Gay-Lussac’s Law PhET Simulation Activity

Simulation Setup

1. Number of gas particles added: 10
   Type of gas particle: Light (initially), then manually changed to 100 Heavy molecules.
2. Temperature units: Kelvin (K).

Observations on Temperature and Pressure

• Effect of increasing temperature: Particles move faster with higher kinetic energy.
• Collisions with container walls: Increase in frequency.
• Pressure change: Pressure increases.

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Trials	Temperature (T)	Pressure (P)	$k_1=P\times T$	$k_2=\frac{P}{T}$
Trial 1	150 K	50 kPa	7500 kPa·K	0.333 kPa/K
Trial 2	300 K	100 kPa	30,000 kPa·K	0.333 kPa/K
Trial 3	450 K	150 kPa	67,500 kPa·K	0.333 kPa/K
Trial 4	600 K	200 kPa	120,000 kPa·K	0.333 kPa/K

• Independent variable: Temperature ($T$).
• Dependent variable: Pressure ($P$).

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Graph Analysis

• Relationship: Pressure is directly proportional to temperature ($P\propto T$) at constant volume.
• At 0 K: Particle motion ceases; pressure approaches zero.

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Consistency in Data

• Constant value: $k_2=\frac{P}{T}$.

Verification of $k_2$:

• Point #1: $P_1=50\text{kPa}$, $T_1=150\text{K}$ $\Rightarrow$ $k_2=0.333\text{kPa/K}$.
• Point #2: $P_2=200\text{kPa}$, $T_2=600\text{K}$ $\Rightarrow$ $k_2=0.333\text{kPa/K}$.

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Gay-Lussac’s Law Equation

$$\frac{P_1}{T_1}=\frac{P_2}{T_2}$$

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Problem Solutions

1. Cooling from 300 K to 100 K:

   $$P_2=1.69\text{atm}\times \frac{100\text{K}}{300\text{K}}=0.56\text{atm}.$$

2. Heating from 500 K to 900 K:

   $$P_2=715\text{mmHg}\times \frac{900\text{K}}{500\text{K}}=1287\text{mmHg}(\approx 1.69\text{atm}).$$

3. Aerosol can heating (22°C to 100°C):

   $$T_1=295\text{K},T_2=373\text{K},P_2=5.9\text{atm}\times \frac{373}{295}\approx 7.45\text{atm}.$$

4. Temperature for 4.75 atm pressure:

   $$T_2=290\text{K}\times \frac{4.75\text{atm}}{1.85\text{atm}}\approx 744.6\text{K}.$$