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Question: The work done during the expansion of a gas from 4 \(dm^{3}\) to 6 \(dm^{3}\) against a constant ext...

The work done during the expansion of a gas from 4 dm3dm^{3} to 6 dm3dm^{3} against a constant external pressure of 3 atm is (1 L atm = 101.32101.32J)

A

-6J

B

– 608 J

C
  • 304 J
D

– 304 J

Answer

– 608 J

Explanation

Solution

: Work is expansion, = PΔVP\Delta V

W = - 3×(6-4) = - 6 L atm

1 L atm = 101.32101.32J

W=−6×101.32=−607.92JW = - 6 \times 101.32 = - 607.92J or – 608 J