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Chapter 9: Problem 24

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?

Short Answer

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

Step by step solution

01

Convert pressure to Pascals

To convert the pressure from atm to Pascals, we use the conversion factor 1 atm = 101325 Pa. \(P_{\text{Pa}} = P_{\text{atm}} \times 101325\) \(P_{\text{Pa}} = 1.01 \ \text{atm} \times 101325 \ \mathrm{\frac{Pa}{atm}} = 102338.25 \ \text{Pa}\)
02

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 \(\mathrm{m^3}\) = \(10^6 \mathrm{mL}\). \(\Delta V = -0.001466 \ \mathrm{m^3} \times 10^6 \ \mathrm{\frac{mL}{m^3}} = -1466 \ \text{mL} \)
03

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.

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