Questions: Table 2 Relationship between Volume and Pressure. Volume of Water Added (mL) Volume of Air in the Erlenmeyer Flask (mL) Pressure Inside the Erlenmeyer Flask (atm) --- --- --- 0 mL type your text here 1.000 atm 10 mL type your text here 1.058 atm 20 mL type your text here 1.123 atm 30 mL type your text here 1.197 atm 40 mL type your text here 1.282 atm 50 mL 1.379 atm

Table 2 Relationship between Volume and Pressure.

Volume of Water Added (mL) Volume of Air in the Erlenmeyer Flask (mL) Pressure Inside the Erlenmeyer Flask (atm)
--- --- ---
0 mL type your text here 1.000 atm
10 mL type your text here 1.058 atm
20 mL type your text here 1.123 atm
30 mL type your text here 1.197 atm
40 mL type your text here 1.282 atm
50 mL 1.379 atm

Transcript text: Table 2 Relationship between Volume and Pressure. \begin{tabular}{|c|c|c|} \hline \begin{tabular}{c} Volume of Water \\ Added (mL) \end{tabular} & \begin{tabular}{c} Volume of Air in the \\ Erlenmeyer Flask (mL) \end{tabular} & \begin{tabular}{c} Pressure Inside the \\ Erlenmeyer Flask (atm) \end{tabular} \\ \hline 0 mL & type your text here & 1.000 atm \\ \hline 10 mL & type your text here & 1.058 atm \\ \hline 20 mL & type your text here & 1.123 atm \\ \hline 30 mL & type your text here & 1.197 atm \\ \hline 40 mL & type your text here & 1.282 atm \\ \hline 50 mL & & 1.379 atm \\ \hline \end{tabular}

Solution

Solution Steps

Step 1: Understanding the Problem

The table provided shows the relationship between the volume of water added to an Erlenmeyer flask and the resulting pressure inside the flask. The task is to determine the volume of air in the flask for each given condition.

Step 2: Applying Boyle's Law

Boyle's Law states that for a given amount of gas at constant temperature, the pressure of the gas is inversely proportional to its volume. Mathematically, it is expressed as:

\[ P_1 V_1 = P_2 V_2 \]

where \(P_1\) and \(V_1\) are the initial pressure and volume, and \(P_2\) and \(V_2\) are the final pressure and volume.

Step 3: Calculating the Volume of Air

Assuming the initial volume of air in the flask is \(V_0\) when no water is added, and the initial pressure is 1.000 atm, we can use Boyle's Law to find the volume of air for each condition:

0 mL Water Added:
- \(P_1 = 1.000 \, \text{atm}\), \(V_1 = V_0\)
- \(P_2 = 1.000 \, \text{atm}\), \(V_2 = V_0\)
- Volume of air = \(V_0\)
10 mL Water Added:
- \(P_1 = 1.000 \, \text{atm}\), \(V_1 = V_0\)
- \(P_2 = 1.058 \, \text{atm}\), \(V_2 = V_0 - 10 \, \text{mL}\)
- Solving \(1.000 \times V_0 = 1.058 \times (V_0 - 10)\)
20 mL Water Added:
- \(P_1 = 1.000 \, \text{atm}\), \(V_1 = V_0\)
- \(P_2 = 1.123 \, \text{atm}\), \(V_2 = V_0 - 20 \, \text{mL}\)
- Solving \(1.000 \times V_0 = 1.123 \times (V_0 - 20)\)
30 mL Water Added:
- \(P_1 = 1.000 \, \text{atm}\), \(V_1 = V_0\)
- \(P_2 = 1.197 \, \text{atm}\), \(V_2 = V_0 - 30 \, \text{mL}\)
- Solving \(1.000 \times V_0 = 1.197 \times (V_0 - 30)\)
40 mL Water Added:
- \(P_1 = 1.000 \, \text{atm}\), \(V_1 = V_0\)
- \(P_2 = 1.282 \, \text{atm}\), \(V_2 = V_0 - 40 \, \text{mL}\)
- Solving \(1.000 \times V_0 = 1.282 \times (V_0 - 40)\)