To determine the work done on the gas during compression, we use the formula for work done in a constant pressure process:

$W = -P \Delta V$

Where:

• $W$ = work done on the gas (J)
• $P$ = pressure (Pa)
• $\Delta V$ = change in volume ($V_f - V_i$) (m³)

Step 1: Convert pressure from atm to Pa

$P = 0.208 \, \text{atm} \times 1.013 \times 10^5 \, \frac{\text{Pa}}{\text{atm}}$ $P = 2.107 \times 10^4 \, \text{Pa}$

Step 2: Convert volume from liters to cubic meters

$1 \, \text{L} = 1 \times 10^{-3} \, \text{m}^3$ $V_i = 7.87 \, \text{L} = 7.87 \times 10^{-3} \, \text{m}^3$ $V_f = 5.54 \, \text{L} = 5.54 \times 10^{-3} \, \text{m}^3$

Step 3: Calculate the change in volume

$\Delta V = V_f - V_i = 5.54 \times 10^{-3} - 7.87 \times 10^{-3} = -2.33 \times 10^{-3} \, \text{m}^3$

Step 4: Calculate the work done

$W = -P \Delta V$ $W = - (2.107 \times 10^4) \times (-2.33 \times 10^{-3})$ $W = 49.1 \, \text{J}$

Final Answer:

The work done on the gas is: $\boxed{49.1 \, \text{J}}$

Would you like further explanation or assistance?
Here are 5 related questions to consider:

1. What is the significance of the negative sign in work calculations?
2. How would the work change if the gas were expanding instead of compressing?
3. Can you calculate the total energy change of the system?
4. What happens to the heat energy in this process?
5. How is the First Law of Thermodynamics applied in this scenario?

Tip: Always check unit consistency when working with physical equations to avoid calculation errors.