To compute the work done ($W$) when a gas expands against a constant pressure, we use the equation:

$W = - P \Delta V$

Where:

• $W$ is the work (in Joules),
• $P$ is the constant pressure (in atm),
• $\Delta V$ is the change in volume ($V_{\text{final}} - V_{\text{initial}}$).

Step 1: Calculate the change in volume ($\Delta V$)

$\Delta V = V_{\text{final}} - V_{\text{initial}} = 21.5 \, \text{L} - 6.6 \, \text{L} = 14.9 \, \text{L}$

Step 2: Convert pressure and volume to appropriate units

• Pressure is given in atm, so we need to convert it to Pascals ($1 \, \text{atm} = 101,325 \, \text{Pa}$): $P = 6.8 \, \text{atm} = 6.8 \times 101,325 \, \text{Pa} = 689,010 \, \text{Pa}$
• Volume must be converted from liters to cubic meters ($1 \, \text{L} = 0.001 \, \text{m}^3$): $\Delta V = 14.9 \, \text{L} = 14.9 \times 0.001 \, \text{m}^3 = 0.0149 \, \text{m}^3$

Step 3: Compute the work

Substitute the values into the formula: $W = - P \Delta V = - (689,010 \, \text{Pa}) \times (0.0149 \, \text{m}^3)$ $W = - 10,272.85 \, \text{J}$

Final Answer:

The work done is: $W = -10,273 \, \text{J} \, (\text{rounded to three significant figures}).$

Would you like further clarification on this calculation?

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Follow-up Questions:

1. Why is work negative in this context?
2. How is the conversion from atm to Pascal performed?
3. What other units could be used for measuring work?
4. How does the sign of work change if the gas is compressed?
5. What assumptions are made when using this formula for work?

Tip: Always ensure consistency in units when applying formulas, especially in physics and chemistry problems.