Boyle's Law PhET Simulation Activity Results

Page 1

1. Graduations on the ruler:
   • Each notch = 1.0 nm (treated as 1.0 L).
2. Gas particles added:
   • Number: 10 particles (after one pump).
   • Type: Light species (e.g., helium).
3. Particle motion:
   • Rapid, random movement with collisions between particles and container walls.
4. Temperature settings:
   • Constant temperature = 300 K
5. Volume reduction effect:
   • Initial temperature increases due to compression.
6. Simulation adjustment:
   • Automatically adds/removes heat to maintain constant temperature.
7. Pressure cause (Kinetic Theory):
   • Collisions of gas particles with container walls.
8. Hypotheses:
   • Smaller volume → Pressure ↑ (more frequent collisions).
   • Larger volume → Pressure ↓ (fewer collisions).

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Page 2

Data Table

Trials	Volume (V)	Pressure (P)	$k_1=\frac{1}{V}$	$k_2=\frac{P}{V}$
1	10.0 L	100 kPa	0.1 L⁻¹	10 kPa/L
2	8.0 L	125 kPa	0.125 L⁻¹	15.6 kPa/L
3	6.0 L	167 kPa	0.167 L⁻¹	27.8 kPa/L
4	5.0 L	200 kPa	0.2 L⁻¹	40 kPa/L

• Independent variable: Volume ($V$).
• Dependent variable: Pressure ($P$).
• Graph relationship: Inverse (as $V\uparrow$, $P\downarrow$).

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Page 3

1. Constant value:
   • Boyle’s Law states $P\cdot V=\text{constant}$.
2. Equation:
   $P_1{V}_1=P_2{V}_2$

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Calculations

1. Problem 1:
   $P_2=\frac{1.75\text{atm}\times 1.25\text{L}}{3.15\text{L}}=\boxed{0.70\text{atm}}$

2. Problem 2:

   • Pressure after compression:
     $P_2=\frac{790\text{mmHg}\times 3.67\text{L}}{2.12\text{L}}=1367\text{mmHg}$
   • Convert to atm:
     $\frac{1367}{760}=\boxed{1.80\text{atm}}$

3. Problem 3:
   $V_2=\frac{4.25\text{atm}\times 5.85\text{L}}{2.75\text{atm}}=\boxed{9.03\text{L}}$

4. Problem 4:

   • Convert $P_2$ to atm:
     $\frac{1460\text{mmHg}}{760}=1.92\text{atm}$
   • New volume:
     $V_2=\frac{5.97\text{atm}\times 2.79\text{L}}{1.92\text{atm}}=\boxed{8.66\text{L}}$