Chapter 8 “Lab” Gas Laws Purpose: To explore the relationship of volume, temperature, pressure, and moles of a gas. Instructions: Click on the link for the gas laws simulation in the Chapter 8 “Lab” Assignment on edulsims/htmlqases: Canvas. Or copy and paste this link: https://phet.colorado.edu/en/simulation/gas-properties. Click on “Intro” on the left and “Laws” on the right. Click on the “Laws” tab. You will see two boxes – to see what everything does, hint: move the box. Play around with the simulation, pump up and down to get some gas particles into the box to start. Activity 1: Pressure and Volume (Boyle’s Law) 1. Click the yellow button in the bottom right corner of the simulation to reset the simulation. 2. Pump some blue gas molecules into the box. 3. Record the pressure of the gas: 4 atm. 4. Now grab the handle on the left side of the box (with your mouse) and pull it to the right. 5. What happened to the pressure of the gas when you decreased the volume? New Pressure: 5.2 atm. Question: When you decreased the size of the box (volume), how did this affect the frequency with which the gas molecules are colliding with the container walls? Let’s assume the volume of the box was 2.00 L when we started and the initial pressure was 3.58 atm. If we decreased the volume to 1.40 L, what would be the new pressure of the gas (assuming the temperature and moles of gas remained constant)? Show your work and don’t forget units.
The Correct Answer and Explanation is:
Boyle’s Law states that for a fixed amount of gas at a constant temperature, the pressure and volume of the gas are inversely related. In other words, when the volume of the gas decreases, the pressure increases, and when the volume increases, the pressure decreases.
From your simulation, when the volume was decreased, the pressure increased from 4 atm to 5.2 atm. This happens because the gas molecules are now in a smaller space, so they collide with the walls more frequently, creating a higher pressure. Since pressure is caused by molecules colliding with the container’s walls, decreasing the volume means the molecules have less space to move around in, and they hit the walls more often, increasing the pressure.
To solve for the new pressure when the volume decreases, we can use Boyle’s Law formula:
P1V1=P2V2P_1 V_1 = P_2 V_2P1V1=P2V2
Where:
- P1P_1P1 is the initial pressure
- V1V_1V1 is the initial volume
- P2P_2P2 is the new pressure (this is what we are solving for)
- V2V_2V2 is the new volume
Given:
- P1=3.58 atmP_1 = 3.58 \, \text{atm}P1=3.58atm
- V1=2.00 LV_1 = 2.00 \, \text{L}V1=2.00L
- V2=1.40 LV_2 = 1.40 \, \text{L}V2=1.40L
Now, we can plug in the known values:3.58 atm×2.00 L=P2×1.40 L3.58 \, \text{atm} \times 2.00 \, \text{L} = P_2 \times 1.40 \, \text{L}3.58atm×2.00L=P2×1.40L7.16=P2×1.407.16 = P_2 \times 1.407.16=P2×1.40
Now, solve for P2P_2P2:P2=7.161.40P_2 = \frac{7.16}{1.40}P2=1.407.16P2=5.11 atmP_2 = 5.11 \, \text{atm}P2=5.11atm
Conclusion:
When the volume decreases from 2.00 L to 1.40 L, the pressure increases to 5.11 atm. This demonstrates Boyle’s Law: as volume decreases, pressure increases, provided the temperature and moles of gas stay constant.
