? K = --- C + 273.15 C = (F - 32) /1.8 Number of gas particles (n) mole

Gas Law Objectives For each of the following pairs of gas properties, (1) describe the relationship between them, (2) describe a simple system that could be used to demonstrate the relationship, and (3) explain the reason for the relationship. V and P when n and T are constant P and T when n and V are constant V and T when n and P are constant n and P when V and T are constant n and V when P and T are constant

Apparatus for Demonstrating Relationships Between Properties of Gases

Boyle’s Law The pressure of an ideal gas is inversely proportional to the volume it occupies if the moles of gas and the temperature are constant.

As the pressure increases, volume decreases

Relationship between P and V

Example 1: 2. 00 L of a gas is at 740. 0 mmHg pressure Example 1: 2.00 L of a gas is at 740.0 mmHg pressure. What is its volume at standard pressure (760 mm Hg)? Answer: this problem is solved by inserting values into P1V1 = P2V2. (740.0 mmHg) (2.00 L) =(760.0 mmHg) (x) x = 1.95 L

Example 2: 5. 00 L of a gas is at 1. 08 atm Example 2: 5.00 L of a gas is at 1.08 atm. What pressure is obtained when the volume is 10.0 L? (1.08 atm) (5.00 L) =(x) (10.0 L) x = 0.54 atm

Charles’ Law For an ideal gas, volume and temperature described in kelvins are directly proportional if moles of gas and pressure are constant.

As the temperature increases, volume increases.

Relationship between T and V

Example 1: A gas is collected and found to fill 2. 85 L at 25. 0°C Example 1: A gas is collected and found to fill 2.85 L at 25.0°C. What will be its volume at standard temperature? Answer: convert 25.0°C to Kelvin and you get 298 K. Standard temperature is 273 K. We plug into our equation like this: x = 2.61 L

Example #2: 4. 40 L of a gas is collected at 50. 0°C Example #2: 4.40 L of a gas is collected at 50.0°C. What will be its volume upon cooling to 25.0°C? First of all, 2.20 L is the wrong answer. Sometimes a student will look at the temperature being cut in half and reason that the volume must also be cut in half. That would be true if the temperature was in Kelvin. However, in this problem the Celsius is cut in half, not the Kelvin. Answer: convert 50.0°C to 323 K and 25.0°C to 298 K. Then plug into the equation and solve for x, like this: x = 4.06 L

Gay-Lussac’s Law The pressure of an ideal gas is directly proportional to the Kelvin temperature of the gas if the volume and moles of gas are constant.

GAY-LUSSAC’S LAW As the temperature increases, the pressure increases.

Relationship between P and T

Example 1: 10. 0 L of a gas is found to exert 97. 0 kPa at 25. 0°C Example 1: 10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What would be the required temperature (in Celsius) to change the pressure to standard pressure? Answer: change 25.0°C to 298.0 K and remember that standard pressure in kPa is 101.325. Insert values into the equation (the ChemTeam will use the left-hand one in the graphic above) and get: x = 311.3 K or 38.3 oC

Example 2: 5. 00 L of a gas is collected at 22. 0°C and 745. 0 mmHg Example 2: 5.00 L of a gas is collected at 22.0°C and 745.0 mmHg. When the temperature is changed to standard, what is the new pressure? Answer: convert to Kelvin and insert: 689 mm Hg