Combined Gas Law and Demonstrations

Aluminum Can Crush Demonstration

  • Initial state: Plain aluminum can with a small amount of water at the bottom.
  • Heating: The can is heated until the water boils.
  • Observation: Adding heat increases the kinetic energy of the water molecules, raising the temperature.
  • Rapid Cooling: The heated can is flipped into an ice bath, causing a sudden temperature change.
  • Result: The can is crushed rapidly due to the change in pressure.

Analysis of the Can Crush

  • Initial Conditions:
    • Air pressure (atmospheric pressure) is present.
    • Water is in the liquid phase at room temperature.
  • Heating Phase:
    • The can is open, allowing air (and thus air pressure) to move in and out.
    • Applying heat causes the water to boil, releasing water vapor (gas molecules) into the can and escaping as steam.
  • Sealing and Cooling:
    • Flipping the can into the ice bath seals the container, preventing steam from escaping.
    • The environment rapidly changes from hot to cold.
    • Water molecules rapidly lose kinetic energy.
  • Pressure Change:
    • The rapid cooling causes a pressure drop inside the can.
    • Initially, atmospheric pressure is balanced by the pressure inside the can (due to the escaping gas particles).
    • After sealing and cooling, the internal pressure decreases.
  • External Pressure:
    • Atmospheric pressure remains constant on the outside of the can.
  • Volume Reduction:
    • The external atmospheric pressure exceeds the internal pressure.
    • This pressure difference causes the can to be crushed, reducing its volume.
    • It's not that water is sucked in but rather external pressure forces the can to collapse.
  • Relationship: The demonstration illustrates the inverse relationship between pressure and volume.

Combined Gas Law

  • The combined gas law integrates all individual gas laws into a single equation.
  • Equation: P<em>1V</em>1n<em>1T</em>1=P<em>2V</em>2n<em>2T</em>2\frac{P<em>1V</em>1}{n<em>1T</em>1} = \frac{P<em>2V</em>2}{n<em>2T</em>2}
  • Variables:
    • PP = Pressure
    • VV = Volume
    • nn = number of moles
    • TT = Temperature
  • Individual Gas Laws:
    • Boyle's Law: P<em>1V</em>1=P<em>2V</em>2P<em>1V</em>1 = P<em>2V</em>2 (at constant nn and TT)
    • Charles's Law: V<em>1T</em>1=V<em>2T</em>2\frac{V<em>1}{T</em>1} = \frac{V<em>2}{T</em>2} (at constant nn and PP)
    • Gay-Lussac's Law: P<em>1T</em>1=P<em>2T</em>2\frac{P<em>1}{T</em>1} = \frac{P<em>2}{T</em>2} (at constant nn and VV)
    • Avogadro's Law: V<em>1n</em>1=V<em>2n</em>2\frac{V<em>1}{n</em>1} = \frac{V<em>2}{n</em>2} (at constant TT and PP)
  • Usage: By canceling out constant variables, the combined gas law can be simplified to solve various gas problems.

Example Problem 1

  • Problem: 3.00 L of a gas is collected at 35°C and 765 mmHg. What is the volume at STP?
  • Givens:
    • V1V_1 = 3.00 L
    • T1T_1 = 35°C = 308 K (35 + 273)
    • P1P_1 = 765 mmHg
    • STP (Standard Temperature and Pressure):
      • T2T_2 = 0°C = 273 K
      • P2P_2 = 1 atm = 760 mmHg
    • Unknown: V2</li></ul></li><li>Sincetheamountofgasisnotchanging,ncanberemovedfromthecombinedgaslaw.</li><li>SimplifiedEquation:V_2</li></ul></li> <li>Since the amount of gas is not changing, n can be removed from the combined gas law.</li> <li>Simplified Equation:\frac{P1V1}{T1} = \frac{P2V2}{T2}</li><li>Pluginthevalues:</li> <li>Plug in the values:\frac{(765)(3.00)}{308} = \frac{(760)V_2}{273}</li><li>Crossmultiplyandsolvefor</li> <li>Cross multiply and solve forV_2<ul><li><ul> <li>(765 \times 3 \times 273) = 626535</li><li></li> <li>(308 \times 760) = 234080 V_2</li><li></li> <li>V_2 = \frac{626535}{234080} = 2.68 L</li></ul></li></ul><h3id="exampleproblem2">ExampleProblem2</h3><ul><li>Problem:Aflexiblecontainercontains0.76molesofheliumat31°Cand16.5L.If0.22molesofheliumisreleased,whatisthenewvolumeatthesametemperatureandpressure?</li><li>Givens:<ul><li></li></ul></li> </ul> <h3 id="exampleproblem2">Example Problem 2</h3> <ul> <li>Problem: A flexible container contains 0.76 moles of helium at 31°C and 16.5 L. If 0.22 moles of helium is released, what is the new volume at the same temperature and pressure?</li> <li>Givens:<ul> <li>n_1=0.76moles</li><li>= 0.76 moles</li> <li>T_1=31°C=304K</li><li>= 31°C = 304 K</li> <li>V_1=16.5L</li><li>= 16.5 L</li> <li>n_2=0.76−0.22=0.54moles(amountleftafterrelease)</li><li>= 0.76 - 0.22 = 0.54 moles (amount left after release)</li> <li>TandandPareconstant.</li><li>Unknown:are constant.</li> <li>Unknown:V_2</li></ul></li><li>SimplifiedEquation(since</li></ul></li> <li>Simplified Equation (sinceTandandPareconstant):are constant):\frac{V1}{n1} = \frac{V2}{n2}</li><li>Pluginthevalues:</li> <li>Plug in the values:\frac{16.5}{0.76} = \frac{V_2}{0.54}</li><li>Crossmultiplyandsolvefor</li> <li>Cross multiply and solve forV_2<ul><li><ul> <li>16.5 \times 0.54 = 8.91</li><li></li> <li>0.76 \times V_2</li><li></li> <li>V_2 = \frac{8.91}{0.76} = 11.7 L \approx 12 L</li></ul></li><li>Result:</li></ul></li> <li>Result:V_2 \approx 12$$ liters (rounded to two significant figures)