A gas is compressed from an initial volume of 10 L to 5 L. The pressure during the compression is constant at 2 atm. Calculate the work done on the gas.
Show Hint
20 L·atm
10 L·atm
- 5 L·atm
- 15 L·atm
The Correct Option is B
Solution and Explanation
Given:
- Initial volume: \( V_1 = 10 \, \text{L} \)
- Final volume: \( V_2 = 5 \, \text{L} \)
- Pressure: \( P = 2 \, \text{atm} \)
Step 1: Use the formula for work done during an isobaric process
The work done on a gas during an isobaric (constant pressure) process is given by the formula: \[ W = -P \Delta V \] where: - \( W \) is the work done on the gas, - \( P \) is the pressure, - \( \Delta V = V_2 - V_1 \) is the change in volume.
Step 2: Substitute the given values into the formula
The change in volume is: \[ \Delta V = 5 \, \text{L} - 10 \, \text{L} = -5 \, \text{L} \] Now, calculate the work: \[ W = -(2 \, \text{atm}) \times (-5 \, \text{L}) = 10 \, \text{Latm} \]
✅ Final Answer:
The work done on the gas is \( \boxed{10 \, \text{Latm}} \).
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