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Boyle's law and charle's law  (Page 6/6)

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Volume (l) He H 2 N 2 O
0 -272.4 -271.8 -275.0
0.25 -245.5 -192.4 -123.5
0.5 -218.6 -113.1 28.1
0.75 -191.8 -33.7 179.6
1.0 -164.9 45.7 331.1
1.5 -111.1 204.4 634.1
2 -57.4 363.1 937.2
2.5 -3.6 521.8 1240.2
3.0 50.2 680.6 1543.2
3.5 103.9 839.3 1846.2
  1. On the same set of axes, draw graphs to show the relationship between temperature and volume for each of the gases.
  2. Describe the relationship you observe.
  3. If you extrapolate the graphs (in other words, extend the graph line even though you may not have the exact data points), at what temperature do they intersect?
  4. What is significant about this temperature?

In the simulation you can verify Charle's law by holding the pressure constant and seeing what happens as you vary the volume and temperature.

Ammonium chloride and calcium hydroxide are allowed to react. The ammonia that is released in the reaction is collected in a gas syringe and sealed in. This gas is allowed to come to room temperature which is 32 ∘ C. The volume of the ammonia is found to be 122 ml. It is now placed in a water bath set at 7 ∘ C. What will be the volume reading after the syringe has been left in the bath for a some time (e.g. 1 hour) (assume the plunger moves completely freely)?

  1. V 1 = 122 ml and V 2 = ?

    T 1 = 32 ∘ C and T 2 = 7 ∘ C

  2. Here, temperature must be converted into Kelvin, therefore:

    T 1 = 32 + 273 = 305 K

    T 2 = 7 + 273 = 280 K

  3. V 1 T 1 = V 2 T 2

    Therefore,

    V 2 = V 1 × T 2 T 1
  4. V 2 = 122 × 280 305 = 112 m l
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Note that here the temperature must be converted to Kelvin (SI) since the change from degrees Celcius involves addition, not multiplication by a fixed conversion ratio (as is the case with pressure and volume.)

At a temperature of 298 K, a certain amount of CO 2 gas occupies a volume of 6 l. What volume will the gas occupy if its temperature is reduced to 273 K?

  1. V 1 = 6 l and V 2 = ?

    T 1 = 298 K and T 2 = 273 K

  2. Temperature data is already in Kelvin, and so no conversions are necessary.

  3. V 1 T 1 = V 2 T 2

    Therefore,

    V 2 = V 1 × T 2 T 1
  4. V 2 = 6 × 273 298 = 5 . 5 l
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The relationship between temperature and pressure

The pressure of a gas is directly proportional to its temperature, if the volume is kept constant ( [link] ). When the temperature of a gas increases, so does the energy of the particles. This causes them to move more rapidly and to collide with each other and with the side of the container more often. Since pressure is a measure of these collisions, the pressure of the gas increases with an increase in temperature. The pressure of the gas will decrease if its temperature decreases.

The relationship between the temperature and pressure of a gas

In the same way that we have done for the other gas laws, we can describe the relationship between temperature and pressure using symbols, as follows:

T ∝ p, therefore p = kT

We can also say that:

p T = k

and that, provided the amount of gas stays the same...

p 1 T 1 = p 2 T 2

More gas laws

  1. A gas of unknown volume has a temperature of 14 ∘ C. When the temperature of the gas is increased to 100 ∘ C, the volume is found to be 5.5 L. What was the initial volume of the gas?
  2. A gas has an initial volume of 2600 mL and a temperature of 350 K.
    1. If the volume is reduced to 1500 mL, what will the temperature of the gas be in Kelvin?
    2. Has the temperature increased or decreased ?
    3. Explain this change, using the kinetic theory of matter.
  3. A cylinder of propane gas at a temperature of 20 ∘ C exerts a pressure of 8 atm. When a cylinder has been placed in sunlight, its temperature increases to 25 ∘ C. What is the pressure of the gas inside the cylinder at this temperature?
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Source:  OpenStax, Siyavula textbooks: grade 11 physical science. OpenStax CNX. Jul 29, 2011 Download for free at http://cnx.org/content/col11241/1.2
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