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

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

Pressure in pascals: \( P = 2 \, \text{atm} \times 1.013 \times 10^5 \, \text{Pa/atm} = 2.026 \times 10^5 \, \text{Pa} \)

Step 2: Calculate work done during expansion at constant pressure

Formula for work: \( W = P \Delta V = P (V_2 - V_1) \)

Step 3: Substitute values and compute work

Volume change: \( \Delta V = 3 \, \text{m}^3 - 1 \, \text{m}^3 = 2 \, \text{m}^3 \)

Work done: \( W = (2.026 \times 10^5 \, \text{Pa}) \times (2 \, \text{m}^3) = 4.052 \times 10^5 \, \text{J} \)

✅ Final Answer:

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