boyle and charles
TRANSCRIPT
Gas laws: Prepared By, JV Temporal
PRESSURE AND VOLUME RELATIONSHIPAT CONSTANT TEMPERATURE
Boyle’s Law1
Gas laws: Prepared By, JV Temporal
Gas particles have a very weak intermolecular force of attraction, hence they move as far as possible from each other. They have the tendency to occupy all the spaces they are contained in. If the pressure is increased, the volume will be decreased forcing the gas particles to move closer to one another.
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V α 1/PThe volume of a gas is inversely proportional to its pressure, if temperature and amount of gas are held constant.
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This proportionality was first stated by Robert Boyle during the 6th century. He performed an experiment wherein he trapped fixed amount of air in the J-tube, he changed the pressure and controlled the temperature and then, he observed its effect to the volume of the air inside the J-tube.
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V α 1/PV= volume; P= pressure
This means that as pressure decreases the volume increases.
As pressure increases the volume decreases.
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V α 1/P
To remove the proportionality symbol (α) we have to introduce a constant, thus, leading us to the equation to the right.
V = k/Pk=VPThe product of volume and pressure is constant.
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According to Boyle’s law VP=kIf we subject the gas to a change in pressure we’ll
have:
V₁P₁ =k
V2P2 =kTherefore,
V₁P₁ =V₂P₂
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V₁P₁ =V₂P₂
1. V₁ = V₂P₂ / P₁ 2. P₁ = V₂P₂ / V₁3. V₂ = V₁P₁ / P₂4. P₂ = V₁P₁ / V₂
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Practical example of this is the Lungs as we exhale and inhale.
Will you please cite another practical example of Boyle’s Law.
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Try this!!!
The inflated balloon that slipped from the hand of Roshan has a volume of 0.50 L at sea level (1.0 atm) and it reached a height of approximately 8km where the atmospheric pressure is approximately 0.33 atm. Assuming that the temperature is constant, compute for the final volume of the balloon.
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Let us analyze the problem!
The inflated balloon that slipped from the hand of Roshan has a volume of 0.50 L at sea level (1.0 atm) and it reached a height of approximately 8km where the atmospheric pressure is approximately 0.33 atm. Assuming that the temperature is constant, compute for the final volume of the balloon.
Given:
V₁ = 0.50 LP₁ = 1 atmV2 = ?
P2 = 0.33 atm
What equation should you use?
Equation #3. Let us try this on the
board.
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Analyze your solution
Notice that the pressure as it goes up decreases, and we know that as gas receives lower pressure it will have an increase in volume.
Did you obtain a higher V2 than V₁ ?Now, try to multiply V₁ and P₁ then, Multiply V2 and P2.
What have you observed with the product of the two?
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Try this on your own!
1. Oxygen gas inside a 1.5 L gas tank has a pressure of 0.95 atm. Provided that the temperature remains constant, how much pressure is needed to reduce its volume to o.75 L.?
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2. Compute for the initial volume of a gas in a tank with initial pressure of 1.5 atm if its final volume is 5L and its pressure is decreased to 1.0 atm.
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Assignment: Answer this on your notebook
1. A scuba diver needs a diving tank in order to provide breathing gas while he is underwater. How much pressure is needed for 6.oo L of gas at 1.01 atm to be compressed in a 3.00 L cylinder.
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TEMPERATURE AND VOLUME RELATIONSHIP
Charles’ Law
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From the properties of gases we have mentioned that the gas particles will expand with the temperature. Meaning, as temperature increases the volume of gas will also increase.
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Vα TThe volume of the gas is
directly proportional to the temperature at constant
pressure.
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This relationship was determined by and named after Jacques Charles. In his experiment, Charles trapped a sample of gas in a cylinder with a movable piston in water bath at different temperatures.
It is important to note that the temperature should always be in Kelvin (k).
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Vα T
V=volume; T=temperature (k)
As the temperature increases the volume of the gas also increases.
As the temperature decreases the volume of the gas also decreases.
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Vα TTo remove the proportionality symbol (α) we have to introduce a constant, refer to the equation to the right.
V=kTK=V/T
The quotient of volume and temperature is constant
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According to the Charles’ law;V/T = k
Then, if we subject a gas to a temperature difference, we’ll get:
V₁/T₁ = kV₂/T₂ = k
Thus,V₁/T₁ = V₂/T₂
k here means constant and not kelvin.
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V₁/T₁ = V₂/T₂ or V₁T₂= V₂T₁
1.V₁ = V₂T₁/T₂2.T₁ = V₁T₂/ V₂ 3.V₂ = V₁T₂/ T₁ 4.T₂ = V₂T₁/ V₁
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Practical example of this is the Hot Air Balloon (HAB) and Sky Lanterns (SL). The flame at the bottom part of the HAB and SL increases the volume thus decreasing the density, and we know that objects with lower density tends to float.
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Why should you not pump large amount of gas inside the tire during summer?
Explain why should aerosol cans must not be placed near a fire?
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Try this!
An inflated balloon with a volume of 0.75 L at 30 ⁰C was placed inside the freezer where the temperature is -10 ⁰C. find out what will happen to the volume of the balloon if the pressure remains constant. Support your answer with computation.
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Analysis of the problem.
An inflated balloon with a volume of 0.75 L at 30 ⁰C was placed inside the freezer where the temperature is -10 ⁰C. find out what will happen to the volume of the balloon if the pressure remains constant. Support your answer with computation.
Given:T₁ = 30 ⁰CV₁ = 0.75 LT₂ = -10 ⁰CV₂ = ? What equation should we
use?Equation #3
Let us try this on the board.
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Let’s analyze it more
Notice that the problem stated that the balloon was placed in a freezer meaning there is a lower temperature in there. And we have mentioned earlier that if the temperature decreases the volume of the gas will also decrease.
Did you obtain Lower V2 than V1?Now, try to divide V1 by T1 then divide V2
by T2.What can you infer with the quotient?
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Try this!!
1. A cylinder with a movable piston contains 250 cm³ air at 10⁰C. if the pressure is kept constant, at what temperature would you expect the volume to be 150 cm³?
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2. A tank (not rigid) contains 2.3 L of helium gas at 25 ⁰C. What will be the volume of the tank after heating it and its content to 40 ⁰C temperature at constant pressure.
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Assignment: answer this in your notebook
1. At 20 ⁰C, the volume of chlorine gas is 0.015L. Compute for the resulting volume if the temperature is adjusted to 318 k provided that the pressure remains the same.
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Reminders!
On your notebook do your EAR!EAR #2 for the Boyle’s Law.EAR #3 for Charles’ Law.
I will check and sign your notebook during your quiz #1 on Friday. Coverage, Properties of gases, Boyle’s Law and Charles’ Law.
-Sir JV.