Charles Law
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UNIVERSITI PENDIDIKAN SULTAN IDRIS
TKU 3023 Information And Communication Technology In Chemistry Assingment : Spreadsheet (Charles’s Law) Lecturer name : Madam Asmayati bt Yahaya Group : B (1.00pm-4.00pm)Thursday NAME
MATRIC NO.
NOR ASILAH BINTI ABD RAFAR
D20081032308
RAFFINE BINTI BAHARUDIN
D20081032309
CHARLES LAW Teacher’s Guide/ Physics Form 4/ Spreadsheets
SCIENTIFIC CONCEPT The relationship between gas volume and temperature was discovered in 1787 by the French scientist Jacques Charles (1746- 1823). Charles found that the volume of a fixed quantity of gas at constant pressure increase linearly with temperature. The Charles law can be stated as follows: The volume of a fixed amount of gas maintained at constant pressure is directly proportional to its absolute temperature (- 273.15 °C). Thus doubling the absolute temperature causes the gas volume to double. Mathematically, Charles law takes the following form:
V = constant × T
or
VT=constant
Temperature (K)
Volume (L)
373
2.98
353
2.87
333
2.68
313
2.53
293
2.33
273
2.21
-293
2.07
- 313
1.88
UNIQUE FEATURE OF ACTIVITY •
The data can be manipulated easily, for example if the volume of gas is change with increasing the temperature, the 1/V can be calculated.
•
Student will discover that a spreadsheet can be used to proess data.
•
Data displayed in a systematic manner, save time in drawing graphs, show relationship of different temperature of reactant against time.
•
Allow the student to do more important things such as analyzing the data or the graph
•
Generate student soft skill.
•
Students will more appreciate to the developer that is Jacques Charles which formulate formula V1/T1=V2/T2.
ENGAGE You are given two different pictures. Observe the pictures carefully.
Picture A
Picture B
Both of the containers in Picture A and Picture B was placed with cold water. Two simillar size balloons are placed as showed in the picture. Balloon in the Picture A is placed outside the container while balloon in the Picture B is placed in the container. Why does the two balloons differ in their size?
EMPOWER 1. Students plan an experiment that related to Charles’s Law. 2. Students are given •
Syringe with cap at one end
•
piston to fit syringe
•
wood block to fit syringe and support masses
•
base block to support the apparatus
•
-40oC to 100oC thermometer
•
Water bath
•
petroleum jelly
•
glass stopper
1. Student have to construct a hypothesis for this experiment.
2. Student should know the effect of volume to the temperature. 3. Helps students by given the figure of the apparatus that must be set up.
4. You may give these instruction to start the activity. •
Open a spreadsheet file.
•
Name the file as volume versus temperature.
•
Measure the diameter of the syringe and record it in the provided space.
•
Insert the thermometer in the glass stopper, and place them in the side opening of the syringe.
•
Place this syringe on the base block with the cap facing down and position the wood block as shown.
•
Place an appropriate mass as shown and record it in the provided space.
•
Adjust the height of the piston so that it is just far enough to support the wood block and the mass by removing the cap temporarily if necessary and/or choosing another mass.
•
This apparatus is immersed in water bath.
•
In the first row of the table below record the air temperature and the volume of trapped air in the syringe.
How to set up the spreadsheet 1. In cell A1 enter the title for the spread sheet volume versus temperature and hit enter. 2. In cell A2 enter the data label temperature (k) and tab to cell B2. Enter the data label Volume (L) in cell B2 and hit enter. This should bring the cursor to cell A3. Enter the data given below for temperature in this cell and tab to cell B3 and enter the corresponding volume. Hit enter and continue entering the rest of the data for volume and temperature. Temperature
Volume (L)
(K) 373
2.98
353
2.87
333
2.68
313
2.53
293
2.33
273
2.21
253
2.07
233
1.88
How to draw a graph with the spreadsheet 1. Highlight the data columns you would like to graph by either putting the cursor in cell A2, then hold down the shift key and click the cursor in the last data cell of column B or by putting the cursor in cell A2 and holding down the left mouse button and dragging to the last data cell in column B. 2. On the menu tool bar click on the "Chart Wizard Icon" a dialogue box will open with two sheet tabs. Select the Standard Types tab. Under Chart Type select XY (Scatter).
3. Select one of the five Chart sub-type, then click the “Press and Hold to View Sample” bar to see what you graph would look like. Select one of the four line graphs that you like and click next. 4. Name the chart 5. Label the x-axis and y-axis
Question 1. Does the volume of a fixed quantity of gas decrease to half its original value when the
temperature is lowered from 1000C to 500C? 2. A sample of gas at 101.3kPa had a volume of 1.2L at 100 oC. What would its volume
be at 0oC at the same pressure? 3. A balloon had a volume of 75L at 25oC. To what does the temperature need to raised
in order for the balloon to have a volume of 100L at the same pressure? Answer: 1. The volume decreases, but it does not decrease to half because the volume is proportional to the temperature on the Kelvin scale but not on the Celcius scale. 2. Vi = 1.2L
Vf = ?
Ti = 100oC = 100 + 273 = 373K
Tf = 0oC = 0 + 273 =273K
1.2/373 = Vf/273 3.22 x 10-3 = Vf/273 Vf = 3.22 x 10-3 x 273 = 0.88L (880mL) 3. Vi = 75L o
Ti = 25 C = 25 + 273 = 298K V i Ti
Vf = 100L Tf = ? (K)
V f Tf
/ = /
75/298 = 100/Tf 0.2517 = 100/Tf Tf = 100/0.2517 = 397K (397-273 = 124oC)
ENHANCE
1. Can you name the object in the above picture? 2. How can the object float and getting down?
Answer: 1. Hot air ballon 2. The balloon is filled with hot air. Once the air in the balloon gets hot enough, the net weight of the balloon plus this hot air is less than the weight of an equivalent volume of cold air and the balloon starts to rise. Helium filled balloons float by the same principle because helium gas is much less dense than air. The hot air filling the balloon reduces its density according to the ideal gas law. When its total density is less than the density of air, it will floats. Getting down can be done in two ways. First, we can either cool the air inside the balloon back down so that more air comes in or we can let more cool air in by opening a small hole in the balloon which lets some hot air escape. It takes a long time for the air in the balloon to cool down all the way, so we usually let air back in. Another way is hot air balloon have a special valve that lets some air come into the balloon without all too much coming in and causing the balloon to fall fast.
TKU 3023 Information And Communication Technology In Chemistry Assingment : Spreadsheet (Charles’s Law) Lecturer name : Madam Asmayati bt Yahaya Group : B (1.00pm-4.00pm)Thursday NAME
MATRIC NO.
NOR ASILAH BINTI ABD RAFAR
D20081032308
RAFFINE BINTI BAHARUDIN
D20081032309
CHARLES LAW Teacher’s Guide/ Physics Form 4/ Spreadsheets
SCIENTIFIC CONCEPT The relationship between gas volume and temperature was discovered in 1787 by the French scientist Jacques Charles (1746- 1823). Charles found that the volume of a fixed quantity of gas at constant pressure increase linearly with temperature. The Charles law can be stated as follows: The volume of a fixed amount of gas maintained at constant pressure is directly proportional to its absolute temperature (- 273.15 °C). Thus doubling the absolute temperature causes the gas volume to double. Mathematically, Charles law takes the following form:
V = constant × T
or
VT=constant
Temperature (K)
Volume (L)
373
2.98
353
2.87
333
2.68
313
2.53
293
2.33
273
2.21
-293
2.07
- 313
1.88
UNIQUE FEATURE OF ACTIVITY •
The data can be manipulated easily, for example if the volume of gas is change with increasing the temperature, the 1/V can be calculated.
•
Student will discover that a spreadsheet can be used to proess data.
•
Data displayed in a systematic manner, save time in drawing graphs, show relationship of different temperature of reactant against time.
•
Allow the student to do more important things such as analyzing the data or the graph
•
Generate student soft skill.
•
Students will more appreciate to the developer that is Jacques Charles which formulate formula V1/T1=V2/T2.
ENGAGE You are given two different pictures. Observe the pictures carefully.
Picture A
Picture B
Both of the containers in Picture A and Picture B was placed with cold water. Two simillar size balloons are placed as showed in the picture. Balloon in the Picture A is placed outside the container while balloon in the Picture B is placed in the container. Why does the two balloons differ in their size?
EMPOWER 1. Students plan an experiment that related to Charles’s Law. 2. Students are given •
Syringe with cap at one end
•
piston to fit syringe
•
wood block to fit syringe and support masses
•
base block to support the apparatus
•
-40oC to 100oC thermometer
•
Water bath
•
petroleum jelly
•
glass stopper
1. Student have to construct a hypothesis for this experiment.
2. Student should know the effect of volume to the temperature. 3. Helps students by given the figure of the apparatus that must be set up.
4. You may give these instruction to start the activity. •
Open a spreadsheet file.
•
Name the file as volume versus temperature.
•
Measure the diameter of the syringe and record it in the provided space.
•
Insert the thermometer in the glass stopper, and place them in the side opening of the syringe.
•
Place this syringe on the base block with the cap facing down and position the wood block as shown.
•
Place an appropriate mass as shown and record it in the provided space.
•
Adjust the height of the piston so that it is just far enough to support the wood block and the mass by removing the cap temporarily if necessary and/or choosing another mass.
•
This apparatus is immersed in water bath.
•
In the first row of the table below record the air temperature and the volume of trapped air in the syringe.
How to set up the spreadsheet 1. In cell A1 enter the title for the spread sheet volume versus temperature and hit enter. 2. In cell A2 enter the data label temperature (k) and tab to cell B2. Enter the data label Volume (L) in cell B2 and hit enter. This should bring the cursor to cell A3. Enter the data given below for temperature in this cell and tab to cell B3 and enter the corresponding volume. Hit enter and continue entering the rest of the data for volume and temperature. Temperature
Volume (L)
(K) 373
2.98
353
2.87
333
2.68
313
2.53
293
2.33
273
2.21
253
2.07
233
1.88
How to draw a graph with the spreadsheet 1. Highlight the data columns you would like to graph by either putting the cursor in cell A2, then hold down the shift key and click the cursor in the last data cell of column B or by putting the cursor in cell A2 and holding down the left mouse button and dragging to the last data cell in column B. 2. On the menu tool bar click on the "Chart Wizard Icon" a dialogue box will open with two sheet tabs. Select the Standard Types tab. Under Chart Type select XY (Scatter).
3. Select one of the five Chart sub-type, then click the “Press and Hold to View Sample” bar to see what you graph would look like. Select one of the four line graphs that you like and click next. 4. Name the chart 5. Label the x-axis and y-axis
Question 1. Does the volume of a fixed quantity of gas decrease to half its original value when the
temperature is lowered from 1000C to 500C? 2. A sample of gas at 101.3kPa had a volume of 1.2L at 100 oC. What would its volume
be at 0oC at the same pressure? 3. A balloon had a volume of 75L at 25oC. To what does the temperature need to raised
in order for the balloon to have a volume of 100L at the same pressure? Answer: 1. The volume decreases, but it does not decrease to half because the volume is proportional to the temperature on the Kelvin scale but not on the Celcius scale. 2. Vi = 1.2L
Vf = ?
Ti = 100oC = 100 + 273 = 373K
Tf = 0oC = 0 + 273 =273K
1.2/373 = Vf/273 3.22 x 10-3 = Vf/273 Vf = 3.22 x 10-3 x 273 = 0.88L (880mL) 3. Vi = 75L o
Ti = 25 C = 25 + 273 = 298K V i Ti
Vf = 100L Tf = ? (K)
V f Tf
/ = /
75/298 = 100/Tf 0.2517 = 100/Tf Tf = 100/0.2517 = 397K (397-273 = 124oC)
ENHANCE
1. Can you name the object in the above picture? 2. How can the object float and getting down?
Answer: 1. Hot air ballon 2. The balloon is filled with hot air. Once the air in the balloon gets hot enough, the net weight of the balloon plus this hot air is less than the weight of an equivalent volume of cold air and the balloon starts to rise. Helium filled balloons float by the same principle because helium gas is much less dense than air. The hot air filling the balloon reduces its density according to the ideal gas law. When its total density is less than the density of air, it will floats. Getting down can be done in two ways. First, we can either cool the air inside the balloon back down so that more air comes in or we can let more cool air in by opening a small hole in the balloon which lets some hot air escape. It takes a long time for the air in the balloon to cool down all the way, so we usually let air back in. Another way is hot air balloon have a special valve that lets some air come into the balloon without all too much coming in and causing the balloon to fall fast.
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