Question

A gas expands from a volume of 2 m\(^3\) to a volume of 5 m\(^3\) at a constant pressure of \( 2 \times 10^5 \, \text{Pa} \). Calculate the work done by the gas.

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

Hint:

For an isobaric process, the work done by the gas is simply the pressure multiplied by the change in volume.

Answer 1

The Correct Option is A

The work done by a gas during an isobaric (constant pressure) expansion is given by: \[ W = P \Delta V \] Where: - \( P = 2 \times 10^5 \, \text{Pa} \) is the pressure, - \( \Delta V = V_f - V_i = 5 - 2 = 3 \, \text{m}^3 \) is the change in volume. Substituting the values: \[ W = 2 \times 10^5 \times 3 = 6 \times 10^5 \, \text{J} \] Thus, the work done by the gas is \( 6 \times 10^5 \, \text{J} \).