Question

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.

• A. 20 L·atm
• B. 10 L·atm
• C. 5 L·atm
• D. 15 L·atm

Hint:

In thermodynamics, work done on a gas during a change in volume is negative for expansion and positive for compression. Be sure to check the sign convention.

Answer 1

The Correct Option is B

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 \( \boxed{10 \, \text{Latm}} \).