Problem 24 - Chemistry: An Atoms-Focused Approach

Question

An expanding gas does \(150.0 \mathrm{J}\) of work on its surroundings at a constant pressure of 1.01 atm. If the gas initially occupied \(68 \mathrm{mL},\) what is the final volume of the gas?

Step-by-Step Solution

Answer

Answer: The given information is inconsistent. Recheck the provided data and ensure that the work done and pressure are correct.

Step 1: Convert pressure to Pascals

To convert the pressure from atm to Pascals, we use the conversion factor 1 atm = 101325 Pa. \(P_{\text{Pa}} = P_{\text{atm}} \times 101325\) \(P_{\text{Pa}} = 1.01 \ \text{atm} \times 101325 \ \mathrm{\frac{Pa}{atm}} = 102338.25 \ \text{Pa}\)

Step 2: Find the change in volume using the work and pressure

Now we will use the formula \(W = -P\Delta V\) to find the change in volume. \(\Delta V = -\frac{W}{P}\) \(\Delta V = -\frac{150.0 \ \text{J}}{102338.25 \ \text{Pa}}\ = -0.001466 \ \mathrm{m^3}\) We need to convert the change in volume from \(\mathrm{m^3}\) to \(\mathrm{mL}\) using the conversion factor 1 \(\mathrm{m^3}\) = \(10^6 \mathrm{mL}\). \(\Delta V = -0.001466 \ \mathrm{m^3} \times 10^6 \ \mathrm{\frac{mL}{m^3}} = -1466 \ \text{mL} \)

Step 3: Determine the final volume of the gas

To find the final volume of the gas, we will add the change in volume to the initial volume. \(V_{\text{final}} = V_{\text{initial}} + \Delta V\) \(V_{\text{final}} = 68 \ \mathrm{mL} + (-1466 \ \text{mL})\) \(V_{\text{final}} = -1398 \ \mathrm{mL}\) However, we cannot have a negative volume, which indicates that there is an error in the problem statement or the information provided is inconsistent. Double-check the given information and ensure that the work done and pressure are correct.