Question

A gas expands from an initial volume of \( V_1 = 1 \, \text{m}^3 \) to a final volume of \( V_2 = 3 \, \text{m}^3 \) under constant pressure of \( P = 2 \, \text{atm} \). What is the work done by the gas during this expansion?

• A. \( 4 \times 10^5 \, \text{J} \)
• B. \( 6 \times 10^5 \, \text{J} \)
• C. \( 2 \times 10^5 \, \text{J} \)
• D. \( 1 \times 10^5 \, \text{J} \)

Hint:

For constant pressure processes, the work done by a gas is calculated using the formula \( W = P \Delta V \), where \( \Delta V \) is the change in volume and \( P \) is the constant pressure.

Answer 1

The Correct Option is A

Given:

• Initial volume \( V_1 = 1 \, \text{m}^3 \)
• Final volume \( V_2 = 3 \, \text{m}^3 \)
• Constant pressure \( P = 2 \, \text{atm} \)

Step 1: Convert pressure from atm to pascals

1 atm = \( 1.013 \times 10^5 \, \text{Pa} \)

\[ P = 2 \, \text{atm} = 2 \times 1.013 \times 10^5 = 2.026 \times 10^5 \, \text{Pa} \]

Step 2: Use the formula for work done during expansion at constant pressure

The work done by the gas is given by the formula: \[ W = P \Delta V = P (V_2 - V_1) \] where \( \Delta V = V_2 - V_1 \) is the change in volume.

Step 3: Substitute the given values into the formula

\[ W = (2.026 \times 10^5 \, \text{Pa}) \times (3 - 1) \, \text{m}^3 \] \[ W = 2.026 \times 10^5 \times 2 = 4.052 \times 10^5 \, \text{J} \]

✅ Final Answer:

The work done by the gas during the expansion is \( \boxed{4 \times 10^5 \, \text{J}} \).