Question:medium

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.

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For an isobaric process, the work done by the gas is simply the pressure multiplied by the change in volume.
Updated On: Nov 26, 2025
  • \( 6 \times 10^5 \, \text{J} \)
  • \( 6 \times 10^4 \, \text{J} \)
  • \( 2 \times 10^6 \, \text{J} \)
  • \( 1 \times 10^5 \, \text{J} \)
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The Correct Option is A

Solution and Explanation

The work performed by a gas during an isobaric expansion (constant pressure) is calculated as: \[ W = P \Delta V \] Given: - Pressure, \( P = 2 \times 10^5 \, \text{Pa} \) - Change in volume, \( \Delta V = V_f - V_i = 5 - 2 = 3 \, \text{m}^3 \) Calculation: \[ W = 2 \times 10^5 \times 3 = 6 \times 10^5 \, \text{J} \] The work done by the gas is \( 6 \times 10^5 \, \text{J} \).
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