To calculate the work done on a gas during compression at constant pressure, we use the formula:

$W = -P \Delta V$

where:

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

Given:

• Initial volume, $V_{\text{initial}} = 65.0 \, \text{L}$
• Final volume, $V_{\text{final}} = 23.0 \, \text{L}$
• Pressure, $P = 63.0 \, \text{atm}$

Step 1: Calculate the Change in Volume

$\Delta V = V_{\text{final}} - V_{\text{initial}} = 23.0 \, \text{L} - 65.0 \, \text{L} = -42.0 \, \text{L}$

Step 2: Calculate Work in Liter-atm

Since $W = -P \Delta V$: $W = -(63.0 \, \text{atm})(-42.0 \, \text{L}) = 2646 \, \text{L·atm}$

Step 3: Convert Work to kJ

To convert from L·atm to kJ, use the conversion factor $1 \, \text{L·atm} = 0.101325 \, \text{kJ}$: $W = 2646 \, \text{L·atm} \times 0.101325 \, \text{kJ/L·atm} = 268.12 \, \text{kJ}$

Step 4: Determine the Sign and Significant Figures

Since the gas is being compressed, work is done on the gas, meaning $W$ should be positive.

With three significant figures in the given values, we round to: $W = 268 \, \text{kJ}$

Final Answer

The work done on the gas mixture is: $\boxed{268 \, \text{kJ}}$

Would you like further details on the calculation steps, or do you have any questions?

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Here are five related questions for practice:

1. What would the work done be if the gas expanded instead of being compressed?
2. How does changing the pressure affect the work done in a similar process?
3. What if the initial volume were increased to 100.0 L—how would that change the result?
4. How is the sign of work determined in thermodynamic processes?
5. How would we calculate work if pressure were not constant?

Tip: Always double-check the units and the sign conventions for work in thermodynamics, as they can vary with different processes and setups.