Question:medium

Work done during the expansion of a gas from a volume of \(4 \, \text{dm}^3\) to \(6 \, \text{dm}^3\) against a constant external pressure of 3 atm is (1 L atm = 101.32 J)

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Work done during expansion or compression of a gas is given by \(W = -P \Delta V\), where \(P\) is the pressure and \(\Delta V\) is the change in volume.
Updated On: Jan 15, 2026
  • \(-608 \, \text{J}\)
  • \(-304 \, \text{J}\)
  • \(+304 \, \text{J}\)
  • \(-6 \, \text{J}\)
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The Correct Option is B

Solution and Explanation

The work done is calculated as: \[ W = -P \Delta V \] Given \(P = 3 \, \text{atm}\) and \(\Delta V = 6 - 4 = 2 \, \text{dm}^3\). Using the conversion \(1 \, \text{atm} = 101.32 \, \text{J/L}\), the work done is \(-304 \, \text{J}\).
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