# Boyle’s Law Problems — Step-by-Step Solutions Boyle’s Law states: \[ P_1 V_1 = P_2 V_2 \] where \(P\) is pressure and \(V\) is volume. Temperature and amount of gas are constant. --- ## 5. Find Ground Level Pressure (in kPa) **Given:** - \(V_1 = 735~\text{mL}\) (ground) - \(V_2 = 1286~\text{mL}\) (altitude) - \(P_2 = 0.8~\text{atm}\) (altitude) - \(P_1 = ?\) (ground, in kPa) ### Step 1: Use Boyle’s Law \[ P_1 V_1 = P_2 V_2 \] \[ P_1 = \frac{P_2 V_2}{V_1} \] ### Step 2: Substitute Values \[ P_1 = \frac{0.8~\text{atm} \times 1286~\text{mL}}{735~\text{mL}} \] \[ P_1 = \frac{1028.8~\text{atm}\cdot\text{mL}}{735~\text{mL}} = 1.399~\text{atm} \] ### Step 3: Convert atm to kPa \(1~\text{atm} = 101.325~\text{kPa}\) \[ P_1 = 1.399~\text{atm} \times 101.325~\text{kPa/atm} = 141.85~\text{kPa} \] ### **Final Answer** \[ \boxed{P_1 = 142~\text{kPa}} \] --- ## 6. New Pressure After Volume Increase **Given:** - \(P_1 = 143.95~\text{kPa}\) - \(V_2 = V_1 + 0.4 V_1 = 1.4 V_1\) - \(P_2 = ?\) ### Step 1: Apply Boyle’s Law \[ P_1 V_1 = P_2 V_2 \] \[ P_2 = \frac{P_1 V_1}{V_2} \] ### Step 2: Substitute \(V_2 = 1.4 V_1\) \[ P_2 = \frac{P_1 V_1}{1.4 V_1} = \frac{P_1}{1.4} \] \[ P_2 = \frac{143.95~\text{kPa}}{1.4} = 102.82~\text{kPa} \] ### **Final Answer** \[ \boxed{P_2 = 102.8~\text{kPa}} \] --- ## 7. Soap Bubble and Pressure Change **Given:** - \(V_1 = 50.0~\text{cm}^3\) - \(P_1 = 101.325~\text{kPa}\) (standard pressure) - \(P_2 = 93.326~\text{kPa}\) - \(V_2 = ?\) - Temperature is constant. ### Step 1: Use Boyle’s Law \[ P_1 V_1 = P_2 V_2 \] \[ V_2 = \frac{P_1 V_1}{P_2} \] ### Step 2: Substitute Values \[ V_2 = \frac{101.325~\text{kPa} \times 50.0~\text{cm}^3}{93.326~\text{kPa}} \] \[ V_2 = \frac{5066.25}{93.326} = 54.31~\text{cm}^3 \] ### Step 3: Find Change in Volume \[ \Delta V = V_2 - V_1 = 54.31 - 50.0 = 4.31~\text{cm}^3 \] ### Step 4: Will the Bubble Get Bigger? **Yes:** Lower pressure → volume increases. ### **Final Answer** - **The bubble will get BIGGER.** - **Change in volume:** \[ \boxed{+4.3~\text{cm}^3} \] --- ## 8. Find Volume of Marble Chips **Given:** - Total initial volume (\(V_{total,1}\)) = \(V_{CO_2,1} + V_{chips}\) = \(100~\text{cm}^3\) - Initial pressure (\(P_1\)) = \(101.33~\text{kPa}\) - Final pressure (\(P_2\)) = \(133.33~\text{kPa}\) - Final total volume (\(V_{total,2}\)) = \(80~\text{cm}^3\) - Temperature constant. **Assume:** Marble chips are solid and do not change volume; only CO₂ gas compresses. Let \(V_{chips}\) = volume of marble chips (constant) Let \(V_{CO_2,1}\) = initial CO₂ volume Let \(V_{CO_2,2}\) = final CO₂ volume \[ V_{CO_2,1} = 100 - V_{chips} \] \[ V_{CO_2,2} = 80 - V_{chips} \] ### Step 1: Boyle’s Law for the Gas Only \[ P_1 V_{CO_2,1} = P_2 V_{CO_2,2} \] \[ 101.33(100 - V_{chips}) = 133.33(80 - V_{chips}) \] ### Step 2: Expand and Solve for \(V_{chips}\) \[ 10133 - 101.33 V_{chips} = 10666.4 - 133.33 V_{chips} \] \[ (133.33 - 101.33)V_{chips} = 10666.4 - 10133 \] \[ 32.00 V_{chips} = 533.4 \] \[ V_{chips} = \frac{533.4}{32.00} = 16.67~\text{cm}^3 \] ### **Final Answer** \[ \boxed{V_{chips} = 16.7~\text{cm}^3} \] --- ## Summary Table | Problem | Final Answer | |---------|--------------------| | 5 | \(142~\text{kPa}\) | | 6 | \(102.8~\text{kPa}\) | | 7 | Bubble gets bigger; \(\Delta V = +4.3~\text{cm}^3\) | | 8 | \(16.7~\text{cm}^3\) (volume of marble chips) |