Pressure Law

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Pressure Law
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Introduction.

• The apparatus opposite has been used to
  investigate the relationship between temperature
  and pressure.

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What Did You Do?

• A volume of air is trapped in a flask, which is
  immersed in water. The water is heated,
  increasing the air temperature.
• The water/air temperature is measured using a
  thermometer. The air pressure is measured by the
  Bourdon Gauge joined by a short length of rubber
  tubing.
• As the water (and hence air) is heated, the
  pressure and temperature are noted.

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Your Results (Graph 1)
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Graph Analysis (Graph 1)

• Note the following points
• The graph is linear. There seems to be a
  relationship between temperature and pressure.
• The graph does not pass through the origin.
• i.e. Pressure is not proportional to the Celsius
  Temperature.
• What now?
• Extrapolate (extend) the graph of Pressure v
  Temperature until it meets the x-axis.

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Your Results (Graph 2)
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Graph Analysis (Graph 2)

• The extended line meets the x-axis at a
  temperature of -2730C.
• At this temperature, the pressure is zero.
• What now?
• We now create a new temperature scale, named
  after Lord Kelvin, On this new scale, we decide
  that -2730C is equal to zero Kelvin, and the size
  of one Kelvin is the same as a Celsius degree.

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Pressure v Temperature (Kelvin)
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Graph Analysis

• The graph of Pressure versus the Kelvin
  Temperature is a straight line passing through
  the origin.
• The Pressure is therefore directly proportional
  to the Temperature in Kelvin
• The lowest temperature possible is zero Kelvin.
  This is called Absolute Zero.

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Some Maths

• P ? T (when T in Kelvin)
• P constant x T
• P/T constant

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Conclusion

• Pressure Law
• the pressure of the gas is directly proportional
  to its temperature in Kelvin. (For a fixed mass
  of gas at constant volume.)

P1 P2 T1 T2
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Example

• A fixed mass of air at atmospheric pressure is
  trapped inside a glass flask at a room
  temperature of 20 oC.
• The flask is gently heated until the air
  temperature is 80 oC.
• Calculate the pressure of the air inside the
  flask at this temperature.

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Solution

• P1 1.01 x 105 Pa
• T1 20 oC 293 K
• P2?
• T2 80 oC 353 K

• P1 P2
• T1 T2

• 1.01 x 105 P2 293 353
• P2 1.22 x 105 Pa