From the experimental results described on the graph p(t) it can be deduced that t = 2730C then p
= 0. Therefore, when t = 2730C is the lowest temperature. Kenvil proposed a temperature scale in which: Choose 2730C as absolute zero corresponding to 00C
A temperature difference of 1K is equal to a difference of 10C.
If you know the temperature in the Kelvin scale (T)
Can you find the temperature in Celsius scale?
0
Kelvin temperature scale, so T = t + 273
Inference: t = 2730C corresponding to 00K combined with the characteristics of
Temperature measured on the Kelvin scale is called absolute temperature.
T = t + 273
INTERPRETATION
From the expression of the law of Charley, we see that when t0 = 2730C, the pressure is 0. This does not exist in reality because the molecules are always in constant motion. Therefore, if he wanted the heat to start from 0 degrees, Kenvil thought of creating a new temperature scale named after him. The Kenvil temperature scale, symbolized by degrees K. In the new temperature scale, Kenvil chose 2730C as the absolute zero corresponding to 00C and the distance of 1 degree K is equal to the distance of 1 degree C.
So the question is: If we know the temperature in the Kelvin scale (T), can we find the temperature in the Celsius scale? We deduce: 0K corresponds to -2730C and the distance of 1 degree K is equal to the distance of 10C. Therefore, each degree K compared to each degree C has a difference of 273. Therefore, T = t + 273.
opposite to.
And temperature
measured in this new scale we call temperature
great
3.2.4 Specific teaching process:
3.2.4.1 Problem 1: Charles's Law
3.2.4.1.1 Action goal orientation:
Δ: For a given amount of gas at constant temperature, what is the relationship between pressure and volume?
HS: p and V are inversely proportional to each other according to the Boyle-Marriot law.
Ο: Today's lesson we continue to study the relationship between pressure and temperature.
pressure and temperature of gas in solid state
Charles's law. Absolute temperature"
constant product. Lesson 46: “
Δ: For a given amount of gas with a fixed mass and constant volume, how does the pressure depend on temperature?
3.2.4.1.2 Task solving orientation
a. Determine the solution:
Δ: To
pay
How to answer this question? Reason or experiment
test?
HS: Inference or experiment.
Δ: How do we reason?
HS: According to the kinetic theory of gases, when the temperature increases
then move
movement of molecules, atoms move faster, number of molecules
The number of atoms per unit area hitting the wall of the container is faster,
more. Leading to increased pressure.
Δ: Can we determine the relationship both qualitatively and quantitatively using only inference?
HS: We have to do experiments to determine the quantitative relationship between p and t.
Ο: In other words, we must determine the mathematical expression describing the relationship between p and t of that amount of gas when V is constant by experiment.
b. Implement the solution:
Δ: What tools are needed to do the experiment?
HS:
A sealed container is needed to contain the gas so that it does not escape.
The vessel is non-expandable to ensure that the volume does not change while the temperature increases.
Pressure gauge, temperature thermometer
To change the temperature, heat directly over fire, or put in a hot water bottle.
(If students cannot answer, the teacher can suggest with the following questions:
What equipment is needed to contain gas? What conditions must the equipment have to ensure that m and V remain constant?
What tools do you use to measure p and t?
How to change the temperature of the gas?) The teacher put the equipment on the table for students to observe.
1
g
(Improved Mattress Experimental Instrument Simulation Diagram):
1.75
ATM
6
7
1.5
1.25
54o C
40o C
27o C
Ο: In addition to this experimental solution, the textbook also provides another solution. The difference is that the textbook uses a water manometer to measure Δp according to Δt. Meanwhile, we use a metal manometer that can measure the value
of pressure p
corresponding to the temperature
t tool
can. In addition, according to the experiment
According to the textbook experiment, to keep the gas volume constant, we must move the liquid tube connecting the two communicating vessels so that the water level in the left tube always remains at its original position.
Δ: How will we proceed with these tools?
HS: Heat the water in the tank, after equal periods of time, read the corresponding values of p and t.
Ο: The teacher coordinates with students to conduct the experiment, asking students to read and write the pressure and temperature values on the board.
HS: Monitor, conduct experiments, and record results in the table.
t (0C)
p (mmHg) |
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Δ: Based on the results table, can we draw any conclusions?
HS: We see that as t increases, p also increases.
Δ: Is p proportional to t? (Does p increase by the same number of times as t increases?)
HS: No
Ο: To find the expression relating pressure (p) and temperature (t0C) in this case we use the method of representing the results
experiment on the system
Coordinates
p – t. Then based on the diagram
market
to find the expression
mathematical formula relating p and t.
Δ: Based on the experimental results, draw a p-t graph and give your comments.
HS: Proceed to draw the p-t graph. The p-t graph looks like a straight line.
Δ: So what rule can be drawn about p – t?
HS: p, t follow the law of first-degree functions in mathematics. The expression is: p = at + b (1). In which a is the slope of the graph, b is the gas pressure at t = 00C, b = p0.
Ο: Let t0 be the temperature of the gas when the pressure is 0. Here, finding the p-t rule is to determine the values of a and b according to p0, t0. In which, determine the value of t0.
Δ: What should we do?
HS: Think and do:
Method 1: From the graph, it can be seen that t0 has an approximate magnitude of 2730. We have:
a p0 , b = p . Substitute into (1) to deduce p
p0 tp
p 1 1 t
t0 0
273
0 0 273
Method 2: Mathematical transformation to determine p0 and t0.
a p t
p2 p1
t2 t1
p2 p0 calculate p, a
0
t
2
but a
p0 should replace p
t
0
0
and a calculates t0
2730C in magnitude. b = p0
Substitute a, b into (1) to get p's law
p0 tp
273 0
p0 1
1 t
273
3.2.4.1.3 Summary and consolidation of results:
Ο: We can calculate t0 = 2730C, Charcoal
conducted many experiments
experiment shows value
this is like
with all gases. And because
So people sign:
1 is the constant pressure gain coefficient.
273
Δ: So please state Charles's law.
3.2.4.2 Problem 2: Absolute temperature
3.2.4.2.1 Action goal orientation:
Δ: From Chargö's law, what is the lowest temperature of a gas? Does that temperature exist in reality?
HS: t = 2730C. In reality, this temperature cannot exist because if it exists, it means the gas pressure is 0. But molecules and atoms are always in chaotic, non-stop motion.
Ο: As
So in theory the temperature
lowest of gas is
– 2730C. Scientist Ken vil found that the lowest temperature should start from 0. So he proposed a temperature scale
new name. Kelvin temperature scale, symbol is degree K. In
New temperature scale, Kenvil chose 0K corresponding to 2730C as absolute zero and the distance of 1 degree K is equal to the distance of 1 degree C.
Δ: If you know the temperature in the Kelvin scale (T), can you find the temperature in the Celsius scale?
3.2.4.2.2 Task solving orientation
Identify solutions and implement solutions:
HS: Yes, by deduction. 00K corresponds to -2730C and the distance of each degree K is equal to the distance of each degree C. Therefore T = t + 273.
opposite to.
Ο: And the temperature measured in this new scale is called the temperature
great
3.2.4.2.3 Generalization and consolidation of results:
Δ: Please define absolute temperature.
Δ: Apply the formula for absolute temperature to represent Charg's law on the Kelvin scale.
HS: Substitute t = T – 273 into the expression:
p = p0 (1 +
γ t) therefore
p p0 1
T 273 273
p 0T
pH /s T
Ο: This is another form of Charg’s law. In this way of writing, we can clearly see that the pressure p and the absolute temperature T are directly proportional to each other. For real gases (not ideal gases), this law and the Boyle-Mariotte law are only approximate.
(The teacher gives out test sheets and asks students to do practice exercises)
Ο: Teacher summarizes the lesson.
3.3 Organizing pedagogical experiments.
3.3.1 Purpose of pedagogical experiment:
Based on the teaching process drafted in Chapter II, I organized a pedagogical experiment to test the scientific hypothesis of the topic and from there supplement and perfect the teaching process to achieve the correct objectives of the topic.
3.3.2 Tasks of pedagogical experiment
Organize teaching according to the prepared process, record teaching hours
Video analysis, phase separation shows teaching processes
Comments, evaluation of experimental advantages and disadvantages
Complete supplementary teaching process
3.3.3 Time and subjects of pedagogical experiment
Teaching time: March 29, 2007, following the curriculum of high school students.
Subject: 10th grade students of TN1, Yen Hoa High School. Subject is
good students of the school, good ability
3.3.4 Evaluation of pedagogical experiment results
3.3.4.1 Disadvantages:
energy
knowledge
Regarding the organization of practical teaching activities in class:
Due to limited time, the experiment was not prepared in time. Therefore, the teacher used a simulation experiment on Power point software.
Number work
experimental material used to
teacher-given survey
based on theoretical calculations so there is no calculation
convincing about the experimental results.
The lesson did not proceed as planned in some places:
Do not organize for students to draw graphs themselves, teachers do it for them.
Since the worksheets have not been prepared yet, in the consolidation phase, students participate in doing exercises directly on the board.
3.3.4.2 Advantages
Besides the above shortcomings, the lecture also achieved
Some advantages:
The lesson has initially implemented the stages of the process of building a physical law through experimentation.
Most of the broadcast
you all pay
Answer the questions prepared
export conditions