Charles and Gay-Lussac laws Chemistry p.106 Relationship between - - PowerPoint PPT Presentation

Charles’ and Gay-Lussac laws

Chemistry p.106

Relationship between Temperature and Volume (Charles’ Law)

2 2 1 1

T V T V k T V T k V T V     

The volume of an enclosed gas is directly proportional to its kelvin temperature, provided the pressure is kept constant.

Typical graph of V versus T

V (cm3) T (in K)

Relationship between Temperature and Pressure (Gay-Lussac)

2 2 1 1

T p T p k T p T k p T p     

Typical graph of p versus T

p

kPa

T(in K)

Kelvin - Celsius

According to the graph the zero point of temperature should be where the gas exerts no pressure, which is

• impossible. A new Kelvin scale was developed for

scientific purposes.

273 K = 0 0C T (Kelvin) = t (Celsius) + 273

Standard Temperature & Pressure

T = 273K p = 101,3 kPa

(p = 1,013 x 105 Pa)

Example

The tyre of a car is pumped to a pressure of 180 kPa at a temperature of 17 0C. Due to the friction with the surface of the road the temperature rises to 37 0C. Assume that the volume of the tyre remains constant and then calculate the pressure of the air in the tyre.

kPa p p T p T p 192 310 290 180

2 2 2 2 1 1

  

=

Use the formula like this:

Homework Exercise 3

• p. 111, no. 2, 4.1, 4.3, 5.1

Exercise 4

• p. 119, no. 6, 9

2.1 Draw a straight line graph of volume versus temperature. Choose an appropriate scale for the axes given

• below. Stipulate the points. Supply

your graph with an appropriate heading.

Graph of V versus T

20 40 60 80 100 V (cm3) T (oC) 60 120

• 60
• 180 -120
• 240

2.2 Extrapolate (extend) your graph and determine at which temperature the graph will intercept the temperature axis.

Graph of V versus T

20 40 60 80 100 V (cm3) T (oC) 60 120

• 60
• 180 -120
• 240
• 273 oC

2.3 Write down in words the relationship between the pressure and Kelvin temperature.

Graph of V versus T

20 40 60 80 100 V (cm3) T (oC) 60 120

• 60
• 180 -120
• 240

Pressure is directly proportional to the temperature in kelvin if the volume remains constant.

2.4 From your graph, determine the volume at 173 K. Show on the graph how you obtained this value

Graph of V versus T

20 40 60 80 100 V (cm3) T (oC) 60 120

• 60
• 180 -120
• 240

44 cm3

2.5 How will the gradient of the graph be affected if a larger mass of gas is used? Write down ONLY increase, decrease

• r stays the same.

Graph of V versus T

20 40 60 80 100 V (cm3) T (oC) 60 120

• 60
• 180 -120
• 240

Increase, if the pressure remains constant

4.1 A gas occupies 250 cm3 at 100 kPa and 40 oC. Calculate the volume which the gas will occupy at the following temperature, if the pressure is kept constant: 100 oC

V2 = cm3

4.3 A gas occupies 250 cm3 at 100 kPa and 40 oC. Calculate the volume which the gas will occupy at the following temperature, if the pressure is kept constant:

• 150 oC

V2 = cm3

5.1 A 32,3 dm3 sample of gas at 35 oC and 1,2 atmospheric pressure is cooled at a constant pressure until the volume is 28,4 dm3. What will the temperature of the gas be now?

T2 = K = - 2,19 oC

6.1 A tyre is filled with air until the pressure gauge reads 220 kPa, before it undertakes a long trip. This gauge shows the pressure in excess of atmospheric pressure (i.e. meter pressure = true pressure - atmospheric pressure). The temperature is then 25 oC. After being driven for a while, the tyre temperature is 70 oC. If it is assumed that the volume of the tyre remains unchanged, what will be the reading on the pressure gauge if it is connected to the tyre at the new temperature of 70 oC? (Hint: Use actual pressure in your calculations.)

p2 = kPa Initial pressure = 220 + 101,3 = 321,3 kPa Meter pressure = 369,8 - 101,3 = 268,5 kPa

6.2 A pressure cooker with a fixed volume contains only water vapour at a pressure of 100 kPa and a temperature of 100 oC. The lid of the pressure cooker is closed, so that no water vapour can

• escape. The temperature of the water vapour is increased to 120 oC.

Calculate the pressure in the pressure cooker at 120 oC.

p2 = kPa

• 9. A learner makes use of the apparatus set up in the diagram below to test the following hypothesis: "The

pressure of a gas is directly proportional to the temperature”. 9.1 What must the learner change to test the hypothesis and how can it be done?

Change the temperature and measure the pressure. Heat the water to increase the pressure.

• 9. A learner makes use of the apparatus set up in the diagram below to test the following hypothesis: "The

pressure of a gas is directly proportional to the temperature”. 9.2 Which variables must be kept constant and how is this done?

Volume, amount of gas in the flask and the type of gas. Completely seal the flask.

• 9. A learner makes use of the apparatus set up in the diagram below to test the following hypothesis: "The

pressure of a gas is directly proportional to the temperature”. 9.3 Which readings must be taken?

Temperature in oC, and pressure

• 9. A learner makes use of the apparatus set up in the diagram below to test the following hypothesis: "The

pressure of a gas is directly proportional to the temperature”. 9.4 Rewrite the hypothesis given in a more correct manner.

If the pressure of a gas increases while the volume is kept constant, the temperature will also increase.

• 9. The learner draws a graph AB of the results

and uses the graph to show how the relationship between the pressure and the temperature of the gas can be deduced. 9.5 What is point S on the temperature axis called?

Absolute zero

S X B A 620 p (kPa) T (oC) 100

• 100
• 300
• 200

• 9. The learner draws a graph AB of the results

and uses the graph to show how the relationship between the pressure and the temperature of the gas can be deduced. 9.6 What is the theoretical pressure of any gas at S?

0 kPa

S X B A 620 p (kPa) T (oC) 100

• 100
• 300
• 200

• 9. The learner draws a graph AB of the results

and uses the graph to show how the relationship between the pressure and the temperature of the gas can be deduced. 9.7 Which temperature in oC is represented by S for pressure versus temperature in Kelvin?

• 273 oC

S X B A 620 p (kPa) T (oC) 100

• 100
• 300
• 200

• 9. The learner draws a graph AB of the results

and uses the graph to show how the relationship between the pressure and the temperature of the gas can be deduced. 9.8 Calculate the value of X, the point of intercept on the pressure axis.

S X B A 620 p (kPa) T (oC) 100

• 100
• 300
• 200

p1 = 453,78 kPa