UNIT 6 SOLUTIONS AND GASES6.8 What is the relationship between P, V, and T of an ideal gas?

May 4, 2011AIM: What are the effects of pressure, volume, and temperature on ideal gases?

DO NOW:1. When you heat a marshmallow in the microwave, the marshmellow

expands. What is the relationship between temperature and volume? (direct or indirect/inverse?)

Direct: as temperature increases, volume increases

2. Why the hot-air balloon can fly. What is happening to the air molecules that cause the balloon to float in the air?

The air molecules are expanding , causing the pressure in the balloon to be less thanthat of outside of the balloon.

LAB

• Complete the Lab.• After you are done with the lab, please fill out

the graphic organizer on the back of you worksheet.– If you are having difficulty filling out the graphic

organizer, please access this PowerPoint (6.8) online!

BOYLE’S LAWPRESSURE AND VOLUME OF A GAS

(at constant TEMPERATURE)

BOYLE’S LAW: TREND

At constant temperature, as pressure increases… VOLUME DECREASES

This is an INDIRECT RELATIONSHIP

MATHEMATICAL EQUATION

P1 V1= P2V2

CHARLES’S LAWVOLUME AND TEMPERATURE OF A GAS

(at constant PRESSURE)

CHARLES’S LAW: TREND

At constant pressure, as volume increases…TEMPERATURE INCREASES

This is an DIRECT RELATIONSHIP

MATHEMATICAL EQUATION

V1 = V2

T1 = T2

How do we solve problems relating to gases?I. BOYLE’S LAW

Example 1A sample of gas in a syringe has a volume of 9.66 mL at a pressure of 64.4 kPa. The plunger is depressed until the pressure is 94.6 kPa. What is the new volume, assuming constant temperature?

Step 1. List the givens.

V1 = V2=

P1= P2=

9.66 ml

64.4 kPa 94.6 kPa

?

Step 2. Write out the formula.P1V1 = P2V2

Step 2. Write out the formula.

P1V1 = P2V2

Step 3. Plug in.

9.66ml x 64.4kPa = 94.6kPa x V2

Step 1. List the givens.

V1 = V2=

P1= P2=

9.66 ml

64.4 kPa 94.6 kPa

?

622.104ml x kPa = 94.6kPa x V2

94.6kPa 94.6kPa

6.576ml = V2

CHARLES’S LAW

Temperature must always be in KELVINS!

Temperature must always be in KELVINS!

How do we solve problems relating to gases?II. CHARLES’S LAW

Example 1The volume of a sample of a gas at 0 C is 200.0 L. If the

volume is decreased to 100.0 L at constant pressure, what will be the new temperature of the gas?

Step 1. List the givens.

V1 = V2=

T1= T2=

200 L

0oC = 273K ?

100 L

Step 2. Write out the formula.V1 = V2

T1 T2

Step 4. Plug in.

200 L = 100 L

273K T2

0.7326 = 100L

T2

136.5K = T2

Step 1. List the givens.V1 = V2=

T1= T2=

200 L

0oC ?

100 L

Step 3. Check the units for temperature. (Convert to Kelvins) 0oC = 273K

Step 2. Write out the formula. V1 = V2

T1 T2

INDEPENDENT PRACTICE

• Complete the Independent Practice • PLEASE COMPLETE THE WORKSHEET FOR

HOMEWORK