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: $\frac{P<em>1V</em>1}{n<em>1T</em>1} = \frac{P<em>2V</em>2}{n<em>2T</em>2}$
• Variables:
  • $P$ = Pressure
  • $V$ = Volume
  • $n$ = number of moles
  • $T$ = Temperature
• Individual Gas Laws:
  • Boyle's Law: $P<em>1V</em>1 = P<em>2V</em>2$ (at constant $n$ and $T$)
  • Charles's Law: $\frac{V<em>1}{T</em>1} = \frac{V<em>2}{T</em>2}$ (at constant $n$ and $P$)
  • Gay-Lussac's Law: $\frac{P<em>1}{T</em>1} = \frac{P<em>2}{T</em>2}$ (at constant $n$ and $V$)
  • Avogadro's Law: $\frac{V<em>1}{n</em>1} = \frac{V<em>2}{n</em>2}$ (at constant $T$ and $P$)
• 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:
  • $V_1$ = 3.00 L
  • $T_1$ = 35°C = 308 K (35 + 273)
  • $P_1$ = 765 mmHg
  • STP (Standard Temperature and Pressure):
    • $T_2$ = 0°C = 273 K
    • $P_2$ = 1 atm = 760 mmHg
  • Unknown: V_2
• Since the amount of gas is not changing, n can be removed from the combined gas law.
• Simplified Equation: \frac{P1V1}{T1} = \frac{P2V2}{T2}
• Plug in the values: \frac{(765)(3.00)}{308} = \frac{(760)V_2}{273}
• Cross multiply and solve for V_2
  • (765 \times 3 \times 273) = 626535
  • (308 \times 760) = 234080 V_2
  • V_2 = \frac{626535}{234080} = 2.68 L

Example Problem 2

• 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?
• Givens:
  • n_1 = 0.76 moles
  • T_1 = 31°C = 304 K
  • V_1 = 16.5 L
  • n_2 = 0.76 - 0.22 = 0.54 moles (amount left after release)
  • T$and$P are constant.
  • Unknown: V_2
• Simplified Equation (since T$and$P$are constant):$\frac{V1}{n1} = \frac{V2}{n2}
• Plug in the values: \frac{16.5}{0.76} = \frac{V_2}{0.54}
• Cross multiply and solve for V_2
  • 16.5 \times 0.54 = 8.91
  • 0.76 \times V_2
  • V_2 = \frac{8.91}{0.76} = 11.7 L \approx 12 L
• Result: V_2 \approx 12$$ liters (rounded to two significant figures)